Let a, b, c be the roots of the equation x³ – 9x² + 15x + 2 = 0. The volume of a parallelopiped with nonparallel sides aî + bĵ + ck̂, bî + cĵ + ak̂ and cî + aĵ + bk̂ is

Volume = |a b c; b c a; c a b| = |3abc – a³ – b³ – c³| = | (a + b + c) (a² + b² + c² – ab – bc – ca)| = | (a + b + c) (a + b + c)² – 3 (ab + bc + ca)| = |9 (81-3 × 15)| = 324

There are two sets A = {a: a ∈ N and –3 ≤ a ≤ 5} and B = {b: b ∈ Z and 0 ≤ b ≤ 4}. The number of elements common in A × B and B × A are

A = {1,2,3,4,5} B = {0,1,2,3,4} No. of elements common in A&B = 4. ∴ No. of elements common in A × B and B × A = 4 × 4 = 16

If x&y are real numbers satisfying the relation x2 + y2 – 6x + 8y + 24 = 0 then minimum value of log2(x2 + y2) is

x² + y² – 6x + 8y + 24 = 0 is circle having centre (3, −4) & r = √ (9+16-24) = 1 √x² + y² min. is min. distance from origin = 4 ∴ minimum value of log? (x² + y²) = log? 16 = 4

Let f(x) = { (sin(2x²/a) + cos(3x²/b))^(ab/x²), x ≠ 0 { e^(2x+3), x = 0 is a continuous function at x = 0, ∀b ∈ R, then 1/a_min is

RHL&LHL lim (x→0) (sin (2x²/a) + cos (3x/b)^ (ab/x²) = e^ (lim (x→0) (sin (2x²/a) + cos (3x/b) - 1) (ab/x²) = e^ (4b²-9a)/2b) f (0) = e³ For continuity at x = 0 Limit = f (0) (4b² - 9a)/2b = 3 ⇒ 4b² – 6b – 9a = 0∀b ∈ R ⇒ D ≥ 0 ⇒ a ≥ -1/4 a? = -1/4 ⇒ |1/a? | = 4.

A vertical pole PS has two marks at Q and R such that portions PQ, PR and PS subtends angle α, β, γ respectively at a point on the ground which is at distance x from the bottom of pole P. If PQ = 1, PR = 2, PS = 3 and α + β + γ = 180°, then x² is

tan α = 1/x, tan β = 2/x, tan γ = 3/x Now α + β + γ = 180° ⇒ tan α + tan β + tan γ = tan α tan β tan γ 1/x + 2/x + 3/x = (123)/ (xxx) ⇒ x² = 1

If a+b+c≠0 and [abcbcacab]=0, then prove that a=b=c .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol. G i v e n t h a t : a + b + c ≠ 0 a n d [ a b c b c a c a b ] = 0 C 1 → C 1 + C 2 + C 3 ⇒ [ a + b + c b c a + b + c c a a + b + c a b ] = 0 ⇒ ( a + b + c ) | 1 b c 1 c a 1 a b | = 0 ( T a k i n g a + b + c c o m m o n f r o m C 1 ) ⇒ a + b + c ≠ 0 ∴ | 1 b c 1 c a 1 a b | = 0 R 1 → R 1 − R 2 a n d R 2 → R 2 − R 3 ⇒ | 0 b − c c − a 0 c − a a − b 1 a b | = 0 E x p a n d i n g a l o n g C 1 ⇒ 1 | b − c c − a c − a a − b | = 0 ⇒ ( b − c ) ( c − a ) − ( c − a ) 2 = 0 ⇒ a b − b 2 − a c + b c − c 2 − a 2 + 2 a c = 0 ⇒ − a 2 − b 2 − c 2 + a b + b c + a c = 0 ⇒ a 2 + b 2 + c 2 − a b − b c − a c = 0 ⇒ &thin

Prove that [bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2] is divisible by a+b+c and find the quotient.

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: L e t Δ = | b c − a 2 c a − b 2 a b − c 2 c a − b 2 a b − c 2 b c − a 2 a b − c 2 b c − a 2 a c − b 2 | C 1 → C 1 + C 2 + C 3 ⇒ [ a b + b c + a c − a 2 − b 2 − c 2 c a − b 2 a b − c 2 a b + b c + a c − a 2 − b 2 − c 2 a b − c 2 b c − a 2 a b + b c + a c − a 2 − b 2 − c 2 b c − a 2 a c − b 2 ] ⇒ ( a b + b c + a c − a 2 − b 2 − c 2 ) | 1 c a − b 2 a b − c 2 1 a b − c 2 b c − a 2 1 b c − a 2 a c − b 2 | ( T a k i n g a b + b c + a c − a 2 − b 2 − c 2 c o m m o n f r o m C 1 ) R 1 → R 1 − R 2 a n d R 2 → R 2 − R 3 ⇒ ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 c a − b 2 − a b + c 2 a b − c 2 − b c + a 2 0 a b − c 2 − b c + a 2 b c − a 2 − a c + b 2 1 b c − a 2 a c − b 2 | ⇒ ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 a ( c − b ) + ( c + b ) ( c − b ) b ( a − c ) + ( a + c ) ( a − c ) 0 b ( a − c ) + ( a + c ) ( a − c ) c ( b − a ) + ( b + a ) ( b − a ) 1 b c − a 2 a c − b 2 | ⇒ ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 ( c − b ) ( a + b + c ) ( a − c ) ( a + b + c ) 0 ( a − c ) ( a + b + c ) ( b − a ) ( a + b + c ) 1 b c − a 2 a c − b 2 | ⇒ ( a b + b c + a c − a 2 − b 2 − c 2 ) ( a + b + c ) ( a + b + c ) | 0 c − b a − c 0 a − c b − a 1 b c − a 2 a c − b 2 | ⇒ ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) | 0 c − b a − c 0 a − c b − a 1 b c − a 2 a c − b 2 | E x p a n d i n g a l o n g C 1 ⇒ ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) [ 1 | c − b a − c a − c b − a | ] ⇒ ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) [ ( c − b ) ( b − a ) − ( a − c ) 2 ] ⇒ ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 − c 2 ) ( b c − c a − b 2 + a b − a 2 − c 2 + 2 a c ) ⇒ ( a + b + c ) 2 ( a b + b c + a c − a 2 − b 2 −

If x+y+z=0 , prove that [xaybzcyczaxbzbxcya]=xyz[abccabbca].

