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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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:

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If then verify that:

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

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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

then verify that:

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

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

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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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