RD Sharma Solutions Class 12 Mathematics Chapter 6 VSA

# RD Sharma Solutions Class 12 Mathematics Chapter 6 VSA

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 02:31 PM IST

RD Sharma provides the best material for maths for CBSE students. They are exam-oriented, extremely detailed, and contain many solved examples. They are preferred even more than NCERT materials as they include more concepts. A majority of CBSE schools in the country use RD Sharma books for maths. This means that the solutions that students will learn from the material might also appear in their exams.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## RD Sharma Class 12 Solutions Chapter6 VSA Adjoint & Inverse of Matrix - Other Exercise

Adjoint and inverse of a Matrix Exercise very short answer type Question 1

$adj A=\left[\begin{array}{rr} -2 & -4 \\ -7 & -3 \end{array}\right]$
Given:
$A=\left[\begin{array}{rr} -3 & 4 \\ 7 & -2 \end{array}\right]$
Hint ; Find the cofactors of the matrix A and then write adj A
Solution ;
$A=\left[\begin{array}{rr} -3 & 4 \\ 7 & -2 \end{array}\right]$
$c_{11}=-2$ $c_{12}=-4$ $c_{21}=-7$ $c_{22}=-3$
$adj A=\left[\begin{array}{rr} -2 & -4 \\ -7 & -3 \end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 2

$Answer: 5$
$Given : A \text{ is a square matrix and }A( adj A)=5 I$
$Hint : \text{you must know the formula of }A( adj A) \text{ to find the answer}$
$Solution:$
$A(a d j A)=5 I-----(1)$
We know that
$A(\operatorname{adj} A)=|A| I-------(2)$
$\text{on comparing (1) and }(2)$
$|A| I=5 I$
$|A|=5$

Adjoint and inverse of a Matrix Exercise very short answer type Question 3

Given : A is a square matrix of order 3 and $\inline \left | A \right |=5$
Hint : use the formula of $\inline adj\left ( A \right )$
solution :
$\inline n=3$ & $\inline \left | A \right |=5$
We know that
\inline \begin{aligned} &|a d j A|=|A|^{n-1} \\ &=(5)^{3-1} \quad(n=3 \&|A|=5) \\ &=5^{2} \\ &|a d j A|=25 \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 5

Answer : $\frac{1}{10}$
Given : A is non-singular square matrix and $\left | A \right |=10$
you must know about the concept of non-singular matrix
Solution : We know that
$\left | A \right |^{-1}=\frac{1}{\left | A \right |}$
$\left | A \right |^{-1}=\frac{1}{10} \ \ \ \ \ \ \ \because \left ( \left | A \right |=10 \right)$

Adjoint and inverse of a Matrix Exercise very short answer type Question 6

Answer: A must be invertible and $\left | A \right |\neq 0$

Given: A, B and C are non null square matrix of same order & AB=AC implies B = C

Hint : you must know about invertible and non-invertible matrix

Solution : AB=AC can be write only when A is invertible$\left | A \right |\neq 0$

$|A| \neq 0$

$A B=A C$

Multiply both sides by $A^{-1}$

$A^{-1} A B=A^{-1} A C$

$I B=I C$

$B=C$

If Ais not invertible then $A^{-1}$ does `nt exist then,

$AB=AC$

$B=C$

Hence, the condition of A is invertible or $\left | A \right |\neq 0$

Adjoint and inverse of a Matrix Exercise very short answer type Question 7

Answer: $\left(A^{T}\right)^{-1}=\left[\begin{array}{ll} 5 & -2 \\ 3 & -1 \end{array}\right]$

Given: A is non singular square matrix and $\left(A\right)^{-1}=\left[\begin{array}{ll} 5 & 3 \\ -2 & -1 \end{array}\right]$

Hint : you must know the formula $\left(A^{T}\right)^{-1}$

Solution : We know that

$\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$

$\left(A^{T}\right)^{-1}=\left[\begin{array}{ll} 5 & 3 \\ -2 & -1 \end{array}\right]^{T}$

$\left(A^{T}\right)^{-1}=\left[\begin{array}{ll} 5 & -2 \\ 3 & -1 \end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 8

Answer : $adj\left ( AB \right )= adjB.adjA$
Given :
Hint : use the formula $adj\left ( AB \right )= adjB.adjA$
Solution : $adj\left ( AB \right )= adjB.adjA$
\begin{aligned} &=\left[\begin{array}{cc} 1 & -2 \\ -3 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 4 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} 2-8 & 3+2 \\ -6+4 & -9-1 \end{array}\right] \end{aligned}
$\left ( adjAB \right )=\left[\begin{array}{cc} -6 & 5 \\ -2 & -10 \end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 9