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: L . H . S . L e t Δ = | x a y b z c y c z a x b z b x c y a | E x p a n d i n g a l o n g R 1 ⇒ x a | z a x b x c y a | − y b | y c x b z b y a | + z c | y c z a z b x c | ⇒ x a ( y z a 2 − x 2 b c ) − y b ( y 2 a c − x z b 2 ) + z c ( x y c 2 − z 2 a b ) ⇒ x y z a 3 − x 3 a b c − y 3 a b c + x y z b 3 + x y z c 3 − z 3 a b c ⇒ x y z ( a 3 + b 3 + c 3 ) − a b c ( x 3 + y 3 + z 3 ) ⇒ x y z ( a 3 + b 3 + c 3 ) − a b c ( 3 x y z ) [ ? ( x + y + z = 0 ) ( ∴ x 3 + y 3 + z 3 = 3 x y z ) ] ⇒ x y z ( a 3 + b 3 + c 3 − 3 a b c ) R . H . S . x y z | a b c c a b b c a | R 1 → R 1 + R 2 + R 3 ⇒ x y z | a + b + c a + b + c a + b + c c a b b c a | ⇒ x y z ( a + b + c ) | 1 1 1 c a b b c a | ( a + b + c c o m m o n f r o m R 1 ) C 1 → C 1 − C 2 a n d C 2 → C 2 − C 3 ⇒ x y z ( a + b + c ) | 0 0 1 c − a a − b b b − c c − a a | E x p a n d i n g a l o n g R 1 ⇒ x y z ( a + b + c ) [ 1 | c − a a − b b − c c − a | ] ⇒ x y z ( a + b + c ) [ ( c − a ) 2 − ( b − c ) ( a − b ) ] ⇒ x y z ( a + b + c ) ( c 2 + a 2 − 2 c a − a b + b 2 + a c − b c ) ⇒ x y z ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) ⇒ x y z ( a 3 + b 3 + c 3 − 3 a b c ) [ a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) ] L . H . S . = R . H . S . H e n c e , p r o v e d .

The determinant [b2−abb−cbc−acb−a2a−bb2−abbc−acc−aab−a2] equals (A) abc(b−c)(c−a)(a−b) (B) (b−c)(c−a)(a−b) (C) (a+b+c)(b−c)(c−a)(a−b) (D) None of these

This is a Objective Type Questions as classified in NCERT Exemplar Sol: L e t Δ = | b 2 − a b b − c b c − a c a b − a 2 a − b b 2 − a b b c − a c c − a a b − a 2 | ⇒ = | b ( b − a ) b − c c ( b − a ) a ( b − a ) a − b b ( b − a ) c ( b − a ) c − a a ( b − a ) | ( T a k i n g ( b − a ) c o m m o n f r o m C 1 a n d C 3 ) ⇒ = ( b − a ) 2 | b b − c c a a − b b c c − a a | C 1 → C 1 − C 3 ⇒ = ( a − b ) 2 | b − c b − c c a − b a − b b c − a c − a a | ( C 1 a n d C 2 a r e i d e n t i c a l c o l u m n s . ) ⇒ = ( a − b ) 2 . 0 = 0 H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

The number of distinct real roots of [sinxcosxcosxcosxsinxcosxcosxcosxsinx]=0 in the interval −π4≤x≤π4 is (A) 0 (B) 2 (C) 1 (D) 3

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t | s i n x c o s x c o s x c o s x s i n x c o s x c o s x c o s x s i n x | = 0 C 1 → C 1 + C 2 + C 3 ⇒ | 2 c o s x + s i n x c o s x c o s x 2 c o s x + s i n x s i n x c o s x 2 c o s x + s i n x c o s x s i n x | = 0 ( T a k i n g 2 c o s x + s i n x c o m m o n f r o m C 1 ) ⇒ ( 2 c o s x + s i n x ) | 1 c o s x c o s x 1 s i n x c o s x 1 c o s x s i n x | = 0 R 1 → R 1 − R 2 , R 2 → R 2 − R 3 ⇒ ( 2 c o s x + s i n x ) | 0 c o s x − s i n x 0 0 s i n x − c o s x c o s x − s i n x 1 c o s x s i n x | = 0 ⇒ ( 2 c o s x + s i n x ) [ 1 | c o s x − s i n x 0 s i n x − c o s x c o s x − s i n x | ] ⇒ ( 2 c o s x + s i n x ) ( c o s x − s i n x ) 2 = 0 2 c o s x + s i n x = 0 a n d ( c o s x − s i n x ) 2 = 0 2 + t a n x = 0 a n d ( c o s x − s i n x ) = 0 ∴ t a n x = − 2 a n d ⇒ t a n x = 1 B u t − π 4 ≤ x ≤ π 4 a n d ⇒ t a n x = t a n π 4 ∴ x = π 4 ∈ [ − π 4 , π 4 ] S o , x &thins

Let f(t|) [costt12sintt2tsinttt], then limt→0 f(t)t2 is equal to (A) 0 (B) –1 (C) 2 (D) 3

This is a Objective Type Questions as classified in NCERT Exemplar Sol: W e h a v e f ( t ) = [ c o s t t 1 2 s i n t t 2 t s i n t t t ] E x p a n d i n g a l o n g R 1 = c o s t | t 2 t t t | − t | 2 s i n t 2 t s i n t t | + 1 | 2 s i n t t s i n t t | = c o s t ( t 2 − 2 t 2 ) − t ( 2 t s i n t − 2 t s i n t ) + ( 2 t s i n t − t s i n t ) = t 2 c o s t + t s i n t ∴ f ( t ) t 2 = − t 2 c o s t + t s i n t t 2 ⇒ f ( t ) t 2 = − c o s t + s i n t t 2 ⇒ l i m t → 0 f ( t ) t 2 = l i m t → 0 ( − c o s t ) + l i m t → 0 s i n t t 2 = − 1 + 1 = 0 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

The maximum value of = [11111+sinθ11+cosθ11] is ( θ is real number) (A) 12 (B) √3/2 (C) √2 (D) 2√34

This is a Objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t Δ = | 1 1 1 1 1 + s i n θ 1 1 + c o s θ 1 1 | C 1 → C 1 − C 2 , C 2 → C 2 − C 3 ⇒ = | 0 0 1 − s i n θ s i n θ 1 c o s θ 0 1 | E x p a n d i n g a l o n g R 1 ⇒ = 1 | − s i n θ s i n θ c o s θ 0 | = − s i n θ c o s θ ⇒ = − 1 2 . 2 s i n θ c o s θ = − 1 2 s i n 2 θ but maximum value of sin2θ=1 ⇒|−12.1|=12 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

Kindly Consider the following [3x−x+y−x+zx−y3yz−yx−zy−z3z]

Kindly Consider the following [x+4xxxx+4xxxx+4]

Kindly Consider the following [a−b−c2a2a2bb−c−a2b2c2cc−a−b]

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: L e t Δ = | a − b − c 2 a 2 a 2 b b − c − a 2 b 2 c 2 c c − a − b | R 1 → R 1 + R 2 + R 3 = | a + b + c a + b + c a + b + c 2 b b − c − a 2 b 2 c 2 c c − a − b | T a k i n g ( a + b + c ) c o m m o n f r o m R 1 = ( a + b + c ) | 1 1 1 2 b b − c − a 2 b 2 c 2 c c − a − b | C 1 → C 1 − C 2 , C 2 → C 2 − C 3 = ( a + b + c ) | 0 0 1 b + c + a − ( b + c + a ) 2 b 0 a + b + c c − a − b | T a k i n g ( b + c + a ) c o m m o n f r o m C 1 a n d C 2 = ( a + b + c ) 3 | 0 0 1 1 − 1 2 b 0 1 c − a − b | E x p a n d i n g a l o n g R 1 = ( a + b + c ) 3 [ 1 | 1 − 1 0 1 | ] = ( a + b + c ) 3 .

