The matrix P = [ 0 0 4 0 4 0 4 0 0 ] is a (A) Square matrix (B) Diagonal matrix (C) Unit matrix (D) None

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t A = [ 0 0 4 0 4 0 4 0 0 ] H e r e , n u m b e r o f c o l u m n s a n d t h e n u m b e r o f r o w s a r e e q u a l i . e . , 3 . S o , A i s a s q u a r e m a t r i x . H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

If [ 2 x + y 4 x 5 x − 7 4 x ] = [ 7 7 y − 1 3 y x + 6 ] , then the value of x + y is (A) x = 3 , y = 1 (B) x = 2 , y = 3 (C) x = 2 , y = 4 (D) x = 3 , y = 3

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : [ 2 x + y 4 x 5 x − 7 4 x ] = [ 7 7 y − 1 3 y x + 6 ] E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s , w e g e t , 2 x + y = 7 … ( i ) a n d 4 x = 7 y − 1 3 … ( i i ) f r o m e q n . ( i i ) 4 x − x = 6 3 x = 6 ∴ x = 2 f r o m e q n . ( i ) 2 × 2 + y = 7 4 + y = 7 ∴ y = 7 − 4 = 3 H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

If A = 1 π [ s i n − 1 ( xπ ) t a n − 1 ( x π ) s i n − 1 ( x π ) c o t − 1 ( π x ) ] , B = 1 π [ − c o s − 1 ( x π ) t a n − 1 ( x π ) s i n − 1 ( x π ) − t a n − 1 ( x π ) ] , then A − B is equal to:

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : A = 1 π [ s i n − 1 ( x π ) t a n − 1 ( x π ) s i n − 1 ( x π ) c o t − 1 ( π x ) ] a n d B = 1 π [ − c o s − 1 ( x π ) t a n − 1 ( x π ) s i n − 1 ( x π ) − t a n − 1 ( π x ) ] A − B = 1 π [ s i n − 1 ( x π ) t a n − 1 ( x π ) s i n − 1 ( x π ) c o t − 1 ( π x ) ] − 1 π [ − c o s − 1 ( x π ) t a n − 1 ( x π ) s i n − 1 ( x π ) − t a n − 1 ( π x ) ] = 1 π [ s i n − 1 ( x π ) + c o s − 1 ( x π ) t a n − 1 ( x π ) − t a n − 1 ( x π ) s i n − 1 ( x π ) − s i n − 1 ( x π ) c o t − 1 ( π x ) + t a n − 1 ( π x ) ] = 1 π [ π 2 0 0 π 2 ] [ ? s i n − 1 x + c o s − 1 x = π 2 t a n − 1 x + c o t − 1 x = π 2 ] = 1 π × π 2 [ 1 0 0 1 ] = 1 2 I H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

Express the matrix [ 2 3 1 1 − 1 2 4 1 2 ] as the sum of a symmetric and a skew-symmetric matrix.

This is a long answer type question as classified in NCERT Exemplar We know that any square matrix can be expressed as the sum of symmetric and skew symmetric matrix i.e. A = 1 2 [ A + A ' ] + 1 2 [ A − A ' ] S o P = 1 2 [ ( 2 3 1 1 − 1 2 4 1 2 ) + ( 2 1 4 3 − 1 1 1 2 2 ) ] = 1 2 [ 2 + 2 3 + 1 1 + 4 1 + 3 − 1 − 1 2 + 1 4 + 1 1 + 2 2 + 2 ] = 1 2 [ 4 4 5 4 − 2 3 5 3 4 ] = [ 2 2 5 2 2 − 1 3 2 5 2 3 2 2 ] P ' = [ 2 2 5 2 2 − 1 3 2 5 2 3 2 2 ] = P A s P ' = P ∴ P i s a s y m m e t r i c m a t r i x . N o w , Q = 1 2 [ A − A ' ] = 1 2 [ ( 2 3 1 1 − 1 2 4 1 2 ) − ( 2 1 4 3 − 1 1 1 2 2 ) ] = 1 2 [ 2 − 2 3 − 1 1 − 4 1 − 3 − 1 + 1 2 − 1 4 − 1 1 − 2 2 − 2 ] = 1 2 [ 0 2 − 3 − 2 0 1 3 − 1 0 ] = [ 0 1 − 3 2 − 1 0 1 2 3 2 − 1 2 0 ] = − [ 0 − 1 3 2 1 0 − 1 2 − 3 2 1 2 0 ] = − Q A s Q = − Q ∴ Q i s a s k e w s y m m e t r i c m a t r i x . S o A = P + Q A = [ 2 2 5 2 2 − 1 3 2 5 2 3 2 2 ] + [ 0 1 − 3 2 − 1 0 1 2 3 2 − 1 2 0 ] = [ 2 + 0 2 + 1 5 2 − 3 2 2 − 1 − 1 + 0 3 2 + 1 2 5 2 + 3 2 3 2 &min

Construct a2 × 2 matrix where (i) a i j = ( i − 2 j ) 2 2 . (ii) a i j = ? − 2 i + 3 j ? .

This is a short answer type question as classified in NCERT Exemplar L e t A = [ a 1 1 a 1 2 a 2 1 a 2 2 ] 2 × 2 ( i ) G i v e n t h a t a i j = ( i − 2 j ) 2 2 a 1 1 = ( 1 − 2 × 1 ) 2 2 = 1 2 ; a 1 2 = ( 1 − 2 × 2 ) 2 2 = 9 2 a 2 1 = ( 2 − 2 × 1 ) 2 2 = 0 ; a 2 2 = ( 2 − 2 × 2 ) 2 2 = 2 H e n c e , t h e m a t r i x A = [ 1 2 9 2 0 2 ] ( i i ) G i v e n t h a t a i j = | − 2 i + 3 j | a 1 1 = | − 2 × 1 + 3 × 1 | = 1 ; a 1 2 = | − 2 × 1 + 3 × 2 | = 4 a 2 1 = | − 2 × 2 + 3 × 1 | = − 1 ; a 2 2 = | − 2 × 2 + 3 × 2 | = 2 H e n c e , t h e m a t r i x A = [ 1 4 − 1 2 ]

Find values of a and b if A = B , where A = [ a + 4 3 b 8 − 6 ] , B = [ 2 a + b b 2 + 2 8 b 2 − 5 b ]

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t A = B ⇒ [ a + 4 3 b 8 − 6 ] = [ 2 a + 2 b 2 + 2 8 b 2 − 5 b ] E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s , w e g e t a + 4 = 2 a + 2 , 3 b = b 2 + 2 a n d b 2 − 5 b = − 6 ⇒ 2 a − a = 2 , b 2 − 3 b + 2 = 0 , b 2 − 5 b + 6 = 0 ∴ a = 2 ∴ b 2 − 3 b + 2 = 0 ∴ b 2 − 5 b + 6 = 0 ⇒ b 2 − 2 b − b + 2 = 0 , ⇒ b 2 − 3 b − 2 b + 6 = 0 ⇒ b ( b − 2 ) − 1 ( b − 2 ) = 0 , ⇒ b ( b − 3 ) − 2 ( b − 3 ) = 0 ⇒ ( b − 1 ) ( b − 2 ) = 0 , ⇒ ( b − 2 ) ( b − 3 ) = 0 ∴ b = 1 , 2 ∴ b = 2 , 3 b u t h e r e 2 i s c o m m o n . H e n c e , t h e v a l u e o f a = 2 a n d b = 2 .

Let x * y = x2 + y3 and (x * 1) * 1 = x * (1 * 1). Then a value of 2 s i n − 1 ( x 4 + x 2 − 2 x 4 + x 2 + 2 ) is.

(x2+1)*1=x*2 ⇒ (x2+1)2+1=x2+8⇒x2=2 2sin−1 (x4+x2−2x4+x2+2)=2sin−1 (12)=π3

Let x,y > 0. If x3y2 = 215, then the least value of 3x + 2y is

x1y>0 and x3y2=215 AM≥GM 3x+2y5≥ (x3y2)15 ⇒3x+2y≥40

Let R1 = { ( a , b ) ∈ N × N : | a − b | ≤ 1 3 } a n d R2 = { ( a , b ) ∈ N × N : | a − b | ≠ 1 3 } . Then on N :

R1 = { ( a , 1 ) ∈ N × N : | a − b | ≤ 1 3 } ( a , a ) ∈ R 1 a s | a − a | ≤ 1 3 ( R e f l e x i v e ) ( a , b ) & ( b , a ) ∈ R 1 a s | a − b | = | b − a | → ( s y m m e t r i c ) . But it is not necessary that if (a,b) & (b, c) ∈R, then (a,c)∈R Eg − ( 2 1 , 1 0 ) ∈ R & ( 1 0 , 1 ) ∈ R b u t ( 2 1 , 1 ) ∉ R 1 R2 = { ( a , b ) ∈ N × N : | a − b | ≠ 1 3 ∴ R 2 → N o t e q u i v a l e n c e s o l u t o i n . } ( a , b ) & ( b , a ) ∈ R 2 a s | a − b | = | b − a | But it is not necessary that if (a, b) & (b, c) ∈ R 2 then (a, c) also ∈ R 2 . Eg – (21, 1) ∈ R 2 & ( 1 , 8 ) ∈ R 2 b u t ( 2 1 , 8 ) ∉ R 2

Maths NCERT Exemplar Solutions Class 12th Chapter Three: Overview, Questions, Preparation

Updated on Jul 23, 2025 08:52 IST

By Payal Gupta, Retainer

Table of content

Matrices Long Answers Type Questions
Matrices Short Answers Type Questions
Matrices Objective Type Questions
Matrices Fill in the blanks Type Questions
Matrices True or False Type Questions
24th June 2022 (second shift)
JEE Mains Solutions 2022,28th june , Maths,Second shift

Matrices Long Answers Type Questions

Q1. If $A B = B A$ for any two square matrices, prove by mathematical induction that ${(A B)}^{n} = A^{n} B^{n}$ .

Sol.

$\begin{array}{l} L e t P (n) : {(A B)}^{n} = A^{n} B^{n} \\ S t e p 1 : P u t n = 1, P (1) : A B = A B w h i c h i s t r u e f o r n = 1 \\ S t e p 2 : P u t n = k, P (k) : {(A B)}^{k} = A^{k} B^{k} L e t i t b e t r u e f o r a n y k \in N \\ S t e p 3 : P u t n = k + 1, P (k + 1) : {(A B)}^{k + 1} = A^{k + 1} B^{k + 1} \\ L . H . S . {(A B)}^{k + 1} = {(A B)}^{k} . A B \\ = A^{k} B^{k} . A B [F r o m s t e p 2] \\ = A^{k + 1} A^{k + 1} R . H . S . \\ H e n c e, i f P (n) i s t r u e f o r P (k) t h e n i t i s t r u e f o r P (k + 1) . \end{array}$

Q2. Find $x$ , $y$ , and $z$ if: $A = [\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$ satisfies $A^{T} = A^{- 1}$ .

Sol.

Commonly asked questions

If $A B = B A$ for any two square matrices, prove by mathematical induction that ${(A B)}^{n} = A^{n} B^{n}$ .

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol.

Find $x$ , $y$ , and $z$ if: $A = [\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$ satisfies $A^{T} = A^{- 1}$ .

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol.

If possible, using elementary row transformations, find the inverse of the following matrices:

(i) $[\begin{matrix} 2 & - 1 & 3 \\ - 5 & 3 & 1 \\ - 3 & 2 & 3 \end{matrix}]$

(ii) $[\begin{matrix} 2 & 3 & - 3 \\ - 1 & - 2 & 2 \\ 1 & 1 & - 1 \end{matrix}]$

(iii) $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol.

Express the matrix $[\begin{matrix} 2 & 3 & 1 \\ 1 & - 1 & 2 \\ 4 & 1 & 2 \end{matrix}]$ as the sum of a symmetric and a skew-symmetric matrix.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t a n y s q u a r e m a t r i x c a n b e expresse d a s t h e s u m o f s y m m e t r i c a n d s k e w s y m m e t r i c m a t r i x i . e . \\ A = \frac{1}{2} [A + A^{'}] + \frac{1}{2} [A - A^{'}] \\ S o P = \frac{1}{2} [(\begin{matrix} 2 & 3 & 1 \\ 1 & - 1 & 2 \\ 4 & 1 & 2 \end{matrix}) + (\begin{matrix} 2 & 1 & 4 \\ 3 & - 1 & 1 \\ 1 & 2 & 2 \end{matrix})] \\ = \frac{1}{2} [\begin{matrix} 2 + 2 & 3 + 1 & 1 + 4 \\ 1 + 3 & - 1 - 1 & 2 + 1 \\ 4 + 1 & 1 + 2 & 2 + 2 \end{matrix}] = \frac{1}{2} [\begin{matrix} 4 & 4 & 5 \\ 4 & - 2 & 3 \\ 5 & 3 & 4 \end{matrix}] \\ = [\begin{matrix} 2 & 2 & \frac{5}{2} \\ 2 & - 1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{matrix}] \\ P^{'} = [\begin{matrix} 2 & 2 & \frac{5}{2} \\ 2 & - 1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{matrix}] = P \\ A s P^{'} = P ∴ P i s a s y m m e t r i c m a t r i x . \\ N o w, Q = \frac{1}{2} [A - A^{'}] \\ = \frac{1}{2} [(\begin{matrix} 2 & 3 & 1 \\ 1 & - 1 & 2 \\ 4 & 1 & 2 \end{matrix}) - (\begin{matrix} 2 & 1 & 4 \\ 3 & - 1 & 1 \\ 1 & 2 & 2 \end{matrix})] \\ = \frac{1}{2} [\begin{matrix} 2 - 2 & 3 - 1 & 1 - 4 \\ 1 - 3 & - 1 + 1 & 2 - 1 \\ 4 - 1 & 1 - 2 & 2 - 2 \end{matrix}] = \frac{1}{2} [\begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 & 1 & \frac{- 3}{2} \\ - 1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{- 1}{2} & 0 \end{matrix}] = - [\begin{matrix} 0 & - 1 & \frac{3}{2} \\ 1 & 0 & \frac{- 1}{2} \\ \frac{- 3}{2} & \frac{1}{2} & 0 \end{matrix}] = - Q \\ A s Q = - Q ∴ Q i s a s k e w s y m m e t r i c m a t r i x . \\ S o A = P + Q \\ A = [\begin{matrix} 2 & 2 & \frac{5}{2} \\ 2 & - 1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{matrix}] + [\begin{matrix} 0 & 1 & \frac{- 3}{2} \\ - 1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{- 1}{2} & 0 \end{matrix}] \\ = [\begin{matrix} 2 + 0 & 2 + 1 & \frac{5}{2} - \frac{3}{2} \\ 2 - 1 & - 1 + 0 & \frac{3}{2} + \frac{1}{2} \\ \frac{5}{2} + \frac{3}{2} & \frac{3}{2} &min \end{matrix} \end{array}$

Matrices Short Answers Type Questions

Q1. If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

Sol. $\begin{array}{l} T h e p o s s i b l e o r d e r s t h a t a m a t r i x h a v i n g 28 e l e m e n t s a r e {28 \times 1, 1 \times 28, 2 \times 14, 14 \times 2, 4 \times 7, 7 \times 4} . \\ T h e p o s s i b l e o r d e r s o f a m a t r i x h a v i n g 13 e l e m e n t s a r e {1 \times 13, 13 \times 1} . \end{array}$

Q2. In the matrix write:

(i) The order of the matrix .

(ii) The number of elements.

(iii) Write elements $a_{23}, a_{31}, a_{12}$ .

Sol.

$\begin{array}{l} (i) T h e o r d e r o f t h e g i v e n m a t r i x A i s 3 \times 3 \\ (i i) T h e n u m b e r o f e l e m e n t s i n a m a t r i x A = 3 \times 3 = 9 \\ (i i i) a_{i j} = t h e e l e m e n t s o f i^{t h} r o w a n d j^{t h} c o l u m n . \\ S o, a_{23} = x^{2} - y, a_{31} = 0, a_{12} = 1 . \end{array}$

Commonly asked questions

If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e p o s s i b l e o r d e r s t h a t a m a t r i x h a v i n g 28 e l e m e n t s a r e {28 \times 1, 1 \times 28, 2 \times 14, 14 \times 2, 4 \times 7, 7 \times 4} . \\ T h e p o s s i b l e o r d e r s o f a m a t r i x h a v i n g 13 e l e m e n t s a r e {1 \times 13, 13 \times 1} . \end{array}$

In the matrix write:

(i) The order of the matrix .

(ii) The number of elements.

(iii) Write elements $a_{23}, a_{31}, a_{12}$ .