Given : A is a symmetric matrix
Hint : A is a symmetric matrix when
$A=A^{T}$
$A^{T}$ is symmetric when $A^{T}=\left ( A^{T} \right )^{T}$
So, we do transpose of $A^{T}$
$\left ( A^{T} \right )^{T}$
$=A^{TT}$
$=A$
$=A^{T}$ $\left ( A=A^{T} \right )$
Hence, $A^{T}$ is a symmetric matrix
$\text { [e.g. } \left.\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]^{\top}=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 10

Answer : $2A$
Given : $n=3$ & $\left | A \right |=2$
Hint : use the formula of $adj\left ( adjA \right )$
Solution : We know that
$adj\left ( adjA \right )=\left | A \right |^{n-2}A$
$=\left ( 2 \right )^{3-2}A\left ( n=3,\left | a \right |=2 \right )$
$adj\left ( adjA \right )=2A$

Adjoint and inverse of a Matrix Exercise very short answer type Question 11

Given : A is square matrix of order 3 such that $\left | A \right |=3$
Hint : use the formula of $adj\left ( adjA \right )$
Solution : $\left | A \right |=3$
$\left | adj\left ( adjA \right ) \right |=\left | A \right |^{\left ( n-1 \right )^{2}}$
$=3^{\left ( 3-1 \right )^{2}}$ (n=3) & $\inline \left ( \left | A \right |=3 \right )$
$\inline =3^{4}$
$\inline \left |adj\left ( adjA \right ) \right |=81$

Adjoint and inverse of a Matrix Exercise very short answer type Question 12

Given : adj(2A) = K adj(A) and A is a square matrix of order 3
Hint : you must know about the concept of square matrix
Solution : We know that
\inline \begin{aligned} &\operatorname{adj}(K A)=K^{n-1} \operatorname{adj}(A) \\ &\operatorname{adj}(2 A)=2^{3-1} \operatorname{adj}(A) \quad(n=3) \\ &\operatorname{adj}(2 A)=4 a d j A \\ &K=4 \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 13

Given : A is a square matrix
Hint : use the formula of $adj\left ( A^{T} \right )$ & $adj\left ( A \right )^{T}$
Solution : We know that
\begin{aligned} &\operatorname{adj}\left(A^{T}\right)=\left|A^{T}\right|\left(A^{T}\right)^{-1}\\ &\operatorname{adj}(A)^{T}=\left(|A| A^{-1}\right)^{T}\\ &\text { Now, we have }\\ &\operatorname{adj}\left(A^{T}\right)-\operatorname{adj}(A)^{T}\\ &=\left|A^{T}\right|\left(A^{T}\right)^{-1}-\left[|A| A^{-1}\right]^{T}\\ &=0 \quad \end{aligned} $\left[\because A^{\top} \cdot A^{-\top}=1\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 14

Given : n=3 and $A\left ( adjA \right )=2I$
Hint : use the formula of $\left | adjA \right |$
Solution : We know that
\begin{aligned} &A(\operatorname{adj} A)=|A| I \\ &A(\operatorname{adj} A)=2 I \\ &|A|=2 \\ &\text { Now, }|\operatorname{adj} A|=|A|^{n-1} \\ &=(2)^{3-1} \\ &=(2)^{2} \\ &=4 \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 15

Given : $A\neq 0$ & $A^{T}=A$
Hint : use the property of $\left ( A^{T} \right )^{-1}=\left ( A^{-1} \right )^{T}$
Solution : $\left ( A^{T} \right )^{-1}=\left ( A^{-1} \right )^{T}$
$\left ( A^{-1} \right )^{T}=A^{-1}\left ( A^{T}=A \right )$
$\left ( A^{-1} \right )^{T}=A^{-1}$
Hence,$A^{-1}$ is a symmetric

Adjoint and inverse of a Matrix Exercise very short answer type Question 16

Given : $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \& A(\operatorname{adj} A)=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right]$
Hint : Find $A^{-1}$ and then multiply both side by A
Solution : $A\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
\begin{aligned} &|A|=\cos ^{2} \theta+\sin ^{2} \theta \\ &=1 \\ &|A| \neq 0 \\ &A^{-1} \text { exists } \\ &A^{-1}=\frac{a d j A}{|A|} \\ &A^{-1}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ &A^{-1}=a d j A \end{aligned}
multiply both side by A
\begin{aligned} &A A^{-1}=A \operatorname{adj} A \\ &I=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right] \\ &{\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right]} \\ &K=1(\text { \text{On comparing }}) \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 17