Kindly Consider the following [y2z2yzy+zz2x2zxz+xx2y2xyx+y] =0

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: L . H . S . = | y 2 z 2 y z y + z z 2 x 2 z x z + x x 2 y 2 x y x + y | Taking R1→xR1, R2→yR2, R3→zR3 and dividing the determinant by xyz. = 1 x y z | x y 2 z 2 x y z x y + z x y z 2 x 2 y z x y z + x y z x 2 y 2 z x y z x + z y | T a k i n g x y z c o m m o n f r o m C 1 a n d C 2 = x y z . x y z x y z | y z 1 x y + z x z x 1 y z + x y x y 1 z x + z y | C 3 → C 3 + C 1 = x y z | y z 1 x y + y z + z x z x 1 x y + y z + z x x y 1 x y + y z + z x | T a k i n g ( x y + y z + z x ) c o m m o n f r o m C 3 = ( x y z ) ( x y + y z + z x ) | y z 1 1 z x 1 1 x y 1 1 | = ( x y z ) ( x y + y z + z x ) | y z 1 1 z x 1 1 x y 1 1 | = 0 [ ? C 2 a n d C 3 a r e i d e n t i c a l ] L . H . S . = R . H . S . H e n c e p r o v e d .

Let α be root of the equation 1 + x2 + x4 = 0. Then the value of α1011 + α2022 - α3033 is equal to

x4 + x2 + 1 = 0 x4 + 2x2 + 1 – x2 = 0 ⇒ (x2+1+x) (x2−x+1)=0 x=±ω, ? ω2 Now, = α1011+α2022−α3033 = ω1011+ω2022−ω3033 = 1 + 1 – 1 = 1

The sum of the infinite series 1+56+1262+2263+3564+5165+7066+.... is equal to

Let S = 1 +56+1262+2263+.... −S6=16+562+...._ 5S6=1+46+762+1063+.... −5S36=16+462+763+....._ 25S36=1+36+362+363+...... 25S36=1+3/61−1/6 25S36=1+3/51 S=288125

The value of limx→1(x2−1)sin2(πx)x4−2x3+2x−1 is equal to

x4−2x3+2x−1= (x−1)2 (x2−1) sinπx=sin (π (1−x) =−sin (sinπ (x−1)) limx→1 (x2−1)⋅sin2πx (x2−1) (x−1)2=limx→1sin2 (π (x−1)) (x−1)2 =limx→1sin2 (π (x−1)) (π (x−1))2⋅π2 = 2

Set of all possible values of k for which f(x) = sin x – cos x – kx + b decreases for all real values of x, is:

(-∞, 1)

The parabola y = x 2 – 9 and y = kx 2 intersect each other at the points A and B. If the length AB is equal to 10 units then the value of k is:

25/16

16/9

Coefficient of variation of two distributions are 50% and 60% and their arithmetic means are 30 and 25 respectively. Difference of standard deviation is:

1.5

2.5

Let α and β are roots of the equation x 2 – 6x + 12 = 0. The value of the [removed]α – 2) 12 + (β−6) 12 /α 12 – 1 is:

2¹⁰

3¹⁰

a, b, c are positive integers forming an increasing G.P. whose common ratio is a natural number. b – a is cube of a natural number and log? a + log 6 b + log e c = 6, then a + b + c =

111

Area of the region bounded by x = 0, y = 0, x = 2, y = 2, y ≤ e? and y ≥ ln x, is:

6 – 4ln 2

4ln 2 – 2

2ln 2 – 4

The statement p → (q → p) is equivalent is:

p → (q ∨ p)

p → (q ∧ p)

p → (p → q)

If z = cos 20° + isin 20, then |z + 2z 2 + 3z 3 + ... + 18z 18 | -1 is:

⅓ sin 10°

½ sin 10°

The tangent at a point where eccentric angle 60° on the ellipse x 2 /a 2 + y 2 /b 2 = 1(a > b) meet the auxiliary circle at L and M. If LM subtends a right angle at the centre, then eccentricity of the ellipse is:

1/√7

1/√3

From first 100 natural numbers, 3 numbers are selected. If these numbers are in A.P., then the probability that these numbers are even is:

2/9

29/56

Let a 2 , b 2 , c 2 be three distinct numbers in A.P. If ab + bc + ca = 1 then (b + c), (c + a) and (a + b) are in:

H.P.

A.P.

G.P.

Maths NCERT Exemplar Solutions Class 12th Chapter Four Determinants

Updated on Sep 11, 2025 10:51 IST

By Pallavi Pathak, Assistant Manager Content

Maths NCERT Exemplar Solutions Class 12th Chapter Four Determinants offers practice questions for Class 12 students who are preparing for the CBSE Board exams and other competitive exams like JEE. The exemplar questions take you beyond the NCERT Solutions Class 12th Chapter Four Determinants, and help you deepen your understanding of the concepts and formulas. It gives extra questions for practice, which include Multiple Choice Questions, Short Answer, and long-answer-type questions.
Students can download the Class 12 Determinants PDF and practice the questions to score high in the examinations. These solutions are given in a step-by-step format, which helps students to understand each step and hence improves their problem-solving skills. Math requires practice, and when one solves the exemplar questions, they are more confident in solving the most complex problems in exams.