This is a short answer type question as classified in NCERT Exemplar

Construct a_{2 × 2} matrix where

(i) $a_{i j} = \frac{{(i - 2 j)}^{2}}{2}$ .
(ii) $a_{i j} = ? - 2 i + 3 j ?$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = {[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]}_{2 \times 2} \\ (i) G i v e n t h a t a_{i j} = \frac{{(i - 2 j)}^{2}}{2} \\ a_{11} = \frac{{(1 - 2 \times 1)}^{2}}{2} = \frac{1}{2}; a_{12} = \frac{{(1 - 2 \times 2)}^{2}}{2} = \frac{9}{2} \\ a_{21} = \frac{{(2 - 2 \times 1)}^{2}}{2} = 0; a_{22} = \frac{{(2 - 2 \times 2)}^{2}}{2} = 2 \\ H e n c e, t h e m a t r i x A = [\begin{matrix} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{matrix}] \\ (i i) G i v e n t h a t a_{i j} = | - 2 i + 3 j | \\ a_{11} = | - 2 \times 1 + 3 \times 1 | = 1; a_{12} = | - 2 \times 1 + 3 \times 2 | = 4 \\ a_{21} = | - 2 \times 2 + 3 \times 1 | = - 1; a_{22} = | - 2 \times 2 + 3 \times 2 | = 2 \\ H e n c e, t h e m a t r i x A = [\begin{matrix} 1 & 4 \\ - 1 & 2 \end{matrix}] \end{array}$

Construct a $3 \times 2$ matrix whose elements are given by $a_{i j} = e^{i x} \cdot s i n (j x)$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = {[\begin{array}{l} a_{11} a_{12} \\ a_{21} a_{22} \\ a_{31} a_{32} \end{array}]}_{3 \times 2} \\ G i v e n t h a t a_{i j} = e^{i x} s i n j x \\ a_{11} = e^{x} s i n x a_{12} = e^{x} s i n 2 x \\ a_{21} = e^{2 x} s i n x a_{22} = e^{2 x} s i n 2 x \\ a_{31} = e^{3 x} s i n x a_{32} = e^{3 x} s i n 2 x \\ H e n c e, t h e m a t r i x A = [\begin{array}{l} e^{x} s i n x e^{x} s i n 2 x \\ e^{2 x} s i n x e^{2 x} s i n 2 x \\ e^{3 x} s i n x e^{3 x} s i n 2 x \end{array}] \end{array}$

Find values of $a$ and $b$ if $A = B$ , where $A = [\begin{matrix} a + 4 & 3 b \\ 8 & - 6 \end{matrix}], B = [\begin{matrix} 2 a + b & b^{2} + 2 \\ 8 & b^{2} - 5 b \end{matrix}]$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = B \\ \Rightarrow [\begin{matrix} a + 4 & 3 b \\ 8 & - 6 \end{matrix}] = [\begin{matrix} 2 a + 2 & b^{2} + 2 \\ 8 & b^{2} - 5 b \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t \\ a + 4 = 2 a + 2, 3 b = b^{2} + 2 a n d b^{2} - 5 b = - 6 \\ \Rightarrow 2 a - a = 2, b^{2} - 3 b + 2 = 0, b^{2} - 5 b + 6 = 0 \\ ∴ a = 2 \\ ∴ b^{2} - 3 b + 2 = 0 ∴ b^{2} - 5 b + 6 = 0 \\ \Rightarrow b^{2} - 2 b - b + 2 = 0, \Rightarrow b^{2} - 3 b - 2 b + 6 = 0 \\ \Rightarrow b (b - 2) - 1 (b - 2) = 0, \Rightarrow b (b - 3) - 2 (b - 3) = 0 \\ \Rightarrow (b - 1) (b - 2) = 0, \Rightarrow (b - 2) (b - 3) = 0 \\ ∴ b = 1, 2 ∴ b = 2, 3 \\ b u t h e r e 2 i s c o m m o n . \\ H e n c e, t h e v a l u e o f a = 2 a n d b = 2 . \end{array}$

If possible, find the sum of the matrices $A$ and $B$ , where

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e o r d e r o f m a t r i x A = 2 \times 2 a n d t h e o r d e r o f m a t r i x B = 2 \times 3 . A d d i t i o n o f m a t r i c e s i s o n l y \\ p o s s i b l e w h e n t h e y h a v e s a m e o r d e r . S o, A + B i s n o t p o s s i b l e . \end{array}$

If find:

(i) $X + Y$ .

(ii) $2 X - 3 Y$ .

(iii) A matrix $Z$ such that $X + Y + Z$ is a zero matrix.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t X = [\begin{matrix} 3 & 1 - 1 \\ 5 & - 2 - 3 \end{matrix}] a n d Y = [\begin{matrix} 2 & 1 - 1 \\ 7 & 2 4 \end{matrix}] \\ (i) X + Y = [\begin{matrix} 3 & 1 - 1 \\ 5 & - 2 - 3 \end{matrix}] + [\begin{matrix} 2 & 1 - 1 \\ 7 & 2 4 \end{matrix}] \\ = [\begin{matrix} 3 + 2 & 1 + 1 - 1 - 1 \\ 5 + 7 & - 2 + 2 - 3 + 4 \end{matrix}] = [\begin{matrix} 5 & 2 - 2 \\ 12 & 0 1 \end{matrix}] \\ (i i) 2 X - 3 Y = 2 [\begin{matrix} 3 & 1 - 1 \\ 5 & - 2 - 3 \end{matrix}] - 3 [\begin{matrix} 2 & 1 - 1 \\ 7 & 2 4 \end{matrix}] \\ = [\begin{matrix} 2 \times 3 & 2 \times 1 - 2 \times 1 \\ 2 \times 5 & - 2 \times 2 - 2 \times 3 \end{matrix}] - [\begin{matrix} 3 \times 2 & 1 \times 3 - 1 \times 3 \\ 3 \times 7 & 3 \times 2 3 \times 4 \end{matrix}] \\ = [\begin{matrix} 6 & 2 - 2 \\ 10 & - 4 - 6 \end{matrix}] - [\begin{matrix} 6 & 3 - 3 \\ 21 & 6 12 \end{matrix}] \\ = [\begin{matrix} 6 - 6 & 2 - 3 - 2 + 3 \\ 10 - 21 & - 4 - 6 - 6 - 12 \end{matrix}] = [\begin{matrix} 0 & - 1 1 \\ - 11 & - 10 - 18 \end{matrix}] \\ (i i i) X + Y + Z = 0 \\ \Rightarrow [\begin{matrix} 3 & 1 - 1 \\ 5 & - 2 - 3 \end{matrix}] + [\begin{matrix} 2 & 1 - 1 \\ 7 & 2 4 \end{matrix}] + [\begin{matrix} a & b c \\ d & e f \end{matrix}] = [\begin{matrix} 0 & 0 0 \\ 0 & 0 0 \end{matrix}] \\ w h e r e Z = [\begin{matrix} a & b c \\ d & e f \end{matrix}] \\ \Rightarrow = [\begin{matrix} 3 + 2 + a & 1 + 1 + b - 1 - 1 + c \\ 5 + 7 + d & - 2 + 2 + e - 3 + 4 + f \end{matrix}] = [\begin{matrix} 0 & 0 0 \\ 0 & 0 & \end{matrix} \end{array}$

Find non-zero values of $x$ satisfying the matrix equation:

x $[\begin{matrix} 2 x & 2 \\ 3 & x \end{matrix}] + 2 [\begin{matrix} 8 & 5 x \\ 4 & 4 x \end{matrix}] = [\begin{matrix} x^{2} + 8 & 24 \\ (10) & 6 x \end{matrix}] .$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n e q u a t i o n c a n b e w r i t t e n a s \\ [\begin{matrix} 2 x^{2} & 2 x \\ 3 x & x^{2} \end{matrix}] + [\begin{matrix} 16 & 10 x \\ 8 & 8 x \end{matrix}] = [\begin{matrix} 2 x^{2} + 16 & 48 \\ 20 & 12 x \end{matrix}] \\ \Rightarrow [\begin{matrix} 2 x^{2} + 16 & a_{12} \\ 3 x + 8 & x^{2} + 8 x \end{matrix}] = [\begin{matrix} 2 x^{2} + 16 & 48 \\ 20 & 12 x \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t \\ 12 x = 48, 3 x + 8 = 20, x^{2} + 8 x = 12 x \\ ∴ x = \frac{48}{12} = 4, 3 x = 20 - 8 = 12, \Rightarrow x^{2} = 12 x - 8 x = 4 x \\ ∴ x = 4, \Rightarrow x^{2} - 4 x = 0 \\ x = 0, x = 4 \\ H e n c e, t h e n o n - z e r o v a l u e s o f x i s 4 . \end{array}$

If $A = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}],$ show that $(A + B) (A - B) \neq A^{2} - B^{2}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \\ A + B = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] + [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \\ \Rightarrow A + B = [\begin{matrix} 0 + 0 & 1 - 1 \\ 1 + 1 & 1 + 0 \end{matrix}] \Rightarrow A + B = [\begin{matrix} 0 & 0 \\ 2 & 1 \end{matrix}] \\ A - B = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] - [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \\ \Rightarrow A - B = [\begin{matrix} 0 - 0 & 1 + 1 \\ 1 - 1 & 1 - 0 \end{matrix}] \Rightarrow A - B = [\begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix}] \\ ∴ (A + B) . (A - B) = [\begin{matrix} 0 & 0 \\ 2 & 1 \end{matrix}] [\begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 4 + 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 5 \end{matrix}] \\ N o w, R . H . S . = A^{2} - B^{2} \\ = A . A - B . B \\ = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] - [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 & 0 + 1 \\ 0 + 1 & 1 + 1 \end{matrix}] - [\begin{matrix} 0 - 1 & 0 + 0 \\ 0 + 0 & - 1 + 0 \end{matrix}] \\ = [\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}] - [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] = [\begin{matrix} 1 + 1 & 1 - 0 \\ 1 - 0 & 2 + 1 \end{matrix}] = [\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix}] \\ H e n c e, [\begin{matrix} 0 & 0 \\ 0 & 5 \end{matrix}] \neq [\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix}] \\ H e n c e, (A + B) . (A - B) \neq A^{2} - B^{2} \end{array}$

Find the value of $x$ if

This is a short answer type question as classified in NCERT Exemplar

Show that $A = [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}]$ satisfies the equation $A^{2} - 3 A - 7 I = O$ , and hence find $A^{- 1}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}] \\ A^{2} = A . A \\ = [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}] [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}] = [\begin{matrix} 25 - 3 & 15 - 6 \\ - 5 + 2 & - 3 + 4 \end{matrix}] = [\begin{matrix} 22 & 9 \\ - 3 & 1 \end{matrix}] \\ A^{2} - 3 A - 7 I = 0 \\ L . H . S . [\begin{matrix} 22 & 9 \\ - 3 & 1 \end{matrix}] - 3 [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}] - 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ \Rightarrow [\begin{matrix} 22 & 9 \\ - 3 & 1 \end{matrix}] - [\begin{matrix} 15 & 9 \\ - 3 & - 6 \end{matrix}] - [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}] \\ \Rightarrow [\begin{matrix} 22 - 15 - 7 & 9 - 9 - 0 \\ - 3 + 3 - 0 & 1 + 6 - 7 \end{matrix}] \Rightarrow [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] R . H . S . \\ W e a r e g i v e n A^{2} - 3 A - 7 I = 0 \\ \Rightarrow A^{- 1} [A^{2} - 3 A - 7 I] = A^{- 1} 0 [P r e - m u l t i p l y i n g b o t h s i d e s b y A^{- 1}] \\ \Rightarrow A^{- 1} A . A - 3 A^{- 1} . A - 7 A^{- 1} I = 0 [A^{- 1} 0 = 0] \\ \Rightarrow I . A - 3 I - 7 A^{- 1} I = 0 \\ \Rightarrow A - 3 I - 7 A^{- 1} = 0 \\ \Rightarrow - 7 A^{- 1} = 3 I - A \\ \Rightarrow A^{- 1} = \frac{1}{- 7} [3 I - A] \\ \Rightarrow A^{- 1} = \frac{1}{- 7} [3 (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) - [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}]] \\ = \frac{1}{- 7} [(\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}) - [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}]] \\ = \frac{1}{- 7} [\begin{matrix} 3 - 5 & 0 - 3 \\ 0 + 1 & 3 + 2 \end{matrix}] = \frac{1}{- 7} [\begin{matrix} - 2 & - 3 \\ 1 & 5 \end{matrix}] \\ H e n c e, A^{- 1} = \frac{1}{- 7} [\begin{matrix} - 2 & - 3 \\ 1 & 5 \end{matrix}] \end{array}$

Find the matrix $A$ satisfying the matrix equation:

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = {[\begin{matrix} a & b \\ c & d \end{matrix}]}_{2 \times 2} \\ {[\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix}]}_{2 \times 2} {[\begin{matrix} a & b \\ c & d \end{matrix}]}_{2 \times 2} {[\begin{matrix} - 3 & 2 \\ 5 & - 3 \end{matrix}]}_{2 \times 2} = {[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]}_{2 \times 2} \\ \Rightarrow {[\begin{matrix} 2 a + c & 2 b + d \\ 3 a + 2 c & 3 b + 2 d \end{matrix}]}_{2 \times 2} {[\begin{matrix} - 3 & 2 \\ 5 & - 3 \end{matrix}]}_{2 \times 2} = {[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]}_{2 \times 2} \\ \Rightarrow [\begin{matrix} - 6 a - 3 c + 10 b + 5 d & 4 a + 2 c - 6 b - 3 d \\ - 9 a - 6 c + 15 b + 10 d & 6 a + 4 c - 9 b - 6 d \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t, \\ - 6 a - 3 c + 10 b + 5 d = 1 \dots (1) \\ - 9 a - 6 c + 15 b + 10 d = 0 \dots (2) \\ 4 a + 2 c - 6 b - 3 d = 0 \dots (3) \\ 6 a + 4 c - 9 b - 6 d = 1 \dots (4) \\ M u l t i p l y i n g e q . (1) b y 2 a n d s u b t r a c t i n g e q . (2), w e g e t, \\ - 12 a - 6 c + 20 b + 10 d = 2 \\ - 9 a - 6 c + 15 b + 10 d = 0 \\ &t \end{array}$

Find $A$ , if Kindly Consider the following

This is a short answer type question as classified in NCERT Exemplar

If then verify ${(B A)}^{2} \neq B^{2} A^{2}$

This is a short answer type question as classified in NCERT Exemplar

If possible, find $B A$ and $A B$ , where

This is a short answer type question as classified in NCERT Exemplar

Show by an example that for $A \neq O, B \neq O$ , $A B = O$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] \\ A B = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] \\ A B = [\begin{matrix} 1 - 1 & 1 - 1 \\ - 1 + 1 & - 1 + 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 \\ H e n c e, A = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] a n d B = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] . \end{array}$

Given Is ${(A B)}^{'} = B^{'} A^{'}$ ?

This is a short answer type question as classified in NCERT Exemplar

Solve for x and y :

Kindly Consider the following

This is a short answer type question as classified in NCERT Exemplar

If X and Y are $2 \times 2$ matrices, then solve the following matrix equations for X and Y:

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}], 3 X + 2 Y = [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] .$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \\ 2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \dots (1) \\ 3 X + 2 Y = [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \dots (2) \\ M u l t i p l y i n g e q . (1) b y 3 a n d e q . (2) b y 2, w e g e t, \\ 3 [2 X + 3 Y] = 3 [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \Rightarrow 6 X + 9 Y = [\begin{matrix} 6 & 9 \\ 12 & 0 \end{matrix}] \dots (3) \\ 2 [3 X + 2 Y] = 2 [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \Rightarrow 6 X + 4 Y = [\begin{matrix} - 4 & 4 \\ 2 & - 10 \end{matrix}] \dots (4) \\ O n s u b t r a c t i n g e q . (4) f r o m e q . (3) w e g e t \\ 5 Y = [\begin{matrix} 6 + 4 & 9 - 4 \\ 12 - 2 & 0 + 10 \end{matrix}] \\ 5 Y = [\begin{matrix} 10 & 5 \\ 10 & 10 \end{matrix}] \Rightarrow Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] \\ N o w, p u t t i n g t h e v a l u e o f Y i n e q u a t i o n (1) w e g e t, \\ 2 X + 3 [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \\ \Rightarrow 2 X + [\begin{matrix} 6 & 3 \\ 6 & 6 \end{matrix}] = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] - [\begin{matrix} 6 & 3 \\ 6 & 6 \end{matrix}] \\ \Rightarrow 2 X = [\begin{matrix} 2 - 6 & 3 - 3 \\ 4 - 6 & 0 - 6 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} - 4 & 0 \\ - 2 & - 6 \end{matrix}] \\ X = \frac{1}{2} [\begin{matrix} - 4 & 0 \\ - 2 & - 6 \end{matrix}] \Rightarrow X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] \\ H e n c e, X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] a n d Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] . \end{array}$

If $A = [3 5], B = [7 3]$ , then find a non-zero matrix $C$ such that $A C = B C$ .

This is a short answer type question as classified in NCERT Exemplar

Give an example of matrices $A$ , $B$ , and $C$ such that $A B = A C$ , where $A$ is a non-zero matrix, but $B \neq C$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], B = [\begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix}] a n d C = [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}] \\ A B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 + 0 & 2 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}] \\ A C = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}] \Rightarrow A C = [\begin{matrix} 1 + 0 & 2 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}] \\ H e n c e, A B = A C f o r m a t r i x A i s n o n - z e r o a n d B \neq C . \end{array}$

If $A = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}], B = [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}], C = [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}],$ verify:

(i) $(A B) C = A (B C)$ .

(ii) $A (B + C) = A B + A C$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}], B = [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}] a n d C = [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}] \\ (i) T o v e r i f y : (A B) C = A (B C) \\ A B = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}] = [\begin{matrix} 2 + 6 & 3 - 8 \\ - 4 + 3 & - 6 - 4 \end{matrix}] = [\begin{matrix} 8 & - 5 \\ - 1 & - 10 \end{matrix}] \\ L . H . S . \\ (A B) C = [\begin{matrix} 8 & - 5 \\ - 1 & - 10 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}] = [\begin{matrix} 8 + 5 & 0 + 0 \\ - 1 + 10 & 0 + 0 \end{matrix}] = [\begin{matrix} 13 & 0 \\ 9 & 0 \end{matrix}] \\ B C = [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}] = [\begin{matrix} 2 - 3 & 0 + 0 \\ 3 + 4 & 0 + 0 \end{matrix}] = [\begin{matrix} - 1 & 0 \\ 7 & 0 \end{matrix}] \\ R . H . S . \\ A (B C) = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} - 1 & 0 \\ 7 & 0 \end{matrix}] = [\begin{matrix} - 1 + 14 & 0 + 0 \\ 2 + 7 & 0 + 0 \end{matrix}] = [\begin{matrix} 13 & 0 \\ 9 & 0 \end{matrix}] \\ L . H . S . = R . H . S . \\ S o, (A B) C = A (B C) \\ (i i) T o v e r i f y : A (B + C) = A B + A C \\ B + C = [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}] + [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}] \\ = [\begin{matrix} 2 + 1 & 3 + 0 \\ 3 - 1 & - 4 + 0 \end{matrix}] = [\begin{matrix} 3 & 3 \\ 2 & - 4 \end{matrix}] \\ L . H . S . A (B + C) = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 3 & 3 \\ 2 & - 4 \end{matrix}] \\ = [\begin{matrix} 3 + 4 & 3 - 8 \\ - 6 + 2 & - 6 - 4 \end{matrix}] = [\begin{matrix} 7 & - 5 \\ - 4 & - 10 \end{matrix}] \\ A B = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 2 & 3 \\ 3 & - 4 \end{matrix}] \\ = [\begin{matrix} 2 + 6 & 3 - 8 \\ - 4 + 3 & - 6 - 4 \end{matrix}] = [\begin{matrix} 8 & - 5 \\ - 1 & - 10 \end{matrix}] \\ A C = [\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 0 \end{matrix}] \\ = [\begin{matrix} 1 - 2 & 0 + 0 \\ - 2 - 1 & 0 + 0 \end{matrix}] = [\begin{matrix} - 1 & 0 \\ - 3 & 0 \end{matrix}] \\ R . H . S . A B + A C = [\begin{matrix} 8 & - 5 \\ - 1 & - 10 \end{matrix}] + [\begin{matrix} - 1 & 0 \\ - 3 & 0 \end{matrix}] = [\begin{matrix} 8 - 1 & - 5 + 0 \\ - 1 - 3 & - 10 + 0 \end{matrix}] \\ = [\begin{matrix} 7 & - 5 \\ - 4 & - 10 \end{matrix}] \\ L . H . S . = R . H . S . \\ H e n c e, A (B + C) = A B + A C \end{array}$

If $P = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] a n d Q = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}],$ prove that $P Q = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] = Q P$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] a n d Q = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] \\ P Q = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] \\ = [\begin{matrix} x a + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + y b + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + z c \end{matrix}] \\ = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] \\ N o w, Q P = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] \\ = [\begin{matrix} x a + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + y b + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + z c \end{matrix}] \\ = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] \\ H e n c e, P Q = Q P . \end{array}$

Kindly Consider the following find A.