Answer : Given : A is an invertible matrix and $\left | A^{-1} \right |=2$
Hint : use the relation between $\left | A^{-1} \right |$ and $\left | A \right |$ of find the answer
Solution : We know that
\begin{aligned} &\left|A^{-1}\right|=\frac{1}{|A|} \\ &2=\frac{1}{|A|}\left(\left|A^{-1}\right|=2\right) \\ &|A|=\frac{1}{2} \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 18

Given: $A(\operatorname{adj} A)=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]$
Hint : Use the formula of A$A\left ( adjA \right )=\left | A \right |I$
Solution : $A(\operatorname{adj} A)=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]$
\begin{aligned} &A(\operatorname{adj} A)=5 I \\ &|A|=5 \\ &N o w|\operatorname{adj} A|=|A|^{n-1} \\ &=(5)^{3-1}(n=3,|A|=5) \\ &=25 \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 19

Answer : $\frac{1}{19}$
Given : $A=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right] \& A^{-1}=K A$
Hint : Find $A^{-1}$ and put it in equation and then compare it.
Solution : $A=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$
\begin{aligned} &|A|=-4-15 \\ &=-19 \\ &|A| \neq 0 \\ &A^{-1} \text { exists } \\ &A^{-1}=\frac{\operatorname{adj} A}{|A|} \end{aligned}
\begin{aligned} &=\frac{-1}{19}\left[\begin{array}{cc} -2 & -3 \\ -5 & 2 \end{array}\right]\\ &=\frac{-1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\text { Now } \mathrm{A}^{-1}=K A\\ &\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=K\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\text { On comparing }\\ &\mathrm{K}=\frac{1}{19} \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 20

Answer : $A^{-1}=I-A$
Given : $A^{2}-A+I=0$
Hint : Use pre-multiplication technique
Solution : $A^{2}-A+I=0$
$A^{2}-A=-I$
pre-multiplication by $A^{-1}$ on both sides
\begin{aligned} &A\left(A A^{-1}\right)-A A^{-1}=-I A^{-1} \\ &A-I=-A^{-1} \\ &A^{-1}=I-A \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 21

Given : $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right]$
Hint : you must know about how to calculate cofactor of an element
Solution : $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right]$
\begin{aligned} &c_{32}=(-1)^{3+2}(8-30) \\ &=-1(-22) \\ &=22 \\ &a_{32}=5 \\ &a_{32} c_{32}=22(5) \\ &=110 \end{aligned}

### Adjoint and inverse of a Matrix Exercise very short answer type Question 22

Answer : $\left[\begin{array}{cc} 5 & 2 \\ 7 & 3 \end{array}\right]$
Given : $\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]$
Hint : you must know about how to find inverse of a matrix
Solution : $\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]$
\begin{aligned} &c_{11}=5, c_{12}=7, c_{21}=2, c_{22}=3 \\ &=\frac{1}{1}\left[\begin{array}{cc} 5 & 7 \\ 2 & 3 \end{array}\right] \\ &=\left[\begin{array}{cc} 5 & 7 \\ 2 & 3 \end{array}\right] \end{aligned}
Now transpose it, we get

Adjoint and inverse of a Matrix Exercise very short answer type Question 23

Answer : $\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
Given : $\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Hint : Find the cofactor and then transpose it
Solution : $\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
\begin{aligned} &c_{11}=\cos \theta, c_{12}=\sin \theta, c_{21}=-\sin \theta, c_{22}=\cos \theta \\ &|A|=\cos ^{2} \theta-\left(-\sin ^{2} \theta\right) \\ &=\cos ^{2} \theta+\sin ^{2} \theta \quad \end{aligned} $\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right] \\$
$=1$
$A^{-1}=\frac{A d j|A|}{|A|}=\frac{1}{1}$$\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Now transpose it, we get
$=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

### Adjoint and inverse of a Matrix Exercise very short answer type Question 24

Answer : $\left[\begin{array}{cc} 0 & 3 \\ -2 & 1 \end{array}\right]$
Given : $\left[\begin{array}{cc} 1 & -3 \\ 2 & 0 \end{array}\right]$
Hint : find the cofactors to find adj A
solution :
$A=\left[\begin{array}{cc} 0 & -2 \\ 3 & 1 \end{array}\right]$
$=\left[\begin{array}{cc} 0 & 3 \\ -2 & 1 \end{array}\right]$ $[ \text{co-factors of element A are, }\left.C_{11}=0, C_{12}=-2, C_{21}=3, C_{22}=1\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 25