Download PDF of NCERT Exemplar Class 12 Maths Chapter Four Determinants

The PDF of the Maths Class 12 Determinants here gives a curated collection of high-quality questions. It follows the CBSE syllabus and exam pattern, and hence it is ideal for exam preparation. Once the student downloads the PDF, they can study it from anywhere and at any time, even without the need for an internet connection. The Maths PDF includes the Short Answer, Long Answer, and MCQ questions, which help students to prepare for the various examinations.
The PDF questions will help students to think logically and deepen their understanding of topics like Cofactors and Minors, Properties of Determinants, Cramer's Rule for solving equations, and Adjoint and Inverse of a Matrix. The students can rely on these solutions and it is available free of cost. They can also take a printout and practice questions on paper.
Also read:
NCERT Solutions

Class 12 Maths Determinants NCERT Exemplar Short Answer Type Questions

SA Questions

Kindly Consider the following

$[\begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2} \\ = | \begin{matrix} x^{2} - 2 x + 2 & x - 1 \\ 0 & x + 1 \end{matrix} | \\ = (x + 1) (x^{2} - 2 x + 2) - 0 \\ = x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2 = x^{3} - x^{2} + 2 \end{array}$

Kindly Consider the following

$[\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{matrix} | = (a + x + y + z) | \begin{matrix} 1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z \end{matrix} | \\ (T a k i n g a + x + y + z c o m m o n) \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (a + x + y + z) | \begin{matrix} 0 & - a & 0 \\ 0 & a & - a \\ 1 & y & a + z \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (a + x + y + z) | 1 (a^{2} - 0) | = a^{2} (a + x + y + z) \end{array}$

Commonly asked questions

Kindly Consider the following

$[\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} y & z y^{2} & 0 \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix} | \\ T a k i n g x,^{2} y^{2} a n d z^{2} c o m m o n f r o m C_{1}, C_{2} a n d C_{3} r e s p e c t i v e l y \\ = x^{2} y^{2} z^{2} | \begin{matrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = x^{2} y^{2} z^{2} [0 | \begin{matrix} 0 & y \\ z & 0 \end{matrix} | - x | \begin{matrix} y & y \\ z & 0 \end{matrix} | + x | \begin{matrix} y & 0 \\ z & z \end{matrix} |] \\ = x^{2} y^{2} z^{2} [- x (0 - y z) + x (y z - 0)] \\ = x^{2} y^{2} z^{2} (x y z + x y z) = x^{2} y^{2} z^{2} (2 x y z) = 2 x^{3} y^{3} z^{3} \end{array}$

Kindly Consider the following

$[\begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} x + y + z & - x + y & - x + z \\ x + y + z & 3 y & z - y \\ x + y + z & y - z & 3 z \end{matrix} | \\ T a k i n g (x + y + z) c o m m o n f r o m C_{1} \\ = (x + y + z) | \begin{matrix} 1 & - x + y & - x + z \\ 1 & 3 y & z - y \\ 1 & y - z & 3 z \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (x + y + z) | \begin{matrix} 0 & - x - 2 y & - x + y \\ 0 & 2 y + z & - y - 2 z \\ 0 & y - z & 3 z \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (x + y + z) [1 | \begin{matrix} - x - 2 y & - x + y \\ 2 y + z & - y - 2 z \end{matrix} |] \\ = (x + y + z) [(- x - 2 y) (- y - 2 z) - (2 y + z) (- x + y)] \\ = (x + y + z) (x y + 2 z x + 2 y^{2} + 4 y z + 2 x y - 2 y^{2} + z x - z y) \\ = (x + y + z) (3 x y + 3 z x + 3 y z) \\ = 3 (x + y + z) (x y + y z + z x) \end{array}$

Kindly Consider the following

$[\begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} 3 x + 4 & x & x \\ 3 x + 4 & x + 4 & x \\ 3 x + 4 & x & x + 4 \end{matrix} | \\ T a k i n g (3 x + 4) c o m m o n f r o m C_{1} \\ = (3 x + 4) | \begin{matrix} 1 & x & x \\ 1 & x + 4 & x \\ 1 & x & x + 4 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (3 x + 4) | \begin{matrix} 0 & - 4 & 0 \\ 0 & 4 & - 4 \\ 1 & x & x + 4 \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (3 x + 4) [1 | \begin{matrix} - 4 & 0 \\ 4 & - 4 \end{matrix} |] \\ = (3 x + 4) (16 - 0) = 16 (3 x + 4) \end{array}$

Kindly Consider the following

$[\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}]$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ = | \begin{matrix} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ T a k i n g (a + b + c) c o m m o n f r o m R_{1} \\ = (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ = (a + b + c) | \begin{matrix} 0 & 0 & 1 \\ b + c + a & - (b + c + a) & 2 b \\ 0 & a + b + c & c - a - b \end{matrix} | \\ T a k i n g (b + c + a) c o m m o n f r o m C_{1} a n d C_{2} \\ = {(a + b + c)}^{3} | \begin{matrix} 0 & 0 & 1 \\ 1 & - 1 & 2 b \\ 0 & 1 & c - a - b \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = {(a + b + c)}^{3} [1 | \begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix} |] \\ = {(a + b + c)}^{3} . \end{array}$

Kindly Consider the following

$[\begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix}]$ =0

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix} | \\ T a k i n g R_{1} \to x R_{1}, R_{2} \to y R_{2}, R_{3} \to z R_{3} a n d d i v i d i n g t h e determinant b y x y z . \\ = \frac{1}{x y z} | \begin{matrix} x y^{2} z^{2} & x y z & x y + z x \\ y z^{2} x^{2} & y z x & y z + x y \\ z x^{2} y^{2} & z x y & z x + z y \end{matrix} | \\ T a k i n g x y z c o m m o n f r o m C_{1} a n d C_{2} \\ = \frac{x y z . x y z}{x y z} | \begin{matrix} y z & 1 & x y + z x \\ z x & 1 & y z + x y \\ x y & 1 & z x + z y \end{matrix} | \\ C_{3} \to C_{3} + C_{1} \\ = x y z | \begin{matrix} y z & 1 & x y + y z + z x \\ z x & 1 & x y + y z + z x \\ x y & 1 & x y + y z + z x \end{matrix} | \\ T a k i n g (x y + y z + z x) c o m m o n f r o m C_{3} \\ = (x y z) (x y + y z + z x) | \begin{matrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{matrix} | \\ = (x y z) (x y + y z + z x) | \begin{matrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{matrix} | = 0 [? C_{2} a n d C_{3} a r e i d e n t i c a l] \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Kindly Consider the following

$[\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}]$ = 4 $x y z$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix} | \\ C_{1} \to C_{1} - (C_{2} + C_{3}) \\ = | \begin{matrix} 0 & z & y \\ - 2 x & z + x & x \\ - 2 x & x & x + y \end{matrix} | \\ T a k i n g - 2 c o m m o n f r o m C_{1} \\ = - 2 | \begin{matrix} 0 & z & y \\ x & z + x & x \\ x & x & x + y \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ = - 2 [x | - z y - z y |] = - 2 (- x y z) = 4 x y z R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Kindly Consider the following

$[\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}]$ = (a – 1)3

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = | \begin{matrix} a^{2} - 1 & a - 1 & 0 \\ 2 a - 2 & a - 1 & 0 \\ 3 & 3 & 1 \end{matrix} | = | \begin{matrix} (a + 1) (a - 1) & a - 1 & 0 \\ 2 (a - 1) & a - 1 & 0 \\ 3 & 3 & 1 \end{matrix} | \\ T a k i n g (a - 1) c o m m o n f r o m C_{1} a n d C_{2} \\ = (a - 1) (a - 1) | \begin{matrix} a + 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ = {(a - 1)}^{2} [1 | \begin{matrix} a + 1 & 1 \\ 2 & 1 \end{matrix} |] \\ = {(a - 1)}^{2} (a + 1 - 2) = {(a - 1)}^{2} (a - 1) = {(a - 1)}^{3} R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Kindly Consider the following