This is a short answer type question as classified in NCERT Exemplar

If verify that $A (B + C) = (A B + A C)$ .

This is a short answer type question as classified in NCERT Exemplar

If $A = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}],$ then verify that $A^{2} + A = A (A + I)$ , where $I$ is a $3 \times 3$ unit matrix.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] \\ A^{2} = A . A \\ = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] \\ = [\begin{matrix} 1 + 0 + 0 & 0 + 0 - 1 & - 1 + 0 - 1 \\ 2 + 2 + 0 & 0 + 1 + 3 & - 2 + 3 + 3 \\ 0 + 2 + 0 & 0 + 1 + 1 & 0 + 3 + 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & - 2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{matrix}] \\ L . H . S . A^{2} + A = [\begin{matrix} 1 & - 1 & - 2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{matrix}] + [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] \\ = [\begin{matrix} 1 + 1 & - 1 + 0 & - 2 - 1 \\ 4 + 2 & 4 + 1 & 4 + 3 \\ 2 + 0 & 2 + 1 & 4 + 1 \end{matrix}] = [\begin{matrix} 2 & - 1 & - 3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{matrix}] \\ R . H . S . A (A + I) = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] [(\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}) + (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})] \\ = [\begin{matrix} 1 & 0 & - 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 2 & 0 & - 1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{matrix}] \\ = [\begin{matrix} 2 + 0 + 0 & 0 + 0 - 1 & - 1 + 0 - 2 \\ 4 + 2 + 0 & 0 + 2 + 3 & - 2 + 3 + 6 \\ 0 + 2 + 0 & 0 + 2 + 1 & 0 + 3 + 2 \end{matrix}] = [\begin{matrix} 2 & - 1 & - 3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{matrix}] \\ L . H . S . = R . H . S \\ A^{2} + A = A (A + I) . H e n c e v e r i f i e d . \end{array}$

then verify that:

(i) ${(A^{'})}^{'} = A$ .

(ii) ${(A B)}^{'} = B^{'} A^{'}$ .

(iii) ${(k A)}^{'} = k (A^{'})$ .

This is a short answer type question as classified in NCERT Exemplar

If then verify that:

(i) ${(2 A + B)}^{'} = 2 A^{'} + B^{'}$ .

(ii) ${(A - B)}^{'} = A^{'} - B^{'}$ .

This is a short answer type question as classified in NCERT Exemplar

Show that $A^{'} A$ and $A A^{'}$ are both symmetric matrices for any matrix $A$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t P = A^{'} A \\ \Rightarrow P^{'} = {(A^{'} A)}^{'} \\ \Rightarrow P^{'} = A^{'} {(A^{'})}^{'} [{(A B)}^{'} = B^{'} A^{'}] \\ \Rightarrow P^{'} = A^{'} A [? {(A^{'})}^{'} = A] \\ \Rightarrow P^{'} = P \\ H e n c e, A^{'} A i s a s y m m e t r i c m a t r i x . \\ N o w, L e t Q = A A^{'} \\ \Rightarrow Q^{'} = {(A A^{'})}^{'} \\ \Rightarrow Q^{'} = {(A^{'})}^{'} A^{'} [{(A B)}^{'} = B^{'} A^{'}] \\ \Rightarrow P^{'} = A A^{'} [? {(A^{'})}^{'} = A] \\ \Rightarrow Q^{'} = Q \\ H e n c e, A^{'} A i s a l s o a s y m m e t r i c m a t r i x . \end{array}$

Let $A$ and $B$ be square matrices of the order $3 \times 3$ . Is ${(A B)}^{2} = A^{2} B^{2}$ ? Give reasons

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A ? ? a n d B a r e t h e m a t r i c e s o f t h e o r d e r 3 \times 3 . \\ {(A B)}^{2} = A B . A B \\ = A A . B B \\ = A^{2} . B^{2} \\ H e n c e, {(A B)}^{2} = A^{2} . B^{2} \end{array}$

Show that if $A$ and $B$ are square matrices such that $A B = B A$ , then ${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o p r o v e t h a t {(A + B)}^{2} = A^{2} + 2 A B + B^{2} \\ L . H . S . {(A + B)}^{2} = (A + B) (A + B) [? A^{2} = A . A] \\ = A . A + A B + B A + B . B \\ = A^{2} + A B + A B + B^{2} [A B = B A] \\ = A^{2} + 2 A B + B^{2} R . H . S . \\ S o, L . H . S . = R . H . S . \end{array}$

Let $A = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}]$ , $B = [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}]$ , $C = [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}]$ , $a = 4$ , $b = - 2$ .

Show that:

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) T o p r o v e t h a t A + (B + C) = (A + B) + C \\ L . H . S . A + (B + C) = (\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}) + [(\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}) + (\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix})] \\ = (\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}) + [\begin{matrix} 4 + 2 & 0 + 0 \\ 1 + 1 & 5 - 2 \end{matrix}] \\ = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}] + [\begin{matrix} 6 & 0 \\ 2 & 3 \end{matrix}] \\ = [\begin{matrix} 1 + 6 & 2 + 0 \\ - 1 + 2 & 3 + 3 \end{matrix}] = [\begin{matrix} 7 & 2 \\ 1 & 6 \end{matrix}] \\ R . H . S . (A + B) + C = [(\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}) + (\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix})] + (\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}) \\ = [\begin{matrix} 1 + 4 & 2 + 0 \\ - 1 + 1 & 3 + 5 \end{matrix}] + [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] \\ = [\begin{matrix} 5 & 2 \\ 0 & 8 \end{matrix}] + [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] \\ = [\begin{matrix} 5 + 2 & 2 + 0 \\ 0 + 1 & 8 - 2 \end{matrix}] = [\begin{matrix} 7 & 2 \\ 1 & 6 \end{matrix}] \\ H e n c e, L . H . S . = R . H . S . \\ A + (B + C) = (A + B) + C H e n c e P r o v e d . \\ (b) T o p r o v e t h a t A (B C) = (A B) C \\ L . H . S . A (B C) = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}] [(\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}) (\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix})] \\ = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 8 + 0 & 0 + 10 \\ 2 + 5 & 0 - 10 \end{matrix}] = [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 8 & 0 \\ 7 & - 10 \end{matrix}] \\ = [\begin{matrix} 8 + 14 & 0 - 20 \end{matrix} \end{array}$

$\begin{array}{l} (c) T o p r o v e t h a t (a + b) B = a B + b B \\ H e r e, a = 4 a n d b = - 2 \\ L . H . S . (a + b) B = (4 - 2) [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}] = 2 [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}] = [\begin{matrix} 8 & 0 \\ 2 & 10 \end{matrix}] \\ R . H . S . a B + b B = 4 [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}] - 2 [\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix}] = [\begin{matrix} 16 & 0 \\ 4 & 20 \end{matrix}] - [\begin{matrix} 8 & 0 \\ 2 & 10 \end{matrix}] \\ = [\begin{matrix} 16 - 8 & 0 - 0 \\ 4 - 2 & 20 - 10 \end{matrix}] = [\begin{matrix} 8 & 0 \\ 2 & 10 \end{matrix}] \\ H e n c e, L . H . S . = R . H . S . \\ (a + b) B = a B + b B H e n c e P r o v e d . \\ (d) T o p r o v e t h a t a (C - A) = a C - a A \\ L . H . S . a (C - A) = 4 [(\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}) - (\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix})] \\ = 4 [\begin{matrix} 2 - 1 & 0 - 2 \\ 1 + 1 & - 2 - 3 \end{matrix}] = 4 [\begin{matrix} 1 & - 2 \\ 2 & - 5 \end{matrix}] = [\begin{matrix} 4 & - 8 \\ 8 & - 20 \end{matrix}] \\ R . H . S . a C - a A = 4 [\begin{matrix} 2 & 0 \\ 1 & - 2 \end{matrix}] - 4 [\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}] \\ &thi \end{array}$

$\begin{array}{l} (g) T o p r o v e t h a t {(A B)}^{T} = B^{T} A^{T} \\ L . H . S . {(A B)}^{T} = {[(\begin{matrix} 1 & 2 \\ - 1 & 3 \end{matrix}) (\begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix})]}^{T} \\ = {[\begin{matrix} 4 + 2 & 0 + 10 \\ - 4 + 3 & 0 + 15 \end{matrix}]}^{T} = {[\begin{matrix} 6 & 10 \\ - 1 & 15 \end{matrix}]}^{T} = [\begin{matrix} 6 & - 1 \end{matrix} \end{array}$

If $A = [\begin{matrix} c o s θ & s i n θ \\ - s i n θ & c o s θ \end{matrix}]$ , then show that $A^{2} = [\begin{matrix} c o s 2 θ & s i n 2 θ \\ - s i n 2 θ & c o s 2 θ \end{matrix}]$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t \\ A = [\begin{matrix} c o s q & s i n q \\ - s i n q & c o s q \end{matrix}] \\ A = A . A = [\begin{matrix} c o s q & s i n q \\ - s i n q & c o s q \end{matrix}] [\begin{matrix} c o s q & s i n q \\ - s i n q & c o s q \end{matrix}] \\ = [\begin{matrix} c o s^{2} q - s i n^{2} q & c o s q s i n q + s i n q c o s q \\ s i n q c o s q - c o s q s i n q & - s i n^{2} q + c o s^{2} q \end{matrix}] \\ = [\begin{matrix} c o s 2 q & s i n 2 q \\ - s i n 2 q & c o s 2 q \end{matrix}] [\begin{array}{l} ? c o s^{2} A - s i n^{2} A = c o s 2 A \\ 2 s i n A c o s A = s i n 2 A \end{array}] \\ H e n c e p r o v e d . \end{array}$

If $A = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}]$ , $B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ , and $x^{2} = - 1$ , then show that ${(A + B)}^{2} = A^{2} + B^{2}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] a n d B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ L . H . S . {(A + B)}^{2} = (A + B) . (A + B) \\ = [(\begin{matrix} 0 & - x \\ x & 0 \end{matrix}) + (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})] . [(\begin{matrix} 0 & - x \\ x & 0 \end{matrix}) + (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})] \\ = [\begin{matrix} 0 + 0 & - x + 1 \\ x + 1 & 0 + 0 \end{matrix}] . [\begin{matrix} 0 + 0 & - x + 1 \\ x + 1 & 0 + 0 \end{matrix}] \\ = [\begin{matrix} 0 + (- x + 1) (x + 1) & 0 + 0 \\ 0 + 0 & (x + 1) (- x + 1) + 0 \end{matrix}] \\ = [\begin{matrix} 1 - x^{2} & 0 \\ 0 & 1 - x^{2} \end{matrix}] \\ P u t x^{2} = - 1 \\ = [\begin{matrix} 1 + 1 & 0 \\ 0 & 1 + 1 \end{matrix}] = [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \\ R . H . S . \\ A^{2} + B^{2} = A . A + B . B \\ = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] . [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] + [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] . [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 - x^{2} & 0 + 0 \\ 0 + 0 & - x^{2} + 0 \end{matrix}] + [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] \\ = [\begin{matrix} - x^{2} & 0 \\ 0 & - x^{2} \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} - x^{2} + 1 & 0 + 0 \\ 0 + 0 & - x^{2} + 1 \end{matrix}] \\ = [\begin{matrix} - x^{2} + 1 & 0 \\ 0 & - x^{2} + 1 \end{matrix}] = [\begin{matrix} 1 + 1 & 0 \\ 0 & 1 + 1 \end{matrix}] [? x^{2} = - 1] \\ = [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \\ H e n c e, L . H . S . = R . H . S . \\ {(A + B)}^{2} = A^{2} + B^{2} \end{array}$

Verify that $A^{2} = I$ when $A = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}]$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] \\ L . H . S . A^{2} = A . A = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] . [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] \\ = [\begin{matrix} 0 + 4 - 3 & 0 - 3 + 3 & 0 + 4 - 4 \\ 0 - 12 + 12 & 4 + 9 - 12 & - 4 - 12 + 16 \\ 0 - 12 + 12 & 3 + 9 - 12 & - 3 - 12 + 16 \end{matrix}] \\ = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = 1 R . H . S . \\ H e n c e, A^{2} = I i s v e r i f i e d . \end{array}$

Prove by Mathematical Induction that ${(A^{'})}^{n} = {(A^{n})}^{'}$ , where $n \in N$ for any square matrix $A$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o p r o v e t h a t {(A^{'})}^{n} = {(A^{n})}^{'} \\ L e t P (n) : {(A^{'})}^{n} = {(A^{n})}^{'} \\ S t e p 1 : P u t n = 1, P (1) : A^{'} = A^{'} w h i c h i s t r u e f o r n = 1 \\ S t e p 2 : P u t n = K, P (K) : {(A^{'})}^{K} = {(A^{K})}^{'} L e t i t b e t r u e f o r n = K \\ S t e p 3 : P u t n = K + 1, P (K + 1) : {(A^{'})}^{K + 1} = {(A^{K + 1})}^{'} \\ L . H . S . {(A^{'})}^{K + 1} = {(A^{'})}^{K} . (A^{'}) \\ = {(A^{K})}^{'} . (A^{'}) [F r o m s t e p 2] \\ = {(A^{K} . A)}^{'} = {(A^{K + 1})}^{'} R . H . S . \\ T h e g i v e n s t a t e m e n t i s t r u e f o r P (K + 1) w h e n e v e r i t i s t r u e f o r P (K), w h e r e K \in N . \end{array}$

Find inverse, by elementary row operations (if possible), of the following matrices:

(i) $[\begin{matrix} 1 & 3 \\ 5 & 7 \end{matrix}]$

(ii) $[\begin{matrix} 1 & - 3 \\ - 2 & 6 \end{matrix}]$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) L e t A = [\begin{matrix} 1 & 3 \\ - 5 & 7 \end{matrix}] \\ | A | = 1 \times 7 - (- 5) \times 3 = 7 + 15 = 22 \neq 0 \\ S o, A i s i n v e r t i b l e . \\ L e t A = I A \\ [\begin{matrix} 1 & 3 \\ - 5 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A \\ R_{2} \to R_{2} + 5 R_{1} \\ \Rightarrow [\begin{matrix} 1 & 3 \\ 0 & 22 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix}] A \\ R_{2} \to \frac{1}{22} R_{2} \\ \Rightarrow [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ \frac{5}{22} & \frac{1}{22} \end{matrix}] A \\ R_{1} \to R_{1} - 3 R_{2} \\ \Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{7}{22} & \frac{- 3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{matrix}] A \\ S o, A^{- 1} = [\begin{matrix} \frac{7}{22} & \frac{- 3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{matrix}] \Rightarrow \frac{1}{22} [\begin{matrix} 7 & - 3 \\ 5 & 1 \end{matrix}] \\ H e n c e, i n v e r s e o f [\begin{matrix} 1 & 3 \\ - 5 & 7 \end{matrix}] i s \frac{1}{22} [\begin{matrix} 7 & - 3 \\ 5 & 1 \end{matrix}] \\ (i i) L e t A = [\begin{matrix} 1 & - 3 \\ - 2 & 6 \end{matrix}] \\ | A | = 1 \times 6 - (- 3) \times (- 2) = 6 - 6 = 0 \Rightarrow | A | = 0 \\ S o, A i s n o t i n v e r t i b l e . \\ H e n c e, i n v e r s e o f [\begin{matrix} 1 & - 3 \\ - 2 & 6 \end{matrix}] i s n o t p o s s i b l e . \end{array}$

If $[\begin{matrix} x y & 4 \\ z + 6 & x + y \end{matrix}] = [\begin{matrix} 8 & w \\ 0 & 6 \end{matrix}]$ , then find the values of $x$ , $y$ , $z$ , and $w$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : [\begin{matrix} x y & 4 \\ z + 6 & x + y \end{matrix}] = [\begin{matrix} 8 & w \\ 0 & 6 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, \\ x y = 8, w = 4, z + 6 = 0 \Rightarrow z = - 6, x + y = 6 \\ N o w, s o l v i n g x + y = 6 \dots (1) \\ a n d x y = 8 \dots (2) \\ F r o m e q n . (1), y = 6 - x \dots (3) \\ P u t t i n g t h e v a l u e o f y i n e q n . (2) w e g e t, \\ x (6 - x) = 8 \Rightarrow 6 x - x^{2} = 8 \\ \Rightarrow x^{2} - 6 x + 8 = 0 \Rightarrow x^{2} - 4 x - 2 x + 8 = 0 \\ \Rightarrow x (x - 4) - 2 (x - 4) = 0 \Rightarrow (x - 4) (x - 2) = 0 \\ ∴ x = 4, 2 \\ F r o m e q n . (3), y = 2, 4 . \\ H e n c e, x = 4 o r 2, y = 2 o r 4, z = - 6 a n d w = 4 \end{array}$

