RD Sharma Solutions Class 12 Mathematics Chapter 6 VSA

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

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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

Answer :
$adj A=\left[\begin{array}{rr} -2 & -4 \\ -7 & -3 \end{array}\right]$
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

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

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

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

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

Answer : symmetric matrix
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

Answer : 81
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) & $\left ( \left | A \right |=3 \right )$
$=3^{4}$
$\left |adj\left ( adjA \right ) \right |=81$

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

Answer : 4
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
$\begin{aligned} &\operatorname{adj}(K A)=K^{n-1} \operatorname{adj}(A) \\ &\operatorname{adj}(2 A)=2^{3-1} \operatorname{adj}(A) \quad(n=3) \\ &\operatorname{adj}(2 A)=4 a d j A \\ &K=4 \end{aligned}$

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

Answer : null matrix
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

Answer : 4
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

Answer : symmetric
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

Answer : 1
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

Answer : 25
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

Answer : 110
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
adj(AB) =adj(B)adj(A)
$=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$adj\left ( AB \right )=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$

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

Answer : 4
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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