If $A + B + C = 0$ , then prove that $[\begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix}] = 0$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ = 1 | \begin{matrix} 1 & c o s A \\ c o s A & 1 \end{matrix} | - c o s C | \begin{matrix} c o s C & c o s B \\ c o s A & 1 \end{matrix} | + c o s B | \begin{matrix} c o s C & c o s B \\ 1 & c o s A \end{matrix} | \\ = 1 (1 - c o s^{2} A) - c o s C (c o s C - c o s A c o s B) + c o s B (c o s A c o s C - c o s B) \\ = s i n^{2} A - c o s^{2} C + c o s A c o s B c o s C + c o s A c o s B c o s C - c o s^{2} B \\ = s i n^{2} A - c o s^{2} B - c o s^{2} C + 2 c o s A c o s B c o s C \\ = - c o s (A + B) . c o s (A - B) - c o s^{2} C + 2 c o s A c o s B c o s C [? s i n^{2} A - c o s^{2} B = - c o s (A + B) . c o s (A - B)] \\ = - c o s (- C) . c o s (A - B) + c o s C + (2 c o s A c o s B - c o s C) [? A + B + C = 0] \\ = - c o s C (c o s A c o s B + s i n A s i n B) + c o s C + (2 c o s A c o s B - c o s C) \\ = - c o s C (c o s A c o s B + s i n A s i n B - 2 c o s A c o s B + c o s C) \\ = - c o s C (- c o s A c o s B + s i n A s i n B + c o s C) \\ = c o s C (c o s A c o s B - s i n A s i n B - c o s C) \\ = c o s C [c o s (A + B) - c o s C] \\ = c o s C [c o s (- C) - c o s C] [? A + B = - C] \\ = c o s C [c o s C - c o s C] = c o s C . 0 = 0 R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Kindly Consider the following

If the coordinates of the vertices of an equilateral triangle with sides of length ‘a’ are $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ , then

${[\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}]}^{2} = \frac{3 a^{2}}{4}$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0 \\ | A | = | \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0 \\ T a k i n g 7 c o m m o n f r o m C_{1} \\ \Rightarrow = 7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} 1 & s i n 3 θ \\ - 7 & - 2 \end{matrix} | + 2 | \begin{matrix} 1 & s i n 3 θ \\ 3 & c o s 2 θ \end{matrix} | = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 (c o s 2 θ - 3 s i n 3 θ) = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 c o s 2 θ - 6 s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 c o s 2 θ + s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 4 s i n^{3} θ - 4 s i n^{2} θ + 3 s i n θ = 0 \\ \Rightarrow - s i n θ (4 s i n^{2} θ + 4 s i n θ - 3) = 0 \\ - s i n θ = 0 o r 4 s i n^{2} θ + 4 s i n θ - 3 = 0 \\ ∴ θ = n π o r 4 s i n^{2} θ + 6 s i n θ - 2 s i n θ - 3 = 0 w h e n n \in I \\ \Rightarrow 2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3) = 0 \\ \Rightarrow (2 s i n θ + 3) (2 s i n θ - 1) = 0 \\ \Rightarrow &thinsp \end{array}$

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] = 0 \\ | A | = | \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow | \begin{matrix} 12 + x & 12 + x & 12 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ T a k i n g (12 + x) c o m m o n f r o m R_{1} \\ \Rightarrow (12 + x) | \begin{matrix} 1 & 1 & 1 \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow (12 + x) | \begin{matrix} 0 & 0 & 1 \\ 2 x & - 2 x & 4 + x \\ 0 & 2 x & 4 - x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (12 + x) [1 . | \begin{matrix} 2 x & - 2 x \\ 0 & 2 x \end{matrix} |] = 0 \\ \Rightarrow (12 + x) (4 x^{2} - 0) = 0 \Rightarrow 12 + x = 0 o r 4 x^{2} = 0 \\ \Rightarrow x = - 12 o r x = 0 \end{array}$

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f a_{1}, a_{2}, a_{3}, \dots a_{r} b e t h e t e r m s o f G . P ., t h e n \\ a_{n} = A R^{n - 1} (w h e r e A i s t h e f i r s t t e r m a n d R i s t h e c o m m o n r a t i o o f t h e G . P .) \\ ∴ a_{r + 1} = A R^{r + 1 - 1} = A R^{r}; a_{r + 5} = A R^{r + 5 - 1} = A R^{r + 4} \\ a_{r + 9} = A R^{r + 9 - 1} = A R^{r + 8}; a_{r + 7} = A R^{r + 7 - 1} = A R^{r + 6} \\ a_{r + 11} = A R^{r + 11 - 1} = A R^{r + 10}; a_{r + 15} = A R^{r + 15 - 1} = A R^{r + 14} \\ a_{r + 17} = A R^{r + 17 - 1} = A R^{r + 16}; a_{r + 21} = A R^{r + 21 - 1} = A R^{r + 20} \\ ∴ T h e determinant b e c o m e s \\ | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} | \\ T a k i n g A R^{r}, A R^{r + 6} a n d A R^{r + 10} c o m m o n f r o m R_{1,} R_{2,} a n d R_{3} r e s p e c t i v e l y . \\ A R^{r} . A R^{r + 6} . A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} | \\ = A R^{r} . A R^{r + 6} . A R^{r + 10} | 0 | [? R_{1} a n d R_{2} a r e i d e n t i c a l r o w s] \\ = 0 \\ H e n c e, t h e g i v e n determinant i s i n d e p e n d e n t o f r . \end{array}$

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f t h e g i v e n points l i e o n a s t r a i g h t l i n e, t h e n t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e \\ points p a i r w i s e i s z e r o . \\ S 0, | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 7 & - 7 & 0 \\ - 2 & 3 & 0 \\ a & a & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ \Rightarrow 1 . | \begin{matrix} 7 & - 7 \\ - 2 & 3 \end{matrix} | = 21 - 14 = 7 u n i t s \\ A s 7 \neq 0 . H e n c e, t h e t h r e e points d o n o t l i e o n a s t r a i g h t l i n e f o r a n y v a l u e o f a . \end{array}$