If $A = [\begin{matrix} 1 & 5 \\ 7 & 12 \end{matrix}]$ and $B = [\begin{matrix} 9 & 1 \\ 7 & 8 \end{matrix}]$ , find a matrix $C$ such that $3 A + 5 B + 2 C$ is a null matrix.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} O r d e r o f m a t r i c e s A a n d B i s 2 \times 2 . \\ ∴ O r d e r o f m a t r i x C m u s t b e 2 \times 2 . \\ L e t C = [\begin{matrix} a & b \\ c & d \end{matrix}] \\ ∴ 3 A + 5 B + 2 C = 0 \\ \Rightarrow 3 [\begin{matrix} 1 & 5 \\ 7 & 12 \end{matrix}] + 5 [\begin{matrix} 9 & 1 \\ 7 & 8 \end{matrix}] + 2 [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ \Rightarrow [\begin{matrix} 3 & 15 \\ 21 & 36 \end{matrix}] + [\begin{matrix} 45 & 5 \\ 35 & 40 \end{matrix}] + [\begin{matrix} 2 a & 2 b \\ 2 c & 2 d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ \Rightarrow [\begin{matrix} 3 + 45 + 2 a & 15 + 5 + 2 b \\ 21 + 35 + 2 c & 36 + 40 + 2 d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ \Rightarrow [\begin{matrix} 48 + 2 a & 20 + 2 b \\ 56 + 2 c & 76 + 2 d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t, \\ 48 + 2 a = 0 \Rightarrow 2 a = - 48 \Rightarrow a = - 24 \\ 20 + 2 b = 0 \Rightarrow 2 b = - 20 \Rightarrow b = - 10 \\ 56 + 2 c = 0 \Rightarrow 2 c = - 56 \Rightarrow c = - 28 \\ 76 + 2 d = 0 \Rightarrow 2 d = - 76 \Rightarrow d = - 38 \\ H e n c e, C = [\begin{matrix} - 24 & - 10 \\ - 28 & - 38 \end{matrix}] \end{array}$

If $A = [\begin{matrix} - 3 & - 5 \\ - 4 & 2 \end{matrix}]$ , then find $A^{2} - 5 A - 14 I$ . Hence, obtain $A^{3}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] \\ A^{2} = A . A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] \\ = [\begin{matrix} 9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4 \end{matrix}] = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] \\ ∴ A^{2} - 5 A - 14 I = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - 5 [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] - 14 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - [\begin{matrix} 15 & - 25 \\ - 20 & 10 \end{matrix}] - [\begin{matrix} 14 & 0 \\ 0 & 14 \end{matrix}] \\ = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] \\ = [\begin{matrix} 29 - 29 & - 25 + 25 \\ - 20 + 20 & 24 - 24 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ H e n c e, A^{2} - 5 A - 14 I = 0 \\ N o w, m u l t i p l y i n g b o t h s i d e s b y A, w e g e t, \\ A^{2} . A - 5 A . A - 14 I A = 0 A \\ \Rightarrow A^{3} - 5 A^{2} - 14 A = 0 \\ \Rightarrow A^{3} = 5 A^{2} + 14 A \\ \Rightarrow A^{3} = 5 [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] + 14 [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] \\ \Rightarrow = [\begin{matrix} 145 & - 125 \\ - 100 & 120 \end{matrix}] + [\begin{matrix} 42 & - 70 \\ - 56 & 28 \end{matrix}] \\ \Rightarrow = [\begin{matrix} 145 + 42 & - 125 - 70 \\ - 100 - 56 & 120 + 28 \end{matrix}] = [\begin{matrix} 187 & - 195 \\ - 156 & 148 \end{matrix}] \\ H e n c e, A^{3} = [\begin{matrix} 187 & - 195 \\ - 156 & 148 \end{matrix}] \end{array}$

Find the values of $a$ , $b$ , $c$ , and $d$ , if

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : 3 [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} a & 6 \\ - 1 & 2 d \end{matrix}] + [\begin{matrix} 4 & a + b \\ c + d & 3 \end{matrix}] \\ \Rightarrow [\begin{matrix} 3 a & 3 b \\ 3 c & 3 d \end{matrix}] = [\begin{matrix} a + 4 & 6 + a + b \\ - 1 + c + d & 2 d + 3 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t, \\ 3 a = a + 4 \Rightarrow 3 a - a = 4 \Rightarrow 2 a = 4 \Rightarrow a = 2 \\ 3 b = 6 + a + b \Rightarrow 3 b - b - a = 6 \Rightarrow 2 b - a = 6 \Rightarrow 2 b - 2 = 6 \\ 2 b = 8 \Rightarrow b = 4 \\ 3 c = - 1 + c + d \Rightarrow 3 c - c - d = - 1 \Rightarrow 2 c - d = - 1 \\ 3 d = 2 d + 3 \Rightarrow 3 d - 2 d = 3 \Rightarrow d = 3 \\ N o w, 2 c - d = - 1 \\ \Rightarrow 2 c - 3 = - 1 \Rightarrow 2 c = 3 - 1 \Rightarrow 2 c = 2 \Rightarrow c = 1 \\ ∴ a = 2, b = 4, c = 1, d = 3 . \end{array}$

Find the matrix $A$ such that

This is a short answer type question as classified in NCERT Exemplar

If $A = [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}]$ , find $A^{2} + 2 A + 7 I$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}] \\ A^{2} = A . A = [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}] = [\begin{matrix} 1 + 8 & 2 + 2 \\ 4 + 4 & 8 + 1 \end{matrix}] = [\begin{matrix} 9 & 4 \\ 8 & 9 \end{matrix}] \\ ∴ A^{2} + 2 A + 7 I = [\begin{matrix} 9 & 4 \\ 8 & 9 \end{matrix}] + 2 [\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}] + 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ = [\begin{matrix} 9 & 4 \\ 8 & 9 \end{matrix}] + [\begin{matrix} 2 & 4 \\ 8 & 2 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}] \\ = [\begin{matrix} 9 + 2 + 7 & 4 + 4 + 0 \\ 8 + 8 + 0 & 9 + 2 + 7 \end{matrix}] = [\begin{matrix} 18 & 8 \\ 16 & 18 \end{matrix}] \\ H e n c e, A^{2} + 2 A + 7 I = [\begin{matrix} 18 & 8 \\ 16 & 18 \end{matrix}] \end{array}$

If $A = [\begin{matrix} c o s α & s i n α \\ - s i n α & c o s α \end{matrix}]$ , and $A^{- 1} = A^{T}$ , find the value of $α$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} H e r e, A = [\begin{matrix} c o s α & s i n α \\ - s i n α & c o s α \end{matrix}] \\ G i v e n t h a t : A^{- 1} = A^{'} \\ Pre - m u l t i p l y i n g b o t h s i d e s b y A \\ A A^{- 1} = A A^{'} \\ \Rightarrow I = A A^{'} [? A A^{- 1} = I] \\ \Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} c o s α & s i n α \\ - s i n α & c o s α \end{matrix}] [\begin{matrix} c o s α & - s i n α \\ s i n α & c o s α \end{matrix}] \\ \Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} c o s^{2} α + s i n^{2} α & - s i n α c o s α + s i n α c o s α \\ - s i n α c o s α + c o s α s i n α & s i n^{2} α + c o s^{2} α \end{matrix}] \\ \Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ H e n c e, i t i s t r u e f o r a l l v a l u e s o f α . \end{array}$

If the matrix $[\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}]$ is a skew-symmetric matrix, find the values of a, b, and c.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}] A^{'} = [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] \\ F o r s k e w s y m m e t r i c m a t r i x, A^{'} = - A \\ [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] = - [\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}] \\ \Rightarrow [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] = [\begin{matrix} 0 & - a & - 3 \\ - 2 & - b & 1 \\ - c & - 1 & 0 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t \\ a = - 2, b = - b \Rightarrow 2 b = 0 \Rightarrow b = 0 a n d c = - 3 \\ H e n c e, a = - 2, b = 0 a n d c = - 3 . \end{array}$

If $P (x) = [\begin{matrix} c o s x & s i n x \\ - s i n x & c o s x \end{matrix}]$ , then show that $P (x) P (y) = P (x + y) = P (y) P (x)$ .

This is a short answer type question as classified in NCERT Exemplar

If $A$ is a square matrix such that $A^{2} = A$ , show that ${(I + A)}^{3} = 7 A + I$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o s h o w t h a t : {(I + A)}^{3} = 7 A + I \\ L . H . S . {(I + A)}^{3} = I^{3} + A^{3} + 3 I^{2} A + 3 I A^{2} \\ \Rightarrow = I + A^{2} . A + 3 I A + 3 I A^{2} \\ \Rightarrow = I + A . A + 3 I A + 3 I A [? A^{2} = A] \\ \Rightarrow = I + A^{2} + 3 I A + 3 I A \\ \Rightarrow = I + A + 3 I A + 3 I A [? A^{2} = A] \\ \Rightarrow = I + A + 3 A + 3 A \Rightarrow 7 A + I R . H . S . \\ L . H . S . = R . H . S . H e n c e, p r o v e d . \end{array}$

If $A, B$ are square matrices of the same order and $B$ is a skew-symmetric matrix, show that $A^{T} B A$ is skew-symmetric

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t B i s a s k e w s y m m e t r i c m a t r i x ∴ B^{'} = - B \\ L e t P = A^{'} B A \\ \Rightarrow P^{'} = {(A^{'} B A)}^{'} \\ \Rightarrow = A^{'} B^{'} {(A^{'})}^{'} [{(A^{'} B)}^{'} = B^{'} A^{'}] \\ \Rightarrow = A^{'} (- B) A \\ \Rightarrow = A^{'} B A = - P \\ S o P^{'} = - P \\ H e n c e, A^{'} B A i s a s k e w s y m m e t r i c m a t r i x . \end{array}$

Matrices Objective Type Questions

Choose the correct answer from the given four options in each of the Exercises 1 to 15:

Q1. The matrix $P = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}]$ is a

(A) square matrix

(B) diagonal matrix

(D) none

Sol:

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}] \\ H e r e, n u m b e r o f c o l u m n s a n d t h e n u m b e r o f r o w s a r e e q u a l i . e ., 3 . S o, A i s a s q u a r e m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Q2. Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is

(A) 9

(B) 27

(D) 512

Sol:

$\begin{array}{l} T o t a l n u m b e r o f p o s s i b l e m a t r i c e s o f o r d e r 3 \times 3 w i t h e a c h e n t r y 0 o r 2 = 2^{3 \times 3} = 2^{9} = 512 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q3. If $[\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}],$ then the value of $x + y$ is

(A) $x = 3, y = 1$

(B) $x = 2, y = 3$

(D) $x = 3, y = 3$

Sol:

$\begin{array}{l} G i v e n t h a t : [\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t, \\ 2 x + y = 7 \dots (i) \\ a n d 4 x = 7 y - 13 \dots (i i) \\ f r o m e q n . (i i) 4 x - x = 6 \\ 3 x = 6 \\ ∴ x = 2 \\ f r o m e q n . (i) 2 \times 2 + y = 7 \\ 4 + y = 7 ∴ y = 7 - 4 = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Q4. If $A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}], B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (x π) \end{matrix}],$ then $A - B$ is equal to:

Sol:

$\begin{array}{l} G i v e n t h a t : A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] \\ a n d B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ A - B = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] - \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) + c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) - t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) - s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) + t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} \frac{π}{2} & 0 \\ 0 & \frac{π}{2} \end{matrix}] [\begin{array}{l} ∵ s i n^{- 1} x + c o s^{- 1} x = \frac{π}{2} \\ t a n^{- 1} x + c o t^{- 1} x = \frac{π}{2} \end{array}] \\ = \frac{1}{π} \times \frac{π}{2} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = \frac{1}{2} I \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q5. If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$ , respectively, and $m = n$ , then the order of the matrix $(5 A - 2 B)$ is

(A) $m \times 3$

(B) $3 \times 3$

(D) $3 \times n$

Sol:

$\begin{array}{l} A s w e k n o w t h a t t h e a d d i t i o n a n d s u b t r a c t i o n o f t w o m a t r i c e s i s o n l y p o s s i b l e w h e n t h e y h a v e \\ s a m e o r d e r . I t i s a l s o g i v e n t h a t m = n . \\ ∴ O r d e r o f (5 A - 2 B) i s 3 \times n . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q6. If $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then $A^{2}$ is equal to

(A) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

(B) $[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]$

(D) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

Sol:

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q7. If matrix $A = [a_{i j}]_{2 \times 2}$ , where

a_{i j} = 1

if $i \neq j$

if $i = j$ ,then $A^{2}$ is equal to

(A) I

(B) A

(D) None of these

Sol:

$\begin{array}{l} G i v e n t h a t A = {[a_{i j}]}_{2 \times 2} \\ L e t A = {[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]}_{2 \times 2} \\ a_{11} = 0 [∵ i = j] \\ a_{12} = 1 [∵ i \neq j] \\ a_{21} = 1 [∵ i \neq j] \\ a_{22} = 0 [∵ i = j] \\ ∴ A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Q8.The matrix $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}]$ is a

(A) Identity matrix

(B) Symmetric matrix

(D) None of these

Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] \\ A^{'} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] = A \\ A^{'} = A, s o A i s a s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Q9. The matrix $[\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}]$ is a

(A) Diagonal matrix

(B) Symmetric matrix

(D) Scalar matrix

Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] \\ A^{'} = [\begin{matrix} 0 & 5 & - 8 \\ - 5 & 0 & - 12 \\ 8 & 12 & 0 \end{matrix}] \\ \Rightarrow A^{'} = - [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] = - A \\ A^{'} = - A, s o A i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Q10. If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $A B^{'}$ and $B^{'} A$ are both defined, then the order of matrix $B$ is

(A) $m \times m$

(B) $n \times n$

(D) $m \times n$

Sol:

$\begin{array}{l} O r d e r o f m a t r i x A = m \times n \\ L e t o r d e r o f m a t r i x B b e K \times P \\ I f A B^{'} i s d e f i n e d t h e n t h e o r d e r o f A B^{'} i s m \times K i f n = P \\ I f B^{'} A i s d e f i n e d t h e n t h e o r d e r o f B^{'} A i s P \times n w h e n K = m \\ N o w, o r d e r o f B^{'} = P \times K \\ ∴ o r d e r o f B^{'} = K \times P \\ = m \times n [∵ K = m, P = n] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q11. If $A$ and $B$ are matrices of the same order, then $A B^{'} - B A^{'}$ is a

(A) Skew symmetric matrix

(B) Null matrix

(D) Unit matrix

Sol:

$\begin{array}{l} L e t P = (A B^{'} - B A^{'}) \\ P^{'} = {(A B^{'} - B A^{'})}^{'} \\ = {(A B^{'})}^{'} - {(B A^{'})}^{'} \\ = {(B^{'})}^{'} A^{'} - {(A^{'})}^{'} B^{'} [∵ {(A B)}^{'} = B^{'} A^{'}] \\ = B A^{'} - A B^{'} \\ = - (A B^{'} - B A^{'}) = - P \\ P^{'} = - P, s o i t i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Q12. If $A$ is a square matrix such that $A^{2} = I$ , then ${(A - I)}^{3} + {(A + I)}^{3} - 7 A$ is equal to

(A) A

(B) I - A

(D) 3A

Sol:

$\begin{array}{l} {(A - I)}^{3} + {(A + I)}^{3} - 7 A = A^{3} - I^{3} - 3 A^{2} I + 3 A I^{2} + A^{3} + I^{3} + 3 A^{2} I + 3 A I^{2} - 7 A \\ = 2 A^{3} + 6 A I^{2} - 7 A \\ = 2 A . A^{2} + 6 A I - 7 A \\ = 2 A I + 6 A I - 7 A [A^{2} = I] \\ = 8 A I - 7 A = 8 A - 7 A = A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Q13. For any two matrices $A$ and $B$ , we have

(A) AB = BA

(B) AB $AB \neq BA$ BA

(D) None of the above

Sol:

$\begin{array}{l} W e k n o w t h a t f o r a n y t w o m t r i c e s A a n d B, w e m a y h a v e A B = B A, A B \neq B A a n d A B = 0, b u t \\ i t i s n o t a l w a y s t r u e . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q14. On using elementary column operations $C_{2} \to C_{2} - 2 C_{1}$ in the following matrix equation: $[\begin{matrix} 1 & - 3 \\ 2 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix}],$ we have:

(A) $[\begin{matrix} 1 & - 5 \\ 0 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ - 2 & 2 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 2 & 0 \end{matrix}]$

(B) $[\begin{matrix} 1 & - 5 \\ 0 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 0 & 2 \end{matrix}]$

(C) $[\begin{matrix} 1 & - 5 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 1 & - 3 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 2 & 4 \end{matrix}]$

(D) $[\begin{matrix} 1 & - 5 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 2 & 0 \end{matrix}]$

Sol:

$\begin{array}{l} G i v e n t h a t : [\begin{matrix} 1 & - 3 \\ 2 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix}] \\ Using C_{2} \to C_{2} - 2 C_{1}, w e g e t \\ [\begin{matrix} 1 & - 5 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 2 & 0 \end{matrix}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Q15. On using elementary row operation $R_{1} \to R_{1} - 3 R_{2}$ in the following matrix equation:

$[\begin{matrix} 4 & 2 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}],$ we have:

(A) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}]$

(B) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} - 1 & - 3 \\ 1 & 1 \end{matrix}]$

(C) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 1 & - 7 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 7 \end{matrix}]$

(D) $[\begin{matrix} 4 & 2 \\ - 5 & - 7 \end{matrix}] = [\begin{matrix} 1 & 2 \\ - 3 & - 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}]$

Sol:

$\begin{array}{l} W e h a v e, [\begin{matrix} 4 & 2 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}] \\ Using e l e m e n t a r y r o w t r a n s f o r m a t i o n R_{1} \to R_{1} - 3 R_{2}, w e g e t \\ [\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & - 7 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Commonly asked questions

The matrix $P = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}]$ is a

(A) Square matrix

(B) Diagonal matrix

(C) Unit matrix

(D) None

This is an Objective Type Questions as classified in NCERT Exemplar

Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is

(A) 9

(B) 27

(C) 81

(D) 512

This is an Objective Type Questions as classified in NCERT Exemplar

If $[\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}],$ then the value of $x + y$ is

(A) $x = 3, y = 1$

(B) $x = 2, y = 3$

(C) $x = 2, y = 4$

(D) $x = 3, y = 3$

This is an Objective Type Questions as classified in NCERT Exemplar

If $A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}], B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (x π) \end{matrix}],$ then $A - B$ is equal to:

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] \\ a n d B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ A - B = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] - \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) + c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) - t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) - s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) + t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} \frac{π}{2} & 0 \\ 0 & \frac{π}{2} \end{matrix}] [\begin{array}{l} ? s i n^{- 1} x + c o s^{- 1} x = \frac{π}{2} \\ t a n^{- 1} x + c o t^{- 1} x = \frac{π}{2} \end{array}] \\ = \frac{1}{π} \times \frac{π}{2} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = \frac{1}{2} I \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$ , respectively, and $m = n$ , then the order of the matrix $(5 A - 2 B)$ is

