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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

Kindly Consider the following find A.

12 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

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 $P Q = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] = Q P$ .

5 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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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} 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}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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

4 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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

Solve for x and y :

Kindly Consider the following

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

5 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

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

4 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

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