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} | \begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A + c o s A \\ - c o s^{2} B - c o s B \end{array} & \begin{array}{l} c o s^{2} B + c o s B \\ - c o s^{2} C - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A - c o s^{2} B \\ + c o s A - c o s B \end{array} & \begin{array}{l} c o s^{2} B - c o s^{2} C \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} (c o s A + c o s B) \times \\ (c o s A - c o s B) \\ + (c o s A - c o s B) \end{array} & \begin{array}{l} (c o s B + c o s C) \times \\ (c o s B - c o s C) \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ T a k i n g (c o s A - c o s B) a n d (c o s B - c o s C) c o m m o n f r o m C_{1} a n d C_{2} r e s p e c t i v e l y . \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & 1 + c o s C \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 & c o s^{2} C + c o s C \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [1 | \begin{matrix} 1 & 1 \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 \end{matrix} |] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [(c o s B + c o s C + 1) - (c o s A + c o s B + 1)] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [c o s B + c o s C + 1 - c o s A - c o s B - 1] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) (c o s C - c o s A) = 0 \\ \Rightarrow c o s A - c o s B = 0 o r c o s B - c o s C = 0 o r c o s C - c o s A = 0 \\ \Rightarrow c o s A = c o s B o r c o s B = c o s C o r c o s C = c o s A \\ \Rightarrow ∠ A = ∠ C o r ∠ B = ∠ C \Rightarrow ∠ A = ∠ B \\ H e n c e, Δ A B C i s a n i s o s c e l e s t r i a n g l e . \end{array}$

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} H e r e, A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ | A | = 0 | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | - 1 | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | \\ = 0 - 1 (0 - 1) + 1 (1 - 0) \\ = 1 + 1 = 2 \neq 0 (n o n - singular m a t r i x) \\ N o w, c o - f a c t o r s, \\ a_{11} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{12} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{13} = + | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1 \\ a_{21} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{22} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{23} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1 \\ a_{31} = + | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1, a_{32} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1, a_{33} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1 \\ A d j (A) = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'} = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} A d j (A) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ H e n c e, A^{2} = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ N o w, w e h a v e t o p r o v e t h a t A^{- 1} = \frac{A^{2} - 3 I}{2} \\ R . H . S . = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}{2} \\ = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]}{2} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ = A^{- 1} = L . H . S . \\ H e n c e, p r o v e d . \end{array}$

NCERT Exemplar Class 12 Maths Chapter Four Long Answer Type Questions

LA Questions

Q1. If $A = [\begin{matrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{matrix}],$ find $A^{- 1}$ . Using $A^{- 1}$ , solve the system of linear equations

$x - 2 y = 10, 2 x - y - z = 8, - 2 y + z = 7 .$

Sol:

Commonly asked questions

Using matrix method, solve the system of equations

$3 x + 2 y - 2 z = 3, x + 2 y + 3 z = 6, 2 x - y + z = 2 .$

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

Sol:

Given $A = [- \begin{matrix} 2 & 2 & - 4 \\ 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], a n d B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}],$ find $B A$ and use this to solve the system of equations:

$y + 2 z = 7, x - y = 3, 2 x + 3 y + 4 z = 17 .$

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

Sol:

If $a + b + c \neq 0$ and $[\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0,$ then prove that $a = b = c$ .

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

Sol.

$\begin{array}{l} G i v e n t h a t : a + b + c \neq 0 a n d [\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow [\begin{matrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{matrix}] = 0 \\ \Rightarrow (a + b + c) | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 (T a k i n g a + b + c c o m m o n f r o m C_{1}) \\ \Rightarrow a + b + c \neq 0 ∴ | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2} a n d R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 1 & a & b \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} b - c & c - a \\ c - a & a - b \end{matrix} | = 0 \\ \Rightarrow (b - c) (c - a) - {(c - a)}^{2} = 0 \\ \Rightarrow a b - b^{2} - a c + b c - c^{2} - a^{2} + 2 a c = 0 \\ \Rightarrow - a^{2} - b^{2} - c^{2} + a b + b c + a c = 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - a c = 0 \\ \Rightarrow &thin \end{array}$

Prove that $[\begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{matrix}]$ is divisible by $a + b + c$ and find the quotient.

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

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & a c - b^{2} \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow [\begin{matrix} a b + b c + a c - a^{2} - b^{2} - c^{2} & c a - b^{2} & a b - c^{2} \\ a b + b c + a c - a^{2} - b^{2} - c^{2} & a b - c^{2} & b c - a^{2} \\ a b + b c + a c - a^{2} - b^{2} - c^{2} & b c - a^{2} & a c - b^{2} \end{matrix}] \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 1 & c a - b^{2} & a b - c^{2} \\ 1 & a b - c^{2} & b c - a^{2} \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | (T a k i n g a b + b c + a c - a^{2} - b^{2} - c^{2} c o m m o n f r o m C_{1}) \\ R_{1} \to R_{1} - R_{2} a n d R_{2} \to R_{2} - R_{3} \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & c a - b^{2} - a b + c^{2} & a b - c^{2} - b c + a^{2} \\ 0 & a b - c^{2} - b c + a^{2} & b c - a^{2} - a c + b^{2} \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & a (c - b) + (c + b) (c - b) & b (a - c) + (a + c) (a - c) \\ 0 & b (a - c) + (a + c) (a - c) & c (b - a) + (b + a) (b - a) \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & (c - b) (a + b + c) & (a - c) (a + b + c) \\ 0 & (a - c) (a + b + c) & (b - a) (a + b + c) \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) (a + b + c) (a + b + c) | \begin{matrix} 0 & c - b & a - c \\ 0 & a - c & b - a \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & c - b & a - c \\ 0 & a - c & b - a \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) [1 | \begin{matrix} c - b & a - c \\ a - c & b - a \end{matrix} |] \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) [(c - b) (b - a) - {(a - c)}^{2}] \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) (b c - c a - b^{2} + a b - a^{2} - c^{2} + 2 a c) \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - \end{array}$

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

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

Sol:

$\begin{array}{l} L . H . S . \\ L e t Δ = | \begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x a | \begin{matrix} z a & x b \\ x c & y a \end{matrix} | - y b | \begin{matrix} y c & x b \\ z b & y a \end{matrix} | + z c | \begin{matrix} y c & z a \\ z b & x c \end{matrix} | \\ \Rightarrow x a (y z a^{2} - x^{2} b c) - y b (y^{2} a c - x z b^{2}) + z c (x y c^{2} - z^{2} a b) \\ \Rightarrow x y z a^{3} - x^{3} a b c - y^{3} a b c + x y z b^{3} + x y z c^{3} - z^{3} a b c \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3}) - a b c (x^{3} + y^{3} + z^{3}) \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3}) - a b c (3 x y z) [? (x + y + z = 0) (∴ x^{3} + y^{3} + z^{3} = 3 x y z)] \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3} - 3 a b c) \\ R . H . S . x y z | \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} | \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow x y z | \begin{matrix} a + b + c & a + b + c & a + b + c \\ c & a & b \\ b & c & a \end{matrix} | \\ \Rightarrow x y z (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{matrix} | (a + b + c c o m m o n f r o m R_{1}) \\ C_{1} \to C_{1} - C_{2} a n d C_{2} \to C_{2} - C_{3} \\ \Rightarrow x y z (a + b + c) | \begin{matrix} 0 & 0 & 1 \\ c - a & a - b & b \\ b - c & c - a & a \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x y z (a + b + c) [1 | \begin{matrix} c - a & a - b \\ b - c & c - a \end{matrix} |] \\ \Rightarrow x y z (a + b + c) [{(c - a)}^{2} - (b - c) (a - b)] \\ \Rightarrow x y z (a + b + c) (c^{2} + a^{2} - 2 c a - a b + b^{2} + a c - b c) \\ \Rightarrow x y z (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3} - 3 a b c) [a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)] \\ L . H . S . = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

NCERT Exemplar Class 12 Maths Determinants Objective Type Questions

Objective Type Questions

Choose the correct answer from given four options in each of the Exercises from 1 to 14.