(A) $m \times 3$

(B) $3 \times 3$

(C) $m \times n$

(D) $3 \times n$

This is an Objective Type Questions as classified in NCERT Exemplar

If $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then $A^{2}$ is equal to

(A) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

(B) $[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]$

(C) $[\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}]$

(D) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

This is an Objective Type Questions as classified in NCERT Exemplar

If matrix $A = [a_{i j}]_{2 \times 2}$ , where $a_{i j} = 1$ if $i \neq j$ if $i = j$ ,then $A^{2}$ is equal to

(A) I

(B) A

(C) 0

(D) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = {[a_{i j}]}_{2 \times 2} \\ L e t A = {[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]}_{2 \times 2} \\ a_{11} = 0 [? i = j] \\ a_{12} = 1 [? i \neq j] \\ a_{21} = 1 [? i \neq j] \\ a_{22} = 0 [? i = j] \\ ∴ A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The matrix $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}]$ is a

(A) Identity matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

The matrix $[\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}]$ is a

(A) Diagonal matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) Scalar matrix

This is an Objective Type Questions as classified in NCERT Exemplar

10. If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $A B^{'}$ and $B^{'} A$ are both defined, then the order of matrix $B$ is

(A) $m \times m$

(B) $n \times n$

(C) $n \times m$

(D) $m \times n$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} O r d e r o f m a t r i x A = m \times n \\ L e t o r d e r o f m a t r i x B b e K \times P \\ I f A B^{'} i s d e f i n e d t h e n t h e o r d e r o f A B^{'} i s m \times K i f n = P \\ I f B^{'} A i s d e f i n e d t h e n t h e o r d e r o f B^{'} A i s P \times n w h e n K = m \\ N o w, o r d e r o f B^{'} = P \times K \\ ∴ o r d e r o f B^{'} = K \times P \\ = m \times n [? K = m, P = n] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If $A$ and $B$ are matrices of the same order, then $A B^{'} - B A^{'}$ is a

(A) Skew symmetric matrix

(B) Null matrix

(C) Symmetric matrix

(D) Unit matrix

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t P = (A B^{'} - B A^{'}) \\ P^{'} = {(A B^{'} - B A^{'})}^{'} \\ = {(A B^{'})}^{'} - {(B A^{'})}^{'} \\ = {(B^{'})}^{'} A^{'} - {(A^{'})}^{'} B^{'} [? {(A B)}^{'} = B^{'} A^{'}] \\ = B A^{'} - A B^{'} \\ = - (A B^{'} - B A^{'}) = - P \\ P^{'} = - P, s o i t i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $A$ is a square matrix such that $A^{2} = I$ , then ${(A - I)}^{3} + {(A + I)}^{3} - 7 A$ is equal to

(A) A

(B) I - A

(C) I + A

(D) 3A

This is an Objective Type Questions as classified in NCERT Exemplar

For any two matrices $A$ and $B$ , we have

(A) AB = BA

(B) AB $AB \neq BA$ BA

(C) AB = O

(D) None of the above

This is an Objective Type Questions as classified in NCERT Exemplar

On using elementary column operations $C_{2} \to C_{2} - 2 C_{1}$ in the following matrix equation: $[\begin{matrix} 1 & - 3 \\ 2 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix}],$ we have:

(A) $[\begin{matrix} 1 & - 5 \\ 0 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ - 2 & 2 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 2 & 0 \end{matrix}]$

(B) $[\begin{matrix} 1 & - 5 \\ 0 & 4 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 0 & 2 \end{matrix}]$

(C) $[\begin{matrix} 1 & - 5 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 1 & - 3 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 2 & 4 \end{matrix}]$

(D) $[\begin{matrix} 1 & - 5 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & - 5 \\ 2 & 0 \end{matrix}]$

This is an Objective Type Questions as classified in NCERT Exemplar

On using elementary row operation $R_{1} \to R_{1} - 3 R_{2}$ in the following matrix equation:

$[\begin{matrix} 4 & 2 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}],$ we have:

(A) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}]$

(B) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} - 1 & - 3 \\ 1 & 1 \end{matrix}]$

(C) $[\begin{matrix} - 5 & - 7 \\ 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 1 & - 7 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 7 \end{matrix}]$

(D) $[\begin{matrix} 4 & 2 \\ - 5 & - 7 \end{matrix}] = [\begin{matrix} 1 & 2 \\ - 3 & - 3 \end{matrix}] [\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}]$

This is an Objective Type Questions as classified in NCERT Exemplar

Matrices Fill in the blanks Type Questions

Q1. ___ matrix is both symmetric and skew symmetric matrix.

Sol.

$N u l l m a t r i x i . e . [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] o r [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] i s b o t h s y m m e t r i c a n d s k e w s y m m e t r i c m a t r i x .$

Q2. Sum of two skew symmetric matrices is always ___ matrix.

Sol.

$\begin{array}{l} L e t A a n d B b e a n y t w o m a t r i c e s \\ ∴ F o r s k e w s y m m e t r i c m a t r i c e s \\ A = - A^{'} \dots (i) \\ a n d B = - B^{'} \dots (i i) \\ A d d i n g (i) a n d (i i) w e g e t \\ A + B = - A^{'} - B^{'} \\ \Rightarrow A + B = - (A^{'} + B^{'}), s o A + B i s s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e s u m o f t w o s k e w s y m m e t r i c m a t r i c e s i s a l w a y s s k e w s y m m e t r i c m a t r i c e s . \end{array}$

Q3. The negative of a matrix is obtained by multiplying it by ___.

Sol.

$\begin{array}{l} L e t A b e a m a t r i x \\ ∴ - A = - 1 . A \\ H e n c e, n e g a t i v e o f a m a t r i x i s o b t a i n e d b y m u l t i p l y i n g i t b y - 1 . \end{array}$

Q4. The product of any matrix by the scalar ___ is the null matrix.

Sol.

$\begin{array}{l} L e t A b e a n y m a t r i x \\ ∴ 0 . A = A . 0 \\ H e n c e, t h e p r o d u c t o f a n y m a t r i x i s b y t h e r s c a l a r 0 i s t h e n u l l m a t r i x . \end{array}$

Q5. A matrix which is not a square matrix is called a ___ matrix.

Sol.

$A m a t r i x w h i c h i s n o t a s q u a r e m a t r i x i s c a l l e d a rectangular m a t r i x .$

Q6. Matrix multiplication is ___ over addition.

Sol:

$\begin{array}{l} M a t r i x m u l t i p l i c a t i o n i s d i s t r i b u t i v e o v e r a d d i t i o n . L e t A, B, a n d C b e a n y m a t r i c e s . \\ S o, (i) A (B + C) = A B + A C \\ (i i) (A + B) C = A C + B C \end{array}$

Q7. If $A$ is a symmetric matrix, then $A^{3}$ is a ___ matrix.

Sol.

$\begin{array}{l} L e t A b e a s y m m e t r i c m a t r i x \\ ∴ A^{'} = A \\ {(A^{3})}^{'} = {(A^{'})}^{3} = A^{3} [∵ {(A^{'})}^{k} = {(A^{k})}^{'}] \\ H e n c e, i f A i s a s y m m e t r i c m a t r i x, t h e n A^{3} i s a s y m m e t r i c m a t r i x . \end{array}$

Q8. If $A$ is a skew symmetric matrix, then $A^{2}$ is a ___.

Sol.

$\begin{array}{l} I f A i s a s k e w s y m m e t r i c m a t r i x, \\ ∴ A^{'} = - A \\ {(A^{2})}^{'} = {(A^{'})}^{2} = {(- A)}^{2} = A^{2} \\ H e n c e, A^{2} i s a s y m m e t r i c m a t r i x . \end{array}$

Q9. If $A$ and $B$ are square matrices of the same order, then

(i) ${(A B)}^{'} =$ ___.
(ii) ${(k A)}^{'} =$ ___ (k is any scalar).
(iii) ${[k (A - B)]}^{'} =$ ___.

Sol.

$\begin{array}{l} (i) {(A B)}^{'} = B^{'} A^{'} \\ (i i) {(k A)}^{'} = k . A^{'} \\ (i i i) {[k (A - B)]}^{'} = k {(A - B)}^{'} = k (A^{'} - B^{'}) \end{array}$

Q10. If $A$ is skew symmetric, then $k A$ is a ___ (k is any scalar).

Sol.

$\begin{array}{l} I f A i s a s k e w s y m m e t r i c m a t r i x \\ ∴ A^{'} = - A \\ {(k A)}^{'} = k A^{'} = k (- A) = - k A \\ H e n c e, k A i s a s k e w s y m m e t r i c m a t r i x . \end{array}$

Q11. If $A$ and $B$ are symmetric matrices, then

(i) $A B - B A$ is a ___.

(ii) $B A - 2 A B$ is a ___.

Sol.

$\begin{array}{l} (i) L e t P = (A B - B A) \\ P^{'} = {(A B - B A)}^{'} \\ = {(A B)}^{'} - {(B A)}^{'} \\ = B^{'} A^{'} - A^{'} B^{'} [∵ {(A B)}^{'} = B^{'} A^{'}] \\ = B A - A B [∵ A^{'} = A a n d B^{'} = B] \\ = - (A B - B A) = - P \\ H e n c e, (A B - B A) i s a s k e w s y m m e t r i c m a t r i x . \\ (i i) L e t Q = (B A - 2 A B) \\ Q^{'} = {(B A - 2 A B)}^{'} \\ = {(B A)}^{'} - {(2 A B)}^{'} \\ = A^{'} B^{'} - 2 {(A B)}^{'} [∵ {(k A)}^{'} = k A^{'}] \\ = A^{'} B^{'} - 2 B^{'} A^{'} \\ = A B - 2 B A [∵ A^{'} = A a n d B^{'} = B] \\ = - (2 B A - A B) \\ H e n c e, (B A - 2 A B) i s n e i t h e r a s y m m e t r i c m a t r i x n o r a s k e w s y m m e t r i c m a t r i x . \end{array}$

Q12. If $A$ is symmetric matrix, then $B^{'} A B$ is ___.

Sol.

$\begin{array}{l} I f A i s a s y m m e t r i c m a t r i x \\ ∴ A^{'} = - A \\ L e t P = B^{'} A B \\ P^{'} = {(B^{'} A B)}^{'} = B^{'} A^{'} {(B^{'})}^{'} [∵ {(A B)}^{'} = B^{'} A^{'}] \\ = B^{'} A B [∵ A^{'} = A a n d {(B^{'})}^{'} = B] \\ P^{'} = P \\ S o, P i s a s y m m e t r i c m a t r i x . \\ H e n c e, B^{'} A B i s a s y m m e t r i c m a t r i x . \end{array}$

Q13. If $A$ and $B$ are symmetric matrices of same order, then $A B$ is symmetric if and only if ___.

Sol.

$\begin{array}{l} (i) G i v e n t h a t A^{'} = A a n d B^{'} = B \\ L e t P = A B \\ P^{'} = {(A B)}^{'} \\ = B^{'} A^{'} \\ = B A [∵ A^{'} = A a n d B^{'} = B] \\ = P \\ H e n c e, A B i s s y m m e t r i c i f a n d o n l y i f A B = B A . \end{array}$

Q14. In applying one or more row operations while finding $A^{- 1}$ by elementary row operations, we obtain all zeros in one or more rows, then $A^{- 1}$ ___.

Sol.

$\begin{array}{l} A^{- 1} d o e s n o t e x i s t i f w e a p p l y o n e o r m o r e r o w o p e r a t i o n s w h i l e f i n d i n g A^{- 1} b y e l e m e n t a r y r o w \\ o p e r a t i o n s, o b t a i n a l l z e r o e s i n o n e o r m o r e r o w s . \end{array}$

Commonly asked questions

___ matrix is both symmetric and skew symmetric matrix.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sum of two skew symmetric matrices is always ___ matrix.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The negative of a matrix is obtained by multiplying it by ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A b e a m a t r i x \\ ∴ - A = - 1 . A \\ H e n c e, n e g a t i v e o f a m a t r i x i s o b t a i n e d b y m u l t i p l y i n g i t b y - 1 . \end{array}$

The product of any matrix by the scalar ___ is the null matrix.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A b e a n y m a t r i x \\ ∴ 0 . A = A . 0 \\ H e n c e, t h e p r o d u c t o f a n y m a t r i x i s b y t h e r s c a l a r 0 i s t h e n u l l m a t r i x . \end{array}$

A matrix which is not a square matrix is called a ___ matrix.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$A m a t r i x w h i c h i s n o t a s q u a r e m a t r i x i s c a l l e d a rectangular m a t r i x .$

Matrix multiplication is ___ over addition.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

If $A$ is a symmetric matrix, then $A^{3}$ is a ___ matrix.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A b e a s y m m e t r i c m a t r i x \\ ∴ A^{'} = A \\ {(A^{3})}^{'} = {(A^{'})}^{3} = A^{3} [? {(A^{'})}^{k} = {(A^{k})}^{'}] \\ H e n c e, i f A i s a s y m m e t r i c m a t r i x, t h e n A^{3} i s a s y m m e t r i c m a t r i x . \end{array}$

If $A$ is a skew symmetric matrix, then $A^{2}$ is a ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

If $A$ and $B$ are square matrices of the same order, then

(i) ${(A B)}^{'} =$ ___.
(ii) ${(k A)}^{'} =$ ___ (k is any scalar).
(iii) ${[k (A - B)]}^{'} =$ ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) {(A B)}^{'} = B^{'} A^{'} \\ (i i) {(k A)}^{'} = k . A^{'} \\ (i i i) {[k (A - B)]}^{'} = k {(A - B)}^{'} = k (A^{'} - B^{'}) \end{array}$

If $A$ is skew symmetric, then $k A$ is a ___ (k is any scalar).

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

If $A$ and $B$ are symmetric matrices, then

(i) $A B - B A$ is a ___.

(ii) $B A - 2 A B$ is a ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) L e t P = (A B - B A) \\ P^{'} = {(A B - B A)}^{'} \\ = {(A B)}^{'} - {(B A)}^{'} \\ = B^{'} A^{'} - A^{'} B^{'} [? {(A B)}^{'} = B^{'} A^{'}] \\ = B A - A B [? A^{'} = A a n d B^{'} = B] \\ = - (A B - B A) = - P \\ H e n c e, (A B - B A) i s a s k e w s y m m e t r i c m a t r i x . \\ (i i) L e t Q = (B A - 2 A B) \\ Q^{'} = {(B A - 2 A B)}^{'} \\ = {(B A)}^{'} - {(2 A B)}^{'} \\ = A^{'} B^{'} - 2 {(A B)}^{'} [? {(k A)}^{'} = k A^{'}] \\ = A^{'} B^{'} - 2 B^{'} A^{'} \\ & \end{array}$

If $A$ is symmetric matrix, then $B^{'} A B$ is ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f A i s a s y m m e t r i c m a t r i x \\ ∴ A^{'} = - A \\ L e t P = B^{'} A B \\ P^{'} = {(B^{'} A B)}^{'} = B^{'} A^{'} {(B^{'})}^{'} [? {(A B)}^{'} = B^{'} A^{'}] \\ = B^{'} A B [? A^{'} = A a n d {(B^{'})}^{'} = B] \\ P^{'} = P \\ S o, P i s a s y m m e t r i c m a t r i x . \\ H e n c e, B^{'} A B i s a s y m m e t r i c m a t r i x . \end{array}$

If $A$ and $B$ are symmetric matrices of same order, then $A B$ is symmetric if and only if ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) G i v e n t h a t A^{'} = A a n d B^{'} = B \\ L e t P = A B \\ P^{'} = {(A B)}^{'} \\ = B^{'} A^{'} \\ = B A [? A^{'} = A a n d B^{'} = B] \\ = P \\ H e n c e, A B i s s y m m e t r i c i f a n d o n l y i f A B = B A . \end{array}$

In applying one or more row operations while finding $A^{- 1}$ by elementary row operations, we obtain all zeros in one or more rows, then $A^{- 1}$ ___.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Matrices True or False Type Questions

Q1. A matrix denotes a number.

Sol.

$\begin{array}{l} F a l s e . \\ A m a t r i x i s a n a r r a y o f e l e m e n t s, n u m b e r s o r f u n c t i o n s h a v i n g r o w s a n d c o l u m n s . \end{array}$

Q2. Matrices of any order can be added.

Sol.

$\begin{array}{l} F a l s e . \\ T h e m a t r i c e s h a v i n g s a m e o r d e r c a n o n l y b e a d d e d . \end{array}$

Q3. Two matrices are equal if they have the same number of rows and the same number of columns.

Sol.

$\begin{array}{l} F a l s e . \\ T h e t w o m a t r i c e s a r e s a i d t o b e e q u a l i f t h e i r c o r r e s p o n d i n g e l e m e n t s a r e s a m e . \end{array}$

Q4. Matrices of different orders cannot be subtracted

Sol.

$\begin{array}{l} T r u e . \\ F o r a d d i t i o n a n d s u b t r a c t i o n, t h e o r d e r o f t w o m a t r i c e s s h o u l d b e s a m e . \end{array}$

Q5. Matrix addition is associative as well as commutative.

Sol.

$\begin{array}{l} T r u e . \\ I f A, B a n d C a r e t h e m a t r i c e s o f a d d i t i o n t h e n \\ A + (B + C) = (A + B) + C (a s s o c i a t i v e) \\ A + B = B + A (c o m m u t a t i v e) \end{array}$

Q6. Matrix multiplication is commutative.

Sol.

$\begin{array}{l} F a l s e . \\ Since A B \neq B A i f A B a n d B A a r e w e l l d e f i n e d . \end{array}$

Q7. A square matrix where every element is unity is called an identity matrix.

Sol.

$\begin{array}{l} F a l s e . \\ Since, i n a n i d e n t i t y m a t r i x a l l t h e e l e m e n t s o f p r i n c i p a l d i a g o n a l a r e u n i t y r e s t a r e z e r o . \\ e . g ., A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I_{3} \end{array}$

Q8. If $A$ and $B$ are two square matrices of the same order, then $A + B = B + A$ .

Sol.