Answer : $\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
Given : $A=\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ &B=$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Hint : find adj A and find adj B and then use the formula of adj (AB)
Solution : $A=\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$
$adj A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$B=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$adj B=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Now we know that
$=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$adj\left ( AB \right )=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 26

Anawer : -11
Given : $A=\left[\begin{array}{cc}3 & 1 \\ 2 & -3\end{array}\right]$
Hint : find $\left | A \right |$ and then use the formula of $\left | adjA \right |$
Solution : $A=\left[\begin{array}{cc}3 & 1 \\ 2 & -3\end{array}\right]$
$\left | A \right |=-9-2$
=-11
Now ,$\left | adjA \right |=\left | A \right |^{n-1}$
$=\left ( -11 \right )^{2-1}$
$=-11$

Adjoint and inverse of a Matrix Exercise very short answer type Question 27

Answer : $A^{-1}=\frac{1}{19}A$
Given : $A.=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$
Hint : to find $A^{-1^{}}$ ,find adj A & $\left | A \right |$
Solution : $A.=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$
$\left | A \right |=-4-15$
$=-19$
$\left | A \right |\neq 0$
So, $A^{-1}$exists
$adjA=\left[\begin{array}{cc} -2 & -5 \\ -3 & 2 \end{array}\right]^{T}$
$=\left[\begin{array}{cc} -2 & -3 \\ -5 & -2 \end{array}\right]$
Now $A^{-1}=\frac{adjA}{\left | A \right |}$
$A^{-1}=-\frac{1}{19}$ $\left[\begin{array}{cc} -2 & -3 \\ -5 & -2 \end{array}\right]$
$A^{-1}=\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$
$A^{-1}=\frac{1}{19}A$

Adjoint and inverse of a Matrix Exercise very short answer type Question 28

Answer : $A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$
Given : $A=\left[\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}\right]$
Hint : find adj A and $\left | A \right |$ and then put in $A^{-1}$ formula
Solution : $A=\left[\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}\right]$
$\left | A \right |=6-5$
$=1$
$\left | A \right |$$\neq 0$
So ,$A^{-1}$ exists
$adjA=\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]^{T}$
$=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$
$A^{-1}=\frac{adjA}{\left | A \right |}$
$A^{-1}=\frac{1}{1}\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$
$A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 29

Answer : $\left[\begin{array}{ll} 2 & 5 \\ 5 & 4 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]$
Given : $\left[\begin{array}{ll} 2 & 1 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right]$
and column operation $C_{2} \rightarrow C_{2}+2 C_{1} \\$
Hint : you must know about the concept of elementary transform
Solution : The Matrix is in the form $X=AB$
Elementary opration can be apply only on X and B
\begin{aligned} &\text { So },\left[\begin{array}{ll} 2 & 1 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right] \\ &C_{2} \rightarrow C_{2}+2 C_{1} \\ &{\left[\begin{array}{ll} 2 & 1+4 \\ 2 & 0+4 \end{array}\right]=\left[\begin{array}{ll} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0+2 \\ -1 & 1-2 \end{array}\right]} \\ &{\left[\begin{array}{ll} 2 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 3 & 1 \\ 2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]} \end{aligned}

Adjoint and inverse of a Matrix Exercise very short answer type Question 30

Answer : $\left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]$
Given : $\left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]$
and elementary operation $R_{2}\rightarrow R_{2}+R_{1}$
Hint : In multiplication row operation is equivalent to left multiplication
Solution :
$\left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]$

By applying operation $R_{2}\rightarrow R_{2}+R_{1}$
$\left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]$

Adjoint and inverse of a Matrix Exercise very short answer type Question 31

Given $\left | A \right |=4$
Hint : use the concept of A adj A
Solution : $\left | A adj A\right |=?$
\begin{aligned} &A(\operatorname{adj} A)=|A| I \\ &A(\operatorname{adj} A)=4 I \quad[|A|=4] \\ &|A(\operatorname{adj} A)|=|4 I| \\ &|A(\operatorname{adj} A)|=4\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &=4 \times 1 \\ &=4 \\ &|A(\operatorname{adj} A)|=4 \end{aligned}

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## RD Sharma Chapter wise Solutions

1. Does this material contain important questions?

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4. Will I be able to solve NCERT questions if I prepare from this material?

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