Q1. If $[\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}] = [\begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix}]$ then the value of $x$ is

(A) 3

(B) ±3

(D) 6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \\ \Rightarrow | \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} | = | \begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix} | \\ \Rightarrow 2 x^{2} - 40 = 18 + 14 \Rightarrow 2 x^{2} = 32 + 40 \\ \Rightarrow 2 x^{2} = 72 \Rightarrow x^{2} = 36 \\ ∴ x \pm 6 \\ H e n c e, t h e c o r r e c t x^{2} o p t i o n i s (c) . \end{array}$

Q2. The value of determinant $[\begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix}]$ is

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 b c$

(C) $a^{3} + b^{3} + c^{3} - 3 a b c$

(D) None of these

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, w e h a v e | \begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix} | \\ C_{2} \to C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} a - b & a + b + c & a \\ b - a & a + b + c & b \\ c - a & a + b + c & c \end{matrix} | \\ \Rightarrow (a + b + c) | \begin{matrix} a - b & 1 & a \\ b - a & 1 & b \\ c - a & 1 & c \end{matrix} | (T a k i n g a + b + c c o m m o n f r o m C_{2}) \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (a + b + c) | \begin{matrix} 2 (a - b) & 0 & a - b \\ b - c & 0 & b - c \\ c - a & 1 & c \end{matrix} | \\ T a k i n g (a - b) a n d (b - c) c o m m o n f r o m R_{1} a n d R_{2} r e s p e c t i v e l y \\ \Rightarrow (a + b + c) (a - b) (b - c) | \begin{matrix} 2 & 0 & 1 \\ 1 & 0 & 1 \\ c - a & 1 & c \end{matrix} | \\ E x p a n d i n g a l o n g C_{2} \\ \Rightarrow (a + b + c) (a - b) (b - c) [- 1 | \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} |] \\ \Rightarrow (a + b + c) (a - b) (b - c) (- 1) \\ \Rightarrow (a + b + c) (a - b) (c - b) \\ H e n c e, t h e c o r r e c t x^{2} o p t i o n i s (d) . \end{array}$

Commonly asked questions

The area of a triangle with vertices $(- 3, 0)$ , $(3, 0)$ , and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(C) –9

(D) 6

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A r e a o f t r i a n g l e w i t h v e r t i c e s (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) w i l l b e : \\ Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | \Rightarrow Δ = \frac{1}{2} | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} | \\ \Rightarrow = \frac{1}{2} [- 3 | \begin{matrix} 0 & 1 \\ k & 1 \end{matrix} | - 0 | \begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} | + 1 | \begin{matrix} 3 & 0 \\ 0 & k \end{matrix} |] \\ \Rightarrow = \frac{1}{2} [- 3 (- k) - 0 + 1 (3 k)] \\ \Rightarrow = \frac{1}{2} (3 k + 3 k) \Rightarrow = \frac{1}{2} (6 k) = 3 k \\ 3 k = 9 \Rightarrow k = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The determinant $[\begin{matrix} b^{2} - a b & b - c & b c - a c \\ b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix}]$ equals

(A) $a b c (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(C) $(a + b + c) (b - c) (c - a) (a - b)$

(D) None of these

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix} | \\ \Rightarrow = | \begin{matrix} b (b - a) & b - c & c (b - a) \\ a (b - a) & a - b & b (b - a) \\ c (b - a) & c - a & a (b - a) \end{matrix} | (T a k i n g (b - a) c o m m o n f r o m C_{1} a n d C_{3}) \\ \Rightarrow = {(b - a)}^{2} | \begin{matrix} b & b - c & c \\ a & a - b & b \\ c & c - a & a \end{matrix} | \\ C_{1} \to C_{1} - C_{3} \\ \Rightarrow = {(a - b)}^{2} | \begin{matrix} b - c & b - c & c \\ a - b & a - b & b \\ c - a & c - a & a \end{matrix} | (C_{1} a n d C_{2} a r e i d e n t i c a l c o l u m n s .) \\ \Rightarrow = {(a - b)}^{2} . 0 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The number of distinct real roots of $[\begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix}] = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(A) 0

(B) 2

(C) 1

(D) 3

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix} | = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} 2 c o s x + s i n x & c o s x & c o s x \\ 2 c o s x + s i n x & s i n x & c o s x \\ 2 c o s x + s i n x & c o s x & s i n x \end{matrix} | = 0 (T a k i n g 2 c o s x + s i n x c o m m o n f r o m C_{1}) \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 1 & s i n x & c o s x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 0 & c o s x - s i n x & 0 \\ 0 & s i n x - c o s x & c o s x - s i n x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ \Rightarrow (2 c o s x + s i n x) [1 | \begin{matrix} c o s x - s i n x & 0 \\ s i n x - c o s x & c o s x - s i n x \end{matrix} |] \\ \Rightarrow (2 c o s x + s i n x) {(c o s x - s i n x)}^{2} = 0 \\ 2 c o s x + s i n x = 0 a n d {(c o s x - s i n x)}^{2} = 0 \\ 2 + t a n x = 0 a n d (c o s x - s i n x) = 0 \\ ∴ t a n x = - 2 a n d \Rightarrow t a n x = 1 \\ B u t - \frac{π}{4} \leq x \leq \frac{π}{4} a n d \Rightarrow t a n x = t a n \frac{π}{4} \\ ∴ x = \frac{π}{4} \in [- \frac{π}{4}, \frac{π}{4}] \\ S o, x &thins \end{array}$

Let f(t|) $[\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}],$ then $\underset{t \to 0}{l i m} \frac{f (t)}{t^{2}}$ is equal to