$\begin{array}{l} T r u e . \\ I f A a n d B a r e s q u a r e m a t r i c e s o f t h e n t h e i r a d d i t i o n i s c o m m u t a t i v e i . e ., A + B = B + A \end{array}$

Q9. If $A$ and $B$ are two matrices of the same order, then $A - B = B - A$ .

Sol.

$\begin{array}{l} F a l s e . \\ Since s u b t r a c t i o n s o f a n y t w o m a t r i c e s o f t h e s a m e o r d e r i s n o t c o m m u t a t i v e i . e ., A - B \neq B - A \end{array}$

Q10. If matrix $A B = O$ , then $A = O$ or $B = O$ or both $A$ and $B$ are null matrices.

Sol.

$\begin{array}{l} F a l s e . \\ Since f o r a n y t w o n o n - z e r o m a t r i c e s A a n d B, w e m a y g e t A B = 0 . \end{array}$

Q11. Transpose of a column matrix is a column matrix.

Sol.

Q12. If $A$ and $B$ are two square matrices of the same order, then $A B = B A$ .

Sol.

$\begin{array}{l} F a l s e . \\ F o r t w o s q u a r e m a t r i c e s A a n d B, A B = B A i s n o t a l w a y s t r u e . \end{array}$

Q13. If each of the three matrices of the same order is symmetric, then their sum is a symmetric matrix.

Sol.

$\begin{array}{l} T r u e . \\ L e t A, B, a n d C b e t h r e e m a t r i c e s o f t h e s a m e o r d e r . \\ G i v e n t h a t A^{'} = A, B^{'} = B, a n d C^{'} = C \\ L e t P = A + B + C \\ \Rightarrow P^{'} = {(A + B + C)}^{'} \\ = A^{'} + B^{'} + C^{'} = A + B + C = P \\ S o, A + B + C i s a l s o a s y m m e t r i c m a t r i x . \end{array}$

Q14. If $A$ and $B$ are any two matrices of the same order, then ${(A B)}^{'} = A^{'} B^{'}$ .

Sol.

$\begin{array}{l} F a l s e . \\ Since {(A B)}^{'} = B^{'} A^{'} . \end{array}$

Q15. If ${(A B)}^{'} = B^{'} A^{'}$ , where $A$ and $B$ are not square matrices, then the number of rows in $A$ is equal to the number of columns in $B$ and the number of columns in $A$ is equal to the number of rows in $B$ .

Sol.

$\begin{array}{l} T r u e . \\ L e t A = {[a_{i j}]}_{m \times n} a n d B = {[b_{i j}]}_{p \times q} \\ A B i s d e f i n e d w h e n n = P \\ ∴ O r d e r o f A B = m \times q \\ \Rightarrow O r d e r o f {(A B)}^{'} = q \times m \\ O r d e r o f B^{'} i s q \times p a n d o r d e r o f A^{'} i s n \times m \\ ∴ B^{'} A^{'} i s d e f i n e d w h e n P = n \\ a n d t h e o r d e r o f B^{'} A^{'} i s q \times m \\ H e n c e, O r d e r o f {(A B)}^{'} = O r d e r o f B^{'} A^{'} i . e ., q \times m . \end{array}$

Q16. If $A$ , $B$ , and $C$ are square matrices of the same order, then $A B = A C$ always implies that $B = C$ .

Sol.

$\begin{array}{l} F a l s e . \\ A = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], B = [\begin{matrix} 0 & 0 \\ 2 & 0 \end{matrix}] a n d C = [\begin{matrix} 0 & 0 \\ 3 & 4 \end{matrix}] \\ ∴ A B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 2 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ ∴ A C = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 3 & 4 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \\ H e r e A B = A C = 0 b u t B \neq C . \end{array}$

Q17. $A A^{'}$ is always a symmetric matrix for any matrix $A$ .

Sol.

$\begin{array}{l} T r u e . \\ L e t P = A A^{'} \\ P^{'} = {(A A^{'})}^{'} \\ = {(A^{'})}^{'} . A^{'} [∵ {(A B)}^{'} = B^{'} A^{'}] \\ = A A^{'} = P \\ S o, P i s a s y m m e t r i c m a t r i x . \\ H e n c e, A A^{'} i s a l w a y s a s y m m e t r i c m a t r i x . \end{array}$

Q18. If

then $A B$ and $B A$ are defined and equal.

Sol.

Q19. If $A$ is a skew symmetric matrix, then $A^{2}$ is a symmetric matrix.

Sol.

$\begin{array}{l} T r u e . \\ {(A^{2})}^{'} = {(A^{'})}^{2} \\ = {[- A]}^{2} [∵ A^{'} = - A] \\ = A^{2} \\ S o, A^{2} i s a s y m m e t r i c m a t r i x . \end{array}$

Q20. ${(A B)}^{- 1} = A^{- 1} B^{- 1}$ , where $A$ and $B$ are invertible matrices satisfying the commutative property with respect to multiplication.

Sol.

$\begin{array}{l} T r u e . \\ I f A a n d B a r e i n v e r t i b l e m a t r i c e s o f t h e s a m e o r d e r . \\ ∴ {(A B)}^{- 1} = {(B A)}^{- 1} [∵ A B = B A] \\ B u t {(A B)}^{- 1} = A^{- 1} B^{- 1} [G i v e n] \\ ∴ {(B A)}^{- 1} = B^{- 1} A^{- 1} \\ A^{- 1} B^{- 1} = B^{- 1} A^{- 1} \\ ∴ A a n d B s a t i s f y c o m m u t a t i v e p r o p e r t y w . r . t m u l t i p l i c a t i o n . \end{array}$

Commonly asked questions

A matrix denotes a number.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ A m a t r i x i s a n a r r a y o f e l e m e n t s, n u m b e r s o r f u n c t i o n s h a v i n g r o w s a n d c o l u m n s . \end{array}$

Matrices of any order can be added.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ T h e t w o m a t r i c e s a r e s a i d t o b e e q u a l i f t h e i r c o r r e s p o n d i n g e l e m e n t s a r e s a m e . \end{array}$

Matrices of different orders cannot be subtracted

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T r u e . \\ F o r a d d i t i o n a n d s u b t r a c t i o n, t h e o r d e r o f t w o m a t r i c e s s h o u l d b e s a m e . \end{array}$

Matrix addition is associative as well as commutative.

This is a True or False Type Questions as classified in NCERT Exemplar

Matrix multiplication is commutative.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ Since A B \neq B A i f A B a n d B A a r e w e l l d e f i n e d . \end{array}$

A square matrix where every element is unity is called an identity matrix.

This is a True or False Type Questions as classified in NCERT Exemplar

Two matrices are equal if they have the same number of rows and the same number of columns.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ T h e t w o m a t r i c e s a r e s a i d t o b e e q u a l i f t h e i r c o r r e s p o n d i n g e l e m e n t s a r e s a m e . \end{array}$

If $A$ and $B$ are two square matrices of the same order, then $A + B = B + A$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T r u e . \\ I f A a n d B a r e s q u a r e m a t r i c e s o f t h e n t h e i r a d d i t i o n i s c o m m u t a t i v e i . e ., A + B = B + A \end{array}$

If $A$ and $B$ are two matrices of the same order, then $A - B = B - A$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ Since s u b t r a c t i o n s o f a n y t w o m a t r i c e s o f t h e s a m e o r d e r i s n o t c o m m u t a t i v e i . e ., A - B \neq B - A \end{array}$

If matrix $A B = O$ , then $A = O$ or $B = O$ or both $A$ and $B$ are null matrices.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ Since f o r a n y t w o n o n - z e r o m a t r i c e s A a n d B, w e m a y g e t A B = 0 . \end{array}$

Transpose of a column matrix is a column matrix.

This is a True or False Type Questions as classified in NCERT Exemplar

If $A$ and $B$ are two square matrices of the same order, then $A B = B A$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ F o r t w o s q u a r e m a t r i c e s A a n d B, A B = B A i s n o t a l w a y s t r u e . \end{array}$

If each of the three matrices of the same order is symmetric, then their sum is a symmetric matrix.

This is a True or False Type Questions as classified in NCERT Exemplar

If $A$ and $B$ are any two matrices of the same order, then ${(A B)}^{'} = A^{'} B^{'}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F a l s e . \\ Since {(A B)}^{'} = B^{'} A^{'} . \end{array}$

If ${(A B)}^{'} = B^{'} A^{'}$ , where $A$ and $B$ are not square matrices, then the number of rows in $A$ is equal to the number of columns in $B$ and the number of columns in $A$ is equal to the number of rows in $B$ .

This is a True or False Type Questions as classified in NCERT Exemplar

If $A$ , $B$ , and $C$ are square matrices of the same order, then $A B = A C$ always implies that $B = C$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$A A^{'}$ is always a symmetric matrix for any matrix $A$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T r u e . \\ L e t P = A A^{'} \\ P^{'} = {(A A^{'})}^{'} \\ = {(A^{'})}^{'} . A^{'} [? {(A B)}^{'} = B^{'} A^{'}] \\ = A A^{'} = P \\ S o, P i s a s y m m e t r i c m a t r i x . \\ H e n c e, A A^{'} i s a l w a y s a s y m m e t r i c m a t r i x . \end{array}$

If then $A B$ and $B A$ are defined and equal.

This is a True or False Type Questions as classified in NCERT Exemplar

If $A$ is a skew symmetric matrix, then $A^{2}$ is a symmetric matrix.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T r u e . \\ {(A^{2})}^{'} = {(A^{'})}^{2} \\ = {[- A]}^{2} [? A^{'} = - A] \\ = A^{2} \\ S o, A^{2} i s a s y m m e t r i c m a t r i x . \end{array}$

${(A B)}^{- 1} = A^{- 1} B^{- 1}$ , where $A$ and $B$ are invertible matrices satisfying the commutative property with respect to multiplication.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T r u e . \\ I f A a n d B a r e i n v e r t i b l e m a t r i c e s o f t h e s a m e o r d e r . \\ ∴ {(A B)}^{- 1} = {(B A)}^{- 1} [? A B = B A] \\ B u t {(A B)}^{- 1} = A^{- 1} B^{- 1} [G i v e n] \\ ∴ {(B A)}^{- 1} = B^{- 1} A^{- 1} \\ A^{- 1} B^{- 1} = B^{- 1} A^{- 1} \\ ∴ A a n d B s a t i s f y c o m m u t a t i v e p r o p e r t y w . r . t m u l t i p l i c a t i o n . \end{array}$

24th June 2022 (second shift)

Commonly asked questions

Let x * y = x² + y³ and (x * 1) * 1 = x * (1 * 1). Then a value of 2 $s i n^{- 1} (\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2})$ is.

$(x^{2} + 1) * 1 = x * 2$

$\Rightarrow {(x^{2} + 1)}^{2} + 1 = x^{2} + 8 \Rightarrow x^{2} = 2$

$2 s i n^{- 1} (\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}) = 2 s i n^{- 1} (\frac{1}{2}) = \frac{π}{3}$

The sum of all the real roots of the equation $(e^{2 x} - 4) (6 e^{2 x} - 5 e^{x} + 1) = 0 i s$

$e^{2 x} - 4 = 0 o r 6 e^{2 x} - 5 e^{x} + 1 = 0$

$\Rightarrow x = l n 2 x = l n (\frac{1}{3}), l n (\frac{1}{2})$

= $- l n 3, - l n 2$

Let the system of linear equations

x + y + $α z = 2$

3x + y + z = 4

x + 2z = 1

have a unique solution $(x^{*}, y^{*}, z^{*}) .$ If $(α, x^{*}), (y^{*}, α) a n d (x^{*}, - y^{*})$ are collinear points, then the sum of absolute values of all possible values of $α$ is

x + y + az = 2………… (i)

3x + y + z = 4 ………… (ii)

x + 2z = 1 ……………. (iii)

x = 1, y = 1, z = 0

(for unique solution $a \neq - 3$

$(α, - 1), (1, α) a n d (1, - 1)$ are collinear.

$\frac{α - 1}{1 - α} = \frac{- 1 - α}{1 - 1} \Rightarrow 1 - α^{2} = 0 \Rightarrow α = \pm 1$

Sum of absolute value = 2

Let x,y > 0. If x³y² = 2¹⁵, then the least value of 3x + 2y is

$x_{1} y > 0 a n d x^{3} y^{2} = 2^{15}$

$A M \geq G M$

$\frac{3 x + 2 y}{5} \geq {(x^{3} y^{2})}^{\frac{1}{5}}$

$\Rightarrow 3 x + 2 y \geq 40$

Let f(x) ${\begin{matrix} \frac{s i n (x - [x])}{x - [x]} & , & x \in (- 2, - 1) \\ m a x {2 x, 3 [| x |]} & , & | x | < 1 \\ 1 & , & o t h e r w i s e \end{matrix}$

where [t] denotes greatest integer $\leq t .$ If m is the number of points where f is not continuous and n is the number of points where f is not differentiable, then the ordered pair (m, n) is

$f (x) = {\begin{cases} \frac{s i n (x + 2)}{x + 2}, x \in (- 2, - 1) \\ 0, x \in (- 1, 0] \\ 2 x, x \in (0, 1) \\ 1, o t h e r w i s e \end{cases}$

$L H D = \underset{h \to 0}{L t} \frac{f (0 - h) - f (0)}{- h} = 0$

$R H D = \underset{h \to 0}{L t} \frac{f (0 + h) - f (0)}{h} = 2$

Hence f (x) is not differentiable at x = 1, 0, 1

$∴ m = 2, n = 3$

The value of the integral $\int_{- π / 2}^{π / 2} \frac{d x}{(1 + e^{x}) (s i n^{6} x + c o s^{6} x)} i s e q u a l t o$

$l = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d x}{(1 + e^{x}) (s i n^{6} x + c o s^{6} x)}$ ……. (i)

$= 2 \int_{0}^{\infty} \frac{d t}{4 + t^{2}} = 2 {(t a n^{- 1} (\frac{t}{2}))}_{0}^{\infty} = π$

$\underset{x \to \infty}{l i m} (\frac{n^{2}}{(n^{2} + 1) (n + 1)} + \frac{n^{2}}{(n^{2} + 4) (n + 2)} + \frac{n^{2}}{(n^{2} + 9) (n + 3)} + . . . + \frac{n^{2}}{(n^{2} + n^{2}) (n + n)}) i s e q u a l t o$

S = $\underset{n \to \infty}{L t} \sum_{r = 1}^{n} \frac{n^{2}}{(n^{2} + r^{2}) (n + r)}$

$\frac{π}{8} + \frac{1}{4} l n 2$

A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with

Equation of tangent at P (x, y) is Y = $\frac{d y}{d x} (X - x)$

$A / q, 2 x - y \frac{d x}{d y} = 0 \Rightarrow 2 \frac{d y}{y} = \frac{d x}{x}$

$\begin{array}{l} \Rightarrow 2 l n y = l n x + l n c \\ y^{2} = x c \end{array}$

It passes through (3, 3), c = 3

$∴ y^{2} = 3 x ∴$ Length of latus rectum = 3

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1, a > 2,$ having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $6 \sqrt[]{3}$ . Then the eccentricity of the ellipse is

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1$

$∴ Δ = \frac{1}{2} \times a (1 + c o s θ) . 4 s i n θ$

$∴ Δ_{m a x}^{Δ} = 6 \sqrt[]{3} \Rightarrow a = 4$

$∴ e = \frac{\sqrt[]{3}}{2}$

Let the area of the triangle with vertices A $(1, α), B (α, 0) C (0, α)$ be 4 sq. units. If the points $(α, - α), (- α, α) a n d (α^{2}, β)$ are collinear, then $β$ is equal to

$Δ = 4 \Rightarrow \frac{1}{2} | \begin{matrix} 1 & α & 1 \\ α & 0 & 1 \\ 0 & α & 1 \end{matrix} | = 4$

$\Rightarrow α = \pm 8$

$(α, - α), (- α, α) a n d (α^{2}, β)$ are collinear.

$∴ | \begin{matrix} α & - α & 1 \\ - α & α & 1 \\ α^{2} & β & 1 \end{matrix} | = 0 \Rightarrow β = - 64$

The number of distinct real roots of the equation x⁷ – 7x – 2 = 0 is

$f (x) = x^{7} - 7 x - 2$

$f' (x) = 7 x^{6} - 7 \Rightarrow f' (x) = 0, x = \pm 1$

$∴ f (1) = < 0 a n d f (- 1) > 0$

Hence number of real roots of f (x) = 0 are 3.

A random variable X has the following probability distribution

$?$ x is a random variable.

$\begin{array}{l} ∴ k + 2 k + 4 k + 6 k + 8 k = 1 \\ ∴ k = \frac{1}{21} \end{array}$

$∴ P ((1 < x < 4) | x \leq 2) = \frac{4 k}{7 k} = \frac{4}{7}$

The number of solutions of the equation $c o s (x + \frac{π}{3}) c o s (\frac{π}{3} - x) = \frac{1}{4} c o s^{2} 2 x, x \in [- 3 π, 3 π] i s :$

$c o s (x + \frac{π}{3}) c o s (\frac{π}{3} - x) = \frac{1}{4} c o s^{2} 2 x$

$x = - 3 π, - 2 π, - π, 0, π, 2 π, 3 π$

$∴$ total number of solution = 7.

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5} i s \frac{1}{\sqrt[]{3}},$ then the sum of all possible values of $λ$ is

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_{1} \times b_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & λ \\ 1 & 4 & 5 \end{matrix} | = \hat{i} (15 - 4 λ) + \hat{j} (λ - 10) + \hat{k} (5)$

$\frac{| (- \hat{i} - 2 \hat{j} - 2 \hat{k}) . {(15 - 4 λ) \hat{i} + (λ - 10) \hat{j} + 5 \hat{k}} |}{\sqrt[]{{(15 - 4 λ)}^{2} + {(λ - 10)}^{2} + 25}} = \frac{1}{\sqrt[]{3}}$

$\Rightarrow λ = 16$

Let the points on the plane P be equidistance from the points (–4, 2, 1) and (2, –2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is

Let P (x, y, z) be any point on plane P₁ then

${(x + 4)}^{2} + {(y - 2)}^{2} + {(z - 1)}^{2} = {(x - 2)}^{2} + {(y + 2)}^{2} + {(z - 3)}^{2}$

$∴ c o s θ = \frac{| 6 - 2 + 3 |}{14} \Rightarrow θ = \frac{π}{3}$

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $| (\hat{a} + \hat{b}) + 2 (\hat{a} \times \hat{b}) |$ = 2. If $θ \in (0, π)$ is the angle between $\hat{a} a n d \hat{b}$ , then among the statements:

$(S 1) : 2 | \hat{a} \times \hat{b} | = | \hat{a} - \hat{b} |$

(S1) : The projection of $\hat{a} o n (\hat{a} + \hat{b}) i s \frac{1}{2}$

$| \vec{a} + \vec{b} + 2 (\vec{a} \times \vec{b}) | = 2, θ \in (0, π)$

squaring on both sides, we get = $\frac{2 π}{3}$

where θ is angle between $\hat{a} a n d \hat{b} .$

$2 | \hat{a} \times \hat{b} | = \sqrt[]{3} = | \hat{a} - \hat{b} |, S_{1}$ is correct.

and projection of $\hat{a} o n \hat{a} + \hat{b} = | \frac{\hat{a} . (\hat{a} + \hat{b})}{| \hat{a} + \hat{b} |} | = \frac{1}{2}$

So (S₂) is correct.

