RD Sharma Class 12 Exercise 6.1 Adjoint and inverse of Matrix Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 6.1 Adjoint and inverse of Matrix Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 02:54 PM IST

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

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (i)

$Adj \left ( A \right )=\left[\begin{array}{cc} 4 & -2 \\ -5 & -3 \end{array}\right]$
Hint:
Here, we use basic concept of adjoint of matrix .
Given:
$Adj A=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]$
Solution:
$A=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]$
\begin{aligned} &|A|=-3 \times 4-5 \times 2 \\ &=-12-10=-22 \\ &C_{11}=4, C_{12}=-2, C_{21}=-5, C_{22}=-3 \end{aligned}
$C_{i j}=\left[\begin{array}{cc} 4 & -2 \\ -5 & -3 \end{array}\right]$
Taking transpose,
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\\ \\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right] \\ \\&=\left[\begin{array}{cc} -12+(-10) & 20-20 \\ 6-6 & -10-12 \end{array}\right] \end{aligned}
$=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]$ (1)
$|A| I=-22\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]$ (2)
\begin{aligned} &A(\operatorname{Adj}(A))=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]\\ \end{aligned}
$=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right]$ (3)
From equation (1), (2) and (3)
$A(\operatorname{Adj}(A))=|A| I=\operatorname{Adj}(A) \times A$

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (ii)

$\operatorname{Adj}(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
$A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$
Solution:
Let
\begin{aligned} &A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\\\ &|A|=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right| \\\\ &|A|=a \times d-c \times b \\\\ &|A|=a d-c b \end{aligned}
Let’s find cofactor
\begin{aligned} &C_{11}=d, C_{12}=-c, C_{21}=-b, C_{22}=a \\ &C_{i j}=\left[\begin{array}{cc} d & -c \\ -b & a \end{array}\right] \end{aligned}
Let’s transpose $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$
Let’s prove this below
$\operatorname{Adj}(A) \times A=|A| I=A \times A d j(A)$
\begin{aligned} &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \times\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \\\\ &=\left[\begin{array}{cc} d a+(-b c) & d b-b d \\ -c a+a c & -c b+a d \end{array}\right] \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} d a-b c & 0 \\ 0 & d a-b c \end{array}\right] \end{aligned} (1)
MHS
\begin{aligned} &|A| I=a d-c b\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \\&=\left[\begin{array}{cc} a d-c b & 0 \\ 0 & a d-c b \end{array}\right] \\\\ &=\left[\begin{array}{cc} d a-b c & 0 \\ 0 & a d-c b \end{array}\right] \end{aligned} (2)
RHS
$A \times A d j(A)=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \times\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$
$=\left[\begin{array}{cc} a d-b c & 0 \\ 0 & d a-b c \end{array}\right]$ (3)
From equation (1), (2) and (3)
$\operatorname{Adj}(A) \times A=|A| I=A \times A d j(A)$

Adjoint and Inverse of a Matrix Exercise 6.1 Question 1 (iii)

$\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$
Hint:
Here, we use basic concept of determinant.
Given:
$A= \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
$|A|=\cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$
Let’s find cofactor
\begin{aligned} &C_{11}=\cos \alpha, C_{12}=-\sin \alpha, C_{21}=-\sin \alpha, C_{22}=\cos \alpha \\ &C_{i j}=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \end{aligned}
Let’s transpose $C_{ij}$
$Adj(A)= \left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$
Let’s prove below
\begin{aligned} &\operatorname{Adj}(A) \times A=|A| \times A=A \times \operatorname{Adj}(A) \\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \times\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \end{aligned}
$=\left[\begin{array}{cc} \cos ^{2} \alpha-\sin ^{2} \alpha & \cos \alpha \sin \alpha-\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \cos ^{2} \alpha-\sin ^{2} \alpha \end{array}\right]$
$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$ (1)
$\left | A \right |I= \cos2 \alpha\times \left[\begin{array}{cc} 1 &0 \\ 0 & 1\end{array}\right]$
$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$ (2)
\begin{aligned} &A \times \operatorname{Adj}(A)=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \times\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \\ &=\left[\begin{array}{cc} \cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha \end{array}\right] \end{aligned}
$= \left[\begin{array}{cc} \cos 2\alpha &0 \\0 & \cos2 \alpha \end{array}\right]$ (3)
From equation (1), (2) and (3)
$A \times \operatorname{Adj}(A)=|A| I=\operatorname{Adj}(A) \times A$

Adjoint and Inverse of a Matrix exercise 6.1 question 1 (iv)

