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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

Sum of two skew symmetric matrices is always ___ matrix.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

___ matrix is both symmetric and skew symmetric matrix.

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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) $m \times n$

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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