If y = tan^-1 (sec x³ tanx³), $\frac{π}{2} < x^{3} < \frac{3 π}{2}, t h e n$

$L e t x^{3} = θ \Rightarrow \frac{θ}{2} \in (\frac{π}{4}, \frac{3 π}{4})$

$∴ y = t a n^{- 1} (s e c θ - t a n θ)$

$t a n^{- 1} (\frac{1 - s i n θ}{c o s θ})$

$\Rightarrow \frac{d y}{d x} = \frac{- 3 x^{2}}{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = - 3 x$

$∴ x^{2} \frac{d^{2} y}{d x^{2}} - 6 y + \frac{3 π}{2} = 0 \Rightarrow x^{2} y^{11} - 6 y + \frac{3 π}{2} = 0$

Consider the following statements:

A : Rishi is a judge.

B : Rishi is honest.

C : Rishi is not arrogant.

The negation of the statement “if Rishi is a judge and he is not arrogant, then he is honest” is

$?$ given statement is

$(A \land C) \to B$ then its negation is ${(A \land C) \to B}$

The slope of normal at any point (x,y), X > 0, y > 0 on the curve y = y(x) is given by $\frac{x^{2}}{x y - x^{2} y^{2} - 1} .$ If the curve passes through the point (1, 1), then e.y (e) is equal to

$? - \frac{d x}{d y} = \frac{x^{2}}{x y - x^{2} y^{2} - 1}$

$∴ \frac{d y}{d x} = \frac{x y - x^{2} y^{2} - 1}{x^{2}}$

$L e t x y = v \Rightarrow x \frac{d y}{d x} + y = \frac{d v}{d x}$

Put x = 1, y = 1 $\Rightarrow t a n^{- 1} = c \Rightarrow c = \frac{π}{4}$

$∴ t a n^{- 1} (x y) = l n x = \frac{π}{4}$

$e (y (e)) = t a n (1 + \frac{π}{4}) = \frac{t a n 1 + 1}{1 - t a n 1}$

Let $λ$ be the largest value of $λ$ for which the function $f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$ is increasing for all $x \in R .$ Then $f_{λ} . (1) + f_{λ} . (- 1)$ is equal to

$? f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$

$f_{λ} (x) = 12 (λ x^{2} - 6 λ x + 3)$

For increasing $f_{λ} (x) \geq 0$

$f_{λ}^{*} (x) = \frac{4}{3} x^{3} - 12 x^{2} + 36 x + 48 ∴ f_{λ}^{*} (1) + f_{λ}^{*} (- 1) = 73 \frac{1}{2} - 1 \frac{1}{2} = 72$

Let S = ${z \in C : | z - 3 | \leq 1 a n d z (4 + 3 i) \leq 24} .$ If $α + i β$ is the point in S which in closest to 4i, then 25 $(α + β)$ is equal to

Circle $| z - 3 | \leq 1 \Rightarrow {(x - 3)}^{2} + y^{2} \leq 1$

and line $z (4 + 3 i) + \bar{z} (4 - 3 i) \leq 24$

$\Rightarrow 4 x - 3 y \leq 12 ∴ s l o p e = t a n θ = \frac{4}{3}$

Let $S = {(\begin{matrix} - 1 & a \\ 0 & b \end{matrix}) a, b \in {1, 2, 3, . . . 100}};$ and let Tn = ${A \in S : A^{n (n + 1)} = I} .$ Then the number of elements in $\cap_{n = 1}^{100} T_{n} i s . . . . . . . . . . . .$

$S = {(\begin{matrix} 1 & a \\ 0 & b \end{matrix}), a, b \in 1, 2, 3, . . . . . 100$

$∴ A = (\begin{matrix} - 1 & a \\ 0 & b \end{matrix})$ then even power of A as A = $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) .$

If b = 1 & $a \in {1, 2, 3 . . . . . 100}$ and n (n + 1) is always even

$∴ T_{1}, T_{2}, T_{3}, . . . . . . . ., T_{n}$ are all 1 for b = 1 and each value of a.

$∴ \cap_{n = 1}^{1000} T_{n} = 100$

The number of 7-digit numbers which are multiples of 1 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$∴$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$∴$ Total possible ways = (4! × 3!) × 4 = 576

The sum of all elements of the set ${α \in {1, 2, . . ., 100} : H F C (α, 24) = 1}$ is

${a \in (1, 2, 3, . . . . . ., 100) : H C F (a, 24) = 1}$

HCF of (a, 24) = 1 $∴$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$∴$ There are four such blocks and a number 97 is there upto 100.

$∴$ complete sum = 96 + (24 × 8 + 96) + (48 × 8 + 96) + (72 × 8 + 96) + 97 = 1633

The remainder on diving 1 + 3 + 3² + 3³ + …. + 3²⁰²¹ by 50 is

S = 1 + 3 + 3² + 3³ + ….+ 3²⁰²¹ $= \frac{3^{2022} - 1}{2} = \frac{1}{2} [a^{1011} - 1]$

^{$= \frac{1}{2} [9^{1011} - 1] = \frac{1}{2} [100 k + 10110 - 1 - 1]$}

= 50k₁ + 4

Remainder = 4

The area (in sq. units) of region enclosed between the parabola y² = 2x and the line x + y = 4 is

Required area

= $\int_{- 4}^{2} (4 - y - \frac{y^{2}}{2}) d y$

= ${(4 y - \frac{y^{2}}{2} - \frac{y^{3}}{6})}_{- 4}^{2} = 18 s q u n i t s$

Let a circle C : ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}, k > 0,$ touch the axis (1,0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to

OM² = OP² = PM²

^{$| \frac{1 + r}{\sqrt[]{2}} | = r^{2} - 1$}

^{$∴ r = 3$}

^$∴$equation of circle is ${(x - 1)}^{2} + {(y - 3)}^{2} = 3^{2}$

$∴ h + k + r = 7$

In an examination, there are 1- true-false type questions. Out of 10, at student can guess the answer of 4 questions correctly with probability $\frac{3}{4}$ and the remaining 6 questions correctly with probability $\frac{1}{4}$ . If the probability that the student guesses the answer of exactly 8 questions correctly out of 10 is $\frac{27 k}{4^{10}},$ then k is equal to

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$∴$ Required possibilities =

$=^{4} C_{4} \times^{6} C_{4} {(\frac{3}{4})}^{4} \times {(\frac{1}{4})}^{4} {(\frac{3}{4})}^{2} +^{4} C_{3} \times^{6} C_{5} {(\frac{3}{4})}^{3} {(\frac{1}{4})}^{5} \times \frac{1}{4} \times \frac{3}{4}$ $+^{4} C_{2} \times^{6} C_{6} \times {(\frac{3}{4})}^{2} {(\frac{1}{4})}^{2} \times {(\frac{1}{4})}^{6}$

$= \frac{27}{4^{10}} [15 \times 27 + 24 \times 3 + 2] = \frac{27 \times 479}{4^{10}}$

Let the hyperbola $H : \frac{x^{2}}{a^{2}} - y^{2} = 1$ and the ellipse E : 3x² + 4y² = 12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If e_H and e_E are the eccentricities of H and E respectively, then the value of $12 (e_{H}^{2} + e_{E}^{2})$ is equal to

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{1} = 1$

Length of latus rectum = $\frac{2}{a}$

$a n d \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

length of latus rectum = $\frac{6}{2}$ = 3

$? \frac{2}{a} = 3 \Rightarrow a = \frac{2}{3}$

$12 (e_{H}^{2} + e_{H}^{2}) = 12 [(1 + \frac{9}{4}) + (1 - \frac{3}{4})] = 12 [\frac{13}{4} + \frac{1}{4}]$ = 12 × $\frac{14}{4} = 42$

Let P₁ be a parabola with vertex (3, 2) and focus (4, 4) and P₂ be its mirror image with respect to the line x + 2y = 6. Then the directrix P₂ is x + 2y =.............

S (4, 4) and V (3, 2)

$∴$ point of intersection of directrix with axis of parabola is A (2, 0)

Image of A (2, 0) with respect to line

$∴ x + 2 y = 6 i s B (x_{2}, y_{2})$

$∴ \frac{x_{2} - 2}{1} = \frac{y_{2} - 0}{2} \Rightarrow \frac{- 2 (2 + 0 - 6)}{5}$

$∴ B (\frac{18}{5}, \frac{16}{5})$

Point B is point of intersection of directrix with axes of parabola P₂.

$∴ x + 2 y = λ$

$B (\frac{18}{5}, \frac{16}{5})$ lies on the line x + 2y = $λ$ $∴ λ = \frac{18}{5} + \frac{32}{5} = 10$

JEE Mains Solutions 2022,28th june , Maths,Second shift

Commonly asked questions

Let R₁ = ${(a, b) \in N \times N : | a - b | \leq 13} a n d$

R₂ = ${(a, b) \in N \times N : | a - b | \neq 13} .$ Then on N :

R₁ = ${(a, 1) \in N \times N : | a - b | \leq 13}$

$(a, a) \in R_{1} a s | a - a | \leq 13 (R e f l e x i v e)$

$(a, b) & (b, a) \in R_{1} a s | a - b | = | b - a | \to (s y m m e t r i c) .$

But it is not necessary that if (a,b) & (b, c) $\in R, t h e n (a, c) \in R$

Eg $- (21, 10) \in R & (10, 1) \in R b u t (21, 1) \notin R_{1}$

R₂ = ${(a, b) \in N \times N : | a - b | \neq 13 ∴ R_{2} \to N o t e q u i v a l e n c e s o l u t o i n .}$

$(a, b) & (b, a) \in R_{2} a s | a - b | = | b - a |$

But it is not necessary that if (a, b) & (b, c) $\in R_{2}$ then (a, c) also $\in R_{2}$ .

Eg – (21, 1) $\in R_{2} & (1, 8) \in R_{2} b u t (21, 8) \notin R_{2}$

let f(x) be a quadratic polynomial such that f(-2) + f(3) = 0. If one of the roots of f(x) = 0 is -1, then the sum of the roots of f(x) = 0 is equal to:

$f (- 2) + f (3) = 0 .$ One root of f (x) = 0 is (-1)

Let’s assume other root to be a

$∴ f (x) = a (x + 1) (x - α)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a = $\frac{14}{3}$

$∴$ sum of roots = $\frac{14}{3} - 1 = \frac{11}{3}$

The number of ways to distribute 30 identical candies among four children C₁, C₂, C₃ and C₄ so that C₂ receives atleast 4 and atmost 7 candies, C₃ receives atleast 2 and atmost 6 candies, is equal to:

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4 $\leq b \leq 7$

$2 \leq c \leq 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$∴ 0 \leq a \leq 24$

$0 \leq d \leq 24$

& a + b + c + d = 30

So, total possible solution to the above equation.

Coefficient of x³⁰ in.

$(x^{0} + x + x^{2} + . . . . + x^{24}) (x^{4} + x^{5} + . . . . + x^{7}) (x^{2} + x^{3} + . . . . + x^{6}) (x^{0} + x + x^{2} + . . . . + x^{24})$

= ${(1 + . . . . + x^{24})}^{2} (x^{4}) (1 + x + . . . . + x^{3}) \times x^{2} (1 + x + . . . . + x^{4})$

$= (x^{56} - 2 x^{31} + x^{6}) \times (x^{9} - x^{4} - x^{5} + 1) \times {(x - 1)}^{- 4}$

x⁵⁶ & x³¹ can never give x³⁰ so we discard them.

$x^{6} \times x^{9} \times {(x - 1)}^{- 4} \to^{15 + 4 - 1} ? C_{4 - 1} =^{18} ? C_{3}$

Coefficient x³⁰ ® ¹⁸C₃ – ²³C₃ – ²²C₃ + ²⁷C₃

= $\frac{18 \times 17 \times 16}{6} - \frac{23 \times 22 \times 21}{6} - \frac{22 \times 21 \times 20}{6} + \frac{27 \times 26 \times 25}{6}$

= 430

The term independent of x in the expansion of $(1 - x^{2} + 3 x^{3}) {(\frac{5}{2} x^{3} - \frac{1}{5 x^{2}})}^{11}, x \neq 0 i s :$

$(1 - x^{2} + 3 x^{3}) {(\frac{5}{2} x^{3} - \frac{1}{5 x^{2}})}^{11}, x \neq 0$

(r + 1)^th term for expansion of ${(\frac{5}{2} x^{3} - \frac{1}{5 x^{2}})}^{11}$

n – r = x₃

r = y $0 \leq x, y \leq 11 s h o u l d b e n a t u r a l n u m b e r .$

= ¹¹C_y $\times {(\frac{5}{2})}^{x} \times {(- \frac{1}{5})}^{y} \times x^{(3 x - 2 y)}$

for term to be independent of x, power of x should be zero.

(3x – 2y) = 0 (when 1 is multiplied).

3x + 3y = 33

y = $\frac{33}{5} (N o s o l u t i o n)$

3x – 2y + 2 = 0 where (-x²) is multiplied).

3x + 3y = 33

y = 7, x = 4

$\begin{matrix} 3 x - 2 y + 3 = 0 & 3 x + 3 y = 33 \end{matrix}}$

$∴$ coefficient of term independent of x.

$(- 1) \times^{11} ? C_{7} \times {(\frac{5}{2})}^{4} \times {(\frac{- 1}{5})}^{7}$

$= \frac{33}{200}$

If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is:

a + a₁……a_n, 100 n arithmetic mean 100

a + n = 33 ……. (i)

$(\frac{a_{1}}{a_{n}} = \frac{1}{7})$

$\frac{(a + d)}{(100 - d)} = \frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d = $\frac{15}{4}$

a = 10

Let f, g : R ® R be functions defined by

$f (x) = {\begin{matrix} [x] & , x < 0 \\ | 1 - x | & , x \geq 0 \end{matrix} a n d g (x) = {\begin{matrix} e^{x} - x & , x < 0 \\ {(x - 1)}^{2} - 1 & , x \geq 0 \end{matrix}$

where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly:

$f (x) = {\begin{matrix} [x], x < 0 & | 1 - x |, x \geq 0 \end{matrix}$ $g (x) = {\begin{matrix} c^{x} - x, x < 0 & {(x - 1)}^{2} - 1, x \geq 0 \end{matrix}$

fog

$f o g (\begin{matrix} \end{matrix} [e^{x} - x], (e^{x} - x < 0) (x < 0) [{(x - 1)}^{2} - 1] {(x - 1)}^{2} - 1 < 0 x \geq 0 | 1 - e^{x} + x |, e^{x} - x \geq 0 x < 0 | 1 - {(x + 1)}^{2} + 1 |, {(x - 1)}^{2} - 1 \geq 0 x \geq 0, x \in (2, \infty) ($

Not possible as of inequalities give $? .$

e^x – x < 0, x < 0 (x – 1)² – 1 < 0

Not possible (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x $\in (0, 2)$

continuous $\forall$ x < 0

$f o g {\begin{matrix} | 1 - e^{x} + x |, x < 0 & | 1 - {(x - 1)}^{2} + 1 |, x = 0 & | 1 - {(x - 1)}^{2} + 1 |, x \geq 2 \end{matrix}$ Discontinuous at 0

continuous $\forall x$

$∴$ fog is discontinuous at 0

Let f : R ® R be a differentiable function such that $f (\frac{π}{4}) = \sqrt[]{2}, f (\frac{π}{4}) = 0 a n d f' (\frac{π}{2}) = 1$ and let $g (x) = \int_{x}^{π / 4} (f' (t) s e c t + s e c t f (t)) d t f o r x \in [\frac{π}{4}, \frac{π}{2}) .$ Then $\underset{x \to {(\frac{π}{2})}^{-}}{l i m} g (x)$ is equal to:

$f (\frac{r}{4}) = \sqrt[]{2}, f (\frac{r}{2}) = 0 & f' (\frac{r}{2}) = 1$

g(x) $\int_{}^{} (f' (t) s e c t + t a n t s e c t f (t)) d t$

$= {[f (t) s e c t]}_{x}^{π / 4}$

= $\sqrt[]{2} \times \sqrt[]{2} - f (x) s e c x$

$= \frac{2 c o s x - f (x)}{c o s x}$

= $\frac{2 s i n x + 1}{s i n x} = 3$

Let f : R® R be a continuous function satisfying f(x) + f(x + k) = n, for all x $\in$ R where k > 0 and n is a positive integer. If $I_{1} = \int_{0}^{4 n k} f (x) d x a n d I_{2} = \int_{- k}^{3 k} f (x) d x, t h e n :$

$f (x) + f (x + k) = n \forall x \in R, & k > 0 . . . . . . . . . . . . (i)$

Replace x by x + k.

$f (x + k) + f (x + 2 k) = n . . . . . . . . . . . . . . . (i i)$

From (i) & (ii), f(x + 2k) = f(x).

$∴ f (x)$ is periodic with period = 2k.