(A) 0

(B) –1

(C) 2

(D) 3

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (t) = [\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}] \\ E x p a n d i n g a l o n g R_{1} \\ = c o s t | \begin{matrix} t & 2 t \\ t & t \end{matrix} | - t | \begin{matrix} 2 s i n t & 2 t \\ s i n t & t \end{matrix} | + 1 | \begin{matrix} 2 s i n t & t \\ s i n t & t \end{matrix} | \\ = c o s t (t^{2} - 2 t^{2}) - t (2 t s i n t - 2 t s i n t) + (2 t s i n t - t s i n t) \\ = t^{2} c o s t + t s i n t \\ ∴ \frac{f (t)}{t^{2}} = \frac{- t^{2} c o s t + t s i n t}{t^{2}} \\ \Rightarrow \frac{f (t)}{t^{2}} = - c o s t + \frac{s i n t}{t^{2}} \\ \Rightarrow \underset{t \to 0}{l i m} \frac{f (t)}{t^{2}} = \underset{t \to 0}{l i m} (- c o s t) + \underset{t \to 0}{l i m} \frac{s i n t}{t^{2}} = - 1 + 1 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The maximum value of = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix}]$ is ( $θ$ is real number)

(A) $\frac{1}{2}$

(B) √3/2

(C) √2

(D) $\frac{2√3}{4}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow = | \begin{matrix} 0 & 0 & 1 \\ - s i n θ & s i n θ & 1 \\ c o s θ & 0 & 1 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow = 1 | \begin{matrix} - s i n θ & s i n θ \\ c o s θ & 0 \end{matrix} | = - s i n θ c o s θ \\ \Rightarrow = - \frac{1}{2} . 2 s i n θ c o s θ = - \frac{1}{2} s i n 2 θ \\ b u t maximum value of sin2 θ =1 \Rightarrow | - \frac{1}{2} . 1 | = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $f (x) = [\begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix}]$ , then

(A) $f (a) = 0$

(B) $f (b) = 0$

(C) $f (0) = 0$

(D) $f (1) = 0$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | \begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix} | \\ f (a) = | \begin{matrix} 0 & 0 & a - b \\ 2 a & 0 & a - c \\ a + b & a + c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = (a - b) | \begin{matrix} 2 a & 0 \\ a + b & a + c \end{matrix} | \\ = (a - b) [2 a (a + c)] = (a - b) . 2 a . (a + c) \neq 0 \\ f (b) = | \begin{matrix} 0 & b - a & 0 \\ b + a & 0 & b - c \\ 2 b & b + c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = - (b - a) | \begin{matrix} b + a & b - c \\ 2 b & 0 \end{matrix} | \\ = - (b - a) [(- 2 b) (b - c)] = 2 b (b - a) (b - c) \neq 0 \\ f (0) = | \begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = a | \begin{matrix} a & - c \\ b & 0 \end{matrix} | - b | \begin{matrix} a & 0 \\ b & c \end{matrix} | \\ = a (b c) - b (a c) = a b c - a b c = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$ , then $A^{- 1}$ exists if

(A) $\begin{matrix} λ \end{matrix}$ $= 2$

(B) $\begin{matrix} λ \end{matrix}$ $\neq 2$

(C) $\begin{matrix} λ \end{matrix}$ $\neq - 2$

(D) None of these

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e, \\ A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}] \Rightarrow | A | = | \begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = 2 | \begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix} | - λ | \begin{matrix} 0 & 5 \\ 1 & 3 \end{matrix} | - 3 | \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} | \\ = 2 (6 - 5) - λ (0 - 5) - 3 (0 - 2) \\ = 2 + 5 λ + 6 = 8 + 5 λ \\ I f A^{- 1} e x i s t s t h e n | A | \neq 0 \\ ∴ 8 + 5 λ \neq 0 S o λ \neq \frac{- 8}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If $A$ and $B$ are invertible matrices, then which of the following is not correct?

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f A a n d B a r e t w o i n v e r t i b l e m a t r i c e s t h e n \\ (a) a d j A = | A | . A^{- 1} i s c o r r e c t \\ (b) d e t {(A)}^{- 1} = {[d e t (A)]}^{- 1} = \frac{1}{d e t (A)} i s c o r r e c t \\ (c) A l s o, {(A B)}^{- 1} = B^{- 1} A^{- 1} i s c o r r e c t \\ (d) {(A + B)}^{- 1} = \frac{1}{| A + B |} . a d j (A + B) \\ ∴ {(A + B)}^{- 1} \neq B^{- 1} + A^{- 1} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If $x, y, z$ are all different from zero and $[\begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix}] = 0,$ then value of

x^–1 + y^–1 + z^–1 is

(A) $x y z$

(B) $x^{- 1} y^{- 1} z^{- 1}$

(C) $- x - y - z$

(D) $- 1$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} | = 0 \\ T a k i n g x, y a n d z c o m m o n f r o m R_{1}, R_{2} a n d R_{3} r e s p e c t i v e l y . \\ \Rightarrow x y z | \begin{matrix} \frac{1}{x} + 1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow x y z | \begin{matrix} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ T a k i n g \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 c o m m o n f r o m R_{1} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) | \begin{matrix} 1 & 1 & 1 \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) | \begin{matrix} 0 & 0 & 1 \\ - 1 & 1 & \frac{1}{y} \\ 0 & - 1 & \frac{1}{z} + 1 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) [1 | \begin{matrix} - 1 & 1 \\ 0 & - 1 \end{matrix} |] = 0 \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) (1) = 0 \\ \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0 a n d x y z \neq 0 (x \neq y \neq z \neq 0) \\ ∴ x^{- 1} + y^{- 1} + z^{- 1} = - 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The value of the determinant $[\begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix}]$

(A) $9 x^{2} (x + y)$

(B) $9 y^{2} (x + y)$

(C) $3 y^{2} (x + y)$

(D) $7 x^{2} (x + y)$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} 3 x + 3 y & x + y & x + 2 y \\ 3 x + 3 y & x & x + y \\ 3 x + 3 y & x + 2 y & x \end{matrix} | \\ = (3 x + 3 y) | \begin{matrix} 1 & x + y & x + 2 y \\ 1 & x & x + y \\ 1 & x + 2 y & x \end{matrix} | [T a k i n g (3 x + 3 y) c o m m o n f r o m C_{1}] \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow = (3 x + 3 y) | \begin{matrix} 0 & y & y \\ 0 & - 2 y & y \\ 1 & x + 2 y & x \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow = 3 (x + y) [1 | \begin{matrix} y & y \\ - 2 y & y \end{matrix} |] \\ \Rightarrow = 3 (x + y) (y^{2} + 2 y^{2}) \Rightarrow 3 (x + y) (3 y^{2}) \Rightarrow 9 y^{2} (x + y) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

There are two values of $a$ which make the determinant $= [\begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix}] = 86,$ then the sum of these numbers is

(A) 4

(B) 5

(C) -4

(D) 9

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: