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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

Find the matrix $A$ such that

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Payal Gupta

Contributor-Level 10

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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.

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

IIIT Nagpur - Indian Institute of Information Technology B.Tech

How many semesters are there in B.Tech at IIIT Nagpur?

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Anushka Bidhi

Contributor-Level 10

The students seeking enrollment in BTech at the IIIT Nagpur should check the eligibility criteria for the program to make sure to meet the requirements. The duration of BTech in IIIT Nagpur is four years which is further divided into eight semesters.

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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 $w$ .

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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

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

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

University of Luxembourg Study Abroad

What are the semesters for the University of Luxembourg Engineering courses?

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Kanika Pandey

Contributor-Level 9

The University of Luxembourg offers only one intake, which is the Winter intake for both bachelor's and Master's programs. Typically, the application period begins in March for UG and in February for PG. For most of the Engineering programs, a Summer semester is unavailable.

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

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

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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

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

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Payal Gupta

Contributor-Level 10

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

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Payal Gupta

Contributor-Level 10

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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9 months 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:

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Payal Gupta

Contributor-Level 10

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

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

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

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

New Question 9 months 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 ${(A + B)}^{2} = A^{2} + 2 A B + B^{2}$ .

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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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9 months 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

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Payal Gupta

Contributor-Level 10

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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9 months ago

University of Luxembourg Engineering Eligibility

What are some of the eligibility requirements to apply for the University of Luxembourg Engineering courses scholarhsips?

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Rachit Arora

Contributor-Level 10

The scholarships for international students at the University of Luxembourg are quite limited, and they are awarded on the basis of merit. Below are some important scholarships for international students.

Guillaume dupaix international master's scholarship eligibility requirements:
- International applicants from both EU and non-EU countries applying for a Master's degree
- Only available for Master's programme that runs their two-year programme in Luxembourg. Master programme that have a mandatory semester or year abroad are not eligible.
- Excellent academic performance during undergraduate studies
- Fulltime students
- Non-beneficiaries

...more

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9 months 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 $A$ .

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Payal Gupta

Contributor-Level 10

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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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If then verify that:

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

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

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Payal Gupta

Contributor-Level 10

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

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9 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

then verify that:

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

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

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

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Payal Gupta

Contributor-Level 10

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

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9 months ago

Accounting & Commerce Science

Is there any possibility to get into any central university for BCom? I have 595 marks out of 1250 in CUET as a general category candidate.

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Trisha Priyadarsini

Beginner-Level 1

You've scored 596 out of 1250 in the CUET UG 2025 — that puts you at roughly 47.68%. Now, on paper, that doesn't sound super strong, especially since most central universities, particularly the well-known ones like DU, BHU, JNU, or JMI, tend to have high cutoffs for B.Com, often upwards of 600–700 (sometimes even higher depending on the category and competition that year).

But here's the real deal:

It's not impossible. It just depends on a few key things:
Your category (General/OBC/SC/ST/EWS)
Which universities you're targeting?
What the cutoff trends are this year (they can shift!)
How many students applied for B.Com and wh

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