$I_{1} = \int_{0}^{4 n k} f (x) d x = 2 x \int_{0}^{2 k} f (x) d x . . . . . . . . . . . . . . . . . . . (i i)$

$I_{2} = \int_{- k}^{3 k} f (x) d x$ put x = t + k

$= \int_{- 2 k}^{2 k} t (t + k) d t = 2 \int_{0}^{2 k} f (t + k) d t$

$= 2 n \int_{0}^{2 k} n d x = 4 n^{2} k .$

The area of the bounded region enclosed by the curve y = 3 $| x - \frac{1}{2} | - | x + 1 |$ - and the x-axis is:

$y = 3 ? | x ? \frac{1}{2} | = | x + 1 |$

Graph

Area = $(\frac{1}{2} * \frac{3}{2} * \frac{3}{4}) * 2 + \frac{3}{2} * \frac{3}{2} = \frac{3}{2} * \frac{3}{2} * \frac{3}{2} = \frac{27}{8}$

Let x = x(y) be the solution of the differential equation 2y $e^{x / y^{2} d x} + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that x(1) = 0. Then, x(e) is equal to:

$2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$

$\Rightarrow 2 e^{2 / y^{2}} (y d x - 2 x d y) + y^{2} d y = 0$

$2 e^{x / y^{2}} d (\frac{x}{y^{2}}) + \frac{d y}{y} = 0$

Integrating $2 e^{x / y^{2}} + I n y = c$

y = 1, n = 0 c = 2

^{$2 e^{x / y^{2}} + I n y = 2$}

$∴ y = e \Rightarrow 2 e^{x / e^{2}} + 1 = 2$

x = -e² In2

Let the lope of the tangent to a curve y = f(x) at (x,y) be given by 2 tan x (cos x – y). If the curve passes through the point $(\frac{π}{4}, 0),$ then the value of $\int_{0}^{π / 2} y d x$ is equal to:

$\frac{d y}{d x} + 2 y t a n x = 2 s i n x$

$I . F . = e^{2 \int t a n x d x} = s e c^{2} x$

$∴$ Solution y sec²x = $2 \int s i n x s e c^{2} x d x = 2 \int s e c x t a n x d x$

y sec² x = 2sec x + c

y = 2 cos x + c cos²x passes $(\frac{π}{4}, 0) = B$ C = $- 2 \sqrt[]{2}$

y = 2 cos x $2 \sqrt[]{2} c o s^{2} x$ $∴ \int_{0}^{π / 2} y d x = 2 - \frac{π}{\sqrt[]{2}}$

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : -4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the nation 1 :3, then the area of the triangle is equal to:

Let a > 0, b > 0. Let a and respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ Let e’ and $l'$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^{2} = \frac{11}{14} l a n d {(e')}^{2} = \frac{11}{8} l',$ then the value of 77a + 44b is equal to

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$e = \sqrt[]{1 + \frac{b^{2}}{a^{2}}}$ $e' = \sqrt[]{1 + \frac{a^{2}}{b^{2}}}$

$l = \frac{2 b^{2}}{a}$ $l' = \frac{2 a^{2}}{b}$

$(1 + \frac{b^{2}}{a^{2}}) = \frac{11}{14} \times \frac{2 b^{2}}{a}$ $(1 + \frac{a^{2}}{b^{2}}) = \frac{11}{8} \times \frac{2 a^{2}}{b}$

$7 \times {{(\frac{7 b}{4})}^{2} + b^{2}} = 11 \times b^{2} \times \frac{7 b}{4} \Rightarrow a \times 77 = 65$

65b² = 44b³

65 = b × 44

$∴ 77 a + 44 b = 65 \times 2 = 130$

Let $\vec{a} = α \hat{i} + 2 \hat{j} - \hat{k} a n d \vec{b} = - 2 \hat{i} + α \hat{j} + \hat{k}, w h e r e α \in R .$ If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a} a n d \vec{b} i s \sqrt[]{15 (α^{2} + 4)},$ then the value of $2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2}$ is equal to:

$\vec{a} = α \hat{i} + 2 \hat{j} - \hat{k} & \vec{b} = - 2 \hat{i} + α \hat{j} + \hat{k}$

$| \vec{a} \times \vec{b} | = \sqrt[]{15 (a^{2} + 4)}$

$2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2} = ?$

$\vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ$

$c o s θ = \frac{- 1}{(a^{2} + 5)}$ $∴ | s i n θ | = \frac{\sqrt[]{{(a^{2} + 5)}^{2} - 1}}{(a^{2} + 5)}$

$| \vec{a} \times \vec{b} | = | \overset{?}{a} | \times | \vec{b} | \times | s i n θ |$

$= (a^{5} + 5) \times \frac{\sqrt[]{{(a^{2} + 5)}^{2} - 1}}{(a^{2} + 5)} = \sqrt[]{15 (a^{2} + 4)}$

${(a^{2} + 5)}^{2} - 1 = 15 (a^{2} + 4)$

$\Rightarrow (α = \pm 3)$

$2 {| \vec{a} |}^{2} + (\vec{a} . \vec{b}) {| \vec{b} |}^{2}$

= $2 (a^{2} + 5) - (a^{2} + 5)$

$(a^{2} + 5) = 14$

If vertex of a parabola is (2, -1) and the equation of its directrix is 4x – 3y = 21, then the length of its latus rectum is:

$a = | \frac{4.2 - 3 (- 1) - 21}{5} |$

$= | \frac{8 + 3 - 21}{5} | = 2$

$∴ L = 4 a = 8$

Let the plane ax + by + cz = d pass through (2, 3, -5) and its perpendicular to the planes

2x + y – 5z = 10 and 3x + 5y – 7z = 12.

If a, b, c, d are integers d > 0 and then the value of a + 7b + c + 20d is equal to:

ax + by + cz = d

2a + 3b – 5c = d 2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$(2 \hat{i} + \hat{j} - 5 \hat{k}), (a \hat{i} + b \hat{j} + c \hat{k}) = 0$

2b = d …………...(i) d > 0

2a + b – 5c = 0 2a + 3b – 5c = d $| a |, | b |, | c |, d \to g . c . d$

= $\frac{α}{2}$

$∴ \frac{α}{2} = 1 \Rightarrow α = 2$

$(3 \hat{i} + 5 \hat{j} - 7 \hat{k}) . (a \hat{i} + b \hat{j} + c \hat{k}) = 0$

3a + 5b – 7c = 0) × $\frac{4}{7} . . . . . . . . . . . . . . . . . . . (i i i)$

$∴ a = - 18, b = 1$

c = -7, d = 2

(iii)…(ii) ® 2a + 3b $\frac{- 33 c}{7} = 0$

2a + 3b = $\frac{33}{7} c$ $c = \frac{- 7}{2} d . . . . . . . . . . . . . . . . . (i v)$

Putting values

a + 7b + c + 20d = 22

The probability that a randomly chosen one-one function from the set ${a, b, c, d}$ to the set ${1, 2, 3, 4, 5}$ satisfies f(a) + 2f(b) – f(c) = f(d) is:

No of one – one functions ® ⁵P₄ = 120

f(a) + 2f(b) – f(c) = f(d) ${1, 2, 3, 4, 5}$

2f(b) = f(d) + f(c) – f(a)

So, f(d) + f(c) – f(a) should be even.

Only possibilities of f(d) f(c) f(a)

Not possible since E E E E - even, O – Odd

function is one one E O O

O O E

O E O

Case (i)

f(d) f(c) f(a)

E O O

2 1 3 similarly we write cases for f(d) = 4

2 1 5 413

2 3 5 431

2 5 1 435

2 3 1 453

2 5 3 415

4512²²²asdf=fhf=kjhfjkdfh45654hdfkjdhjh

Case (ii)

O O E O O E

1 5 2 1 5 4

5 1 2 5 1 4

3 1 2 1 3 4

1 3 2 3 1 4

3 5 2 5 3 4

5 3 2 &nb

The value of $\underset{n \to \infty}{l i m} 6 t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{r^{2} + 3 r + 3})}$ is equal to:

$\underset{n \to \infty}{l i m} 6 t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{r^{2} + 3 r + 3})}$

$= \underset{n \to \infty}{l i m} 6 t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{(r + 2) - (r + 1)}{1 + (r + 2) (r + 1)})}$

$= \underset{n \to \infty}{l i m} 6 t a n {\frac{r}{2} - t a n^{- 1} (2)}$

$6 \times \frac{1}{2} = 3$

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k} .$ If $\vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k},$ then the projection of the vector $\vec{a}$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is:

$\vec{a} ⊥ t o 3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$ $\vec{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$

$\vec{a} \cdot (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}) = 0$ $(x i + y \hat{j} + z \hat{k}) \times (2 \hat{i} + \hat{k})$

Let $\vec{a} = (x \hat{i} + y \hat{j} + 2 \hat{k})$

$(x \hat{i} + y \hat{j} + z \hat{k}) . (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}) = 0$ $= | \begin{matrix} i & j & k \\ x & y & z \\ 2 & 0 & 1 \end{matrix} |$

$3 x + \frac{y}{2} + 2 z = 0$ $= i (y - 0) - j (x - 2 z) + k (x . 0 - 2 y)$

4x – 12 = 0 y = 2

x = 3

z = -5

x -2z = 13

$∴ \vec{a} = (3 \hat{i} + 2 \hat{j} - 5 \hat{k})$

$\frac{\vec{a} \cdot \vec{b}}{| b |} = | a c o s θ |$ $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ $\vec{a} \cdot \hat{i} = 6 + 4 - 5 = 5$

$\vec{a} = 3 \hat{i} + 2 \hat{j} - 5 k$

$\Rightarrow (\frac{5}{3}) = | a c o s θ |$

If cot a = 1 and sec b = $- \frac{5}{3}$ , where $π < α < \frac{3 π}{2} a n d \frac{π}{2} < β < π,$ then the value of tan(a + b) and the quadrant in which a + b lies, respectively are:

cota = 1 & secβ = $\frac{- 5}{3}$

< < $\frac{}{}$ $\frac{3 π}{2}$ secβ = $\frac{- 5}{3}$

$α = (π + \frac{π}{4})$ cosβ = $\frac{- 3}{5} = c o s (180 - 53)$

tanb = $\frac{- 4}{3}$

tan(α + β) = $\frac{t a n α + t a n β}{1 - t a n α t a n β}$

$∴ A - \frac{1}{4} & 4 t h q u a d r a n t .$

$= \frac{1 - \frac{4}{3}}{1 + \frac{4}{3}} = \frac{- 1}{7}$

$∴ - \frac{1}{4} & 4 t h q u a d r a n t .$

Let the image of the point P(1, 2, 3) in the line $L : \frac{x - 6}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ be Q $L e t R (α, β, γ)$ . be a point that divides internally the line segment PQ in the ration 1 : 3. Then the value of $22 (α + β + γ)$ is equal to………….

Any point on line L is

M (3r + 6, 2r + 1, 3r + 2)

$\vec{P M} ⊥ r t o L$

$∴ 3 (3 r + 5) + 2 (2 r - 1) + 3 (3 r - 1) = 0$

$r = \frac{- 5}{11}$

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is…………..

$\frac{1}{7} \sum_{i = 1}^{7} {(x_{i} - 62)}^{2} = 20$ $\bar{x} = 62$

$\sum_{i = 1}^{7} {(x_{i} - 62)}^{2}$ = 140……………… (i)

If any one student get less 50 marks then ${(x_{i} - 62)}^{2} \geq 144$

but $\sum_{i = 1}^{7} {(x_{i} - 62)}^{2} = 140$

$∴$ both condition cannot satisfy together hence no students fails.

If one of the diameters of the circle $x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$ is a chord of circle. ${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2},$ then the value of r² is equal to…………….

$x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$

centre $(\sqrt[]{2}, 3 \sqrt[]{2})$

radius ${({(\sqrt[]{2})}^{2} + {(3 \sqrt[]{2})}^{2} - 14)}^{1 / 2}$

$= {(2 + 18 - 14)}^{1 / 2} = (\sqrt[]{6})$

${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2}$

centre $(2 \sqrt[]{2}, 2 \sqrt[]{2})$

OA = $\sqrt[]{{(2 \sqrt[]{2} - \sqrt[]{2})}^{2} + {(2 \sqrt[]{2} - 3 \sqrt[]{2})}^{2}} = \sqrt[]{2 + 2} = 2$

r² = ${(\sqrt[]{6})}^{2} + {(2)}^{2} = 6 + 4 = 10$

If $\underset{x \to 1}{l i m} \frac{s i n (3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = - 2,$ then the value of (a – b) is equal to………….

$\underset{x \to 1}{l i m} \frac{s i n (3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = - 2$

For this limit to be defined 2x³ – 7x² + ax + b should also trend to 0 or x ® 1.

$∴ 2 . {(1)}^{3} - 7 {(1)}^{2} + a \cdot 1 + b = 0$

2 – 7 + (a + b) = 0

(a + b) = 5 …………….(i)

Now this becomes % form $∴$ we apply L’lopital rule

$\underset{x \to 1}{l i m} \frac{(3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = \underset{x \to 1}{l i m} \frac{c o s (3 x^{2} - 4 x + 1) (6 x - 4) - 2 x}{6 x^{2} - 14 x + a}$

Now the numerator again ® 0 as x = 1

$∴$ 6x² – 14x + a ® 0 as x = 1

6 . (1)² – 14 + a = 0

a = 8 …………….(ii)

a + b = 5 $∴ a - b = 8 - (- 3) = 11$

(b = -3) ® from (i) & (ii)

Let for n = 1, 2,……, 50, S_n be the sum of the infinite geometric progression whose first term is n² and whose common ratio is Then the value of $\frac{1}{26} + \sum_{n = 1}^{50} (S_{n} + \frac{2}{n + 1} - n - 1)$ is equal to…………..

$S_{n} = \frac{n^{2}}{1 - \frac{1}{{(n + 1)}^{2}}} = \frac{n {(n + 1)}^{2}}{(n + 2)} = {(n + 1)}^{2} - 2 (n + 1) + 2 - \frac{2}{(n + 2)}$

$\frac{1}{26} + \sum_{n = 1}^{50} (S_{n} + \frac{2}{(n + 1)} - (n + 1)) = \frac{1}{26} + \sum_{n = 1}^{50} {(n + 1)}^{2} - 3 (n + 1) + 2 + 2 (\frac{1}{(n + 1)} - \frac{1}{(n + 2)})$

$= \frac{1}{26} + 45525 - 3 \times 1325 + 2 \times 50 + 2 (\frac{1}{2} - \frac{1}{52})$

$= \frac{1}{26} + 41550 + 100 + 1 - \frac{1}{26} = 41651$

If the system of linear equations

2x – 3y = $γ$ = 5,

aX + 5y = b + 1, where $α, β, γ \in R$ has infinitely many solutions, then the value of $| 9 α + 3 β + 5 γ |$ is equal to…………..

$\begin{matrix} 2 x - 3 y = γ + 5 & α x + 5 y = (β + 1) \end{matrix}} i n f i n i t e l y m a n y s o l u t i o n$

$∴ \frac{2}{α} = \frac{- 3}{5} = (\frac{γ + 5}{β + 1})$

(i) $\frac{2}{α} = \frac{- 3}{5} = \frac{γ + 5}{β + 1}$

$α = \frac{5 \times 2}{- 3}$ 5x + 25 = -3β - 3

^{$5 γ + 3 β = - 28$}

^{$| 9 α + 5 γ + 3 β | = | 9 \times \frac{- 10}{3} - 28 |$}

=^{$| - 30 - 28 | = 58$}

Let A = $(\begin{matrix} 1 + i & 1 \\ - i & 0 \end{matrix})$ where i $i = \sqrt[]{- 1} .$ Then, the number of elements in the set ${n \in {1, 2, . . . ., 100} : A^{n} = A}$ is…………….

$(\begin{matrix} 1 + i & 1 \\ - i & 0 \end{matrix})$

$A^{2} = (\begin{matrix} 1 + i & 1 \\ - i & 0 \end{matrix}) (\begin{matrix} 1 + i & 1 \\ - i & 0 \end{matrix}) = (\begin{matrix} i & 1 + i \\ 1 - i & - i \end{matrix})$

$A^{4} = (\begin{matrix} i & 1 + i \\ 1 - i & - i \end{matrix}) (\begin{matrix} i & 1 + i \\ 1 - i & - i \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$

$∴ A^{5} = A$

A⁹ = A

$∴ f o r n = 1, 5, 9, . . . . ., 97$

total possible values of n = 25

Sum of squares of modulus of all the complex numbers z satisfying $\bar{z} = i z^{2} + z^{2} - z$ is equal to…………..

$\tilde{z} = i z^{2} + z^{2} - z$

$z + \bar{Z} = z^{2} (i + 1)$

$z + \bar{Z} = z^{2} (i + 1)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2 x = (x^{2} - y^{2} + 2 i x y) (i + 1)$

Equating the real & in eg part.

$(x^{2} - y^{2} + 2 i x y) = 0 . . . . . . . . . (i)$

$(x^{2} - y^{2} - 2 x y) = (2 x) . . . . . . . . . . . . . . (i i)$

(i) & (ii)

4xy = -2x Þ x = 0 or y = $(\frac{- 1}{2})$

(for x = 0, y = 0)

For y = $\frac{- 1}{2}$

x² $- \frac{1}{4} + 2 (\frac{- 1}{2}) x = 0$

x = $\frac{4 \pm \sqrt[]{16 + 16}}{2.4}$

= $(\frac{1 + \sqrt[]{2}}{2}) o r (\frac{1 - \sqrt[]{2}}{2})$

$∴ s u m$ of ${| z |}^{2} = {(\frac{1 + \sqrt[]{2}}{2})}^{2} + \frac{1}{4} + {(\frac{1 - \sqrt[]{2}}{2})}^{2} + \frac{1}{4} + 0^{2} + O^{2}$

= $\frac{3}{4} + \frac{\sqrt[]{2}}{2} + \frac{1}{4} + \frac{3}{4} - \frac{\sqrt[]{2}}{2} + \frac{1}{4} = \frac{3}{2} + \frac{1}{2} = 2$

Let S = ${1, 2, 3, 4} .$ Then the number of elements in the set is……………….

${f : S \times S : f i s o n t o a n d f (a, b) = f (b, a) \geq a \forall (a, b) \in S \times S}$

$(1, 1) (1, 4) (4, 1) (2, 4) (4, 2) (3, 4) (4, 3) (4, 4)$ all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$∴$ total function = 3 × 3 × 2 × 2 × 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 × 2 × 1 × 1 × 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$∴$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

The maximum number of compound propositions, out of $p \lor r \lor s, p \lor r \lor \sim s, p \lor \sim q \lor s, \sim p \lor \sim r \lor s, \sim p \lor \sim r \lor \sim s, \sim p \lor q \lor \sim s,$ $q \lor r \lor \sim s, q \lor \sim r \lor \sim s, \sim p \lor \sim q \lor \sim s$ that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to………….

p q r s

F T F

Maths NCERT Exemplar Solutions Class 12th Chapter Three Exam

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