$Adj\left ( A \right )=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$
Hint:
Here, we use basic concept of determinant.
Given:
$A=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$
Solution:
Let’s find $|A|$
$|A|=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$
\begin{aligned} &|A|=1-\left[\left(-\tan \frac{\alpha}{2}\right)\left(\tan \frac{\alpha}{2}\right)\right] \\ &|A|=1+\tan ^{2} \frac{\alpha}{2} \\ &|A|=\sec ^{2} \frac{\alpha}{2} \end{aligned}
Let’s find cofactor
$C_{11}=1, C_{12}=\tan \frac{\alpha}{2}, C_{21}=-\tan \frac{\alpha}{2}, C_{22}=1$
$C_{ij}=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]$
Take transpose of $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]$
Let’s prove below
\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{cc} 1+\tan ^{2} \frac{\alpha}{2} & \tan \frac{\alpha}{2}-\tan \frac{\alpha}{2}\\ \\ \tan \frac{\alpha}{2}-\tan \frac{\alpha}{2} & 1+\tan ^{2} \frac{\alpha}{2} \end{array}\right] \\ \\&=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned} (1)
\begin{aligned} &|A| I=\sec ^{2} \frac{\alpha}{2} \times\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned} (2)
$A \times A d j(A)=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]$
\begin{aligned} &=\left[\begin{array}{cc} 1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2} \end{array}\right] \\\\ &=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned} (3)
From equation (1), (2) and 3
$\operatorname{Adj}(A) \times A=|A| \times I=A \times \operatorname{Adj}(A)$
Hence proved

Adjoint and Inverse of a Matrix exercise 6.1 question 2 (i)

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right]$
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} 1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
\begin{aligned} &|A|=1\left|\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right|-2\left|\begin{array}{ll} 2 & 2 \\ 2 & 1 \end{array}\right|+2\left|\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right| \\\\ &=1(1-4)-2(2-4)+2(4-2) \\\\ &=(-3)+4+4 \\ \\&=5 \end{aligned}
Let’s find cofactors
\begin{aligned} &C_{11}=+(1-4)=-3 \\\\ &C_{12}=-(2-4)=2 \\\\ &C_{13}=+(4-2)=2 \\\\ &C_{21}=-(2-4)=2 \\\\ &C_{22}=+(1-4)=-3 \end{aligned}

\begin{aligned} &C_{23}=-(2-4)=2 \\\\ &C_{31}=+(4-2)=2 \\\\ &C_{32}=-(2-4)=2 \\\\ &C_{33}=+(1-4)=-3 \\\\ &C_{i j}=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \end{aligned}
Let’s transpose it
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right]$
Let’s verify below
\begin{aligned} \operatorname{Adj}(A) \times A &=|A| I=A \times \operatorname{Adj}(A) \\\\ \operatorname{Adj}(A) \times A &=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \times\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \end{aligned}
$=\left[\begin{array}{ccc} -3+4+4 & 0 & 0 \\ 0 & 4-3+4 & 0 \\ 0 & 0 & 4+4-3 \end{array}\right]$
$=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]$ (1)
$|A| I=5 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]$ (2)
\begin{aligned} &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \times\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} -3+4+4 & 0 & 0 \\ 0 & -3+4+4 & 0 \\ 0 & 0 & -3+4+4 \end{array}\right] \end{aligned}
$=\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]$ (3)
From equation (1), (2) and (3)
$A \times A d j(A)=|A| I=\operatorname{Adj}(A) \times A$

Adjoint and Inverse of a Matrix exercise 6.1 question 2 (ii)

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \end{array}\right]$
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
\begin{aligned} &|A|=1\left|\begin{array}{ll} 3 & 1 \\ 1 & 1 \end{array}\right|-2\left|\begin{array}{cc} 2 & 1 \\ -1 & 1 \end{array}\right|+5\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right| \\\\ &=1(3-1)-2(2+1)+5(2+3) \\\\ &=2-6+25=21 \end{aligned}
Let’s find cofactors
\begin{aligned} &C_{11}=+(3-1)=2 \\ &C_{12}=-(2+1)=-3 \\ &C_{13}=+(2+3)=5 \\ &C_{21}=-(2-5)=3 \\ &C_{22}=+(1+6)=6 \end{aligned}
\begin{aligned} &C_{23}=-(1+2)=-3 \\ &C_{31}=+(2-15)=-13 \\ &C_{32}=-(1-10)=9 \\ &C_{33}=+(3-4)=-1 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 6 & -3 \\ -13 & 9 & -1 \end{array}\right]$
Let’s take transpose of $C_{i j}$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \end{array}\right]$
Let’s verify below
\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right] \end{aligned} (1)
$|A| I=21 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right]$ (2)
$A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & 1 \end{array}\right]=\left[\begin{array}{ccc} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{array}\right]$ (3)
From equation (1), (2) and (3)
$\operatorname{Adj}(A) \times A=|A| I=A \times A d j(A)$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
$A=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right|$
Solution:
Let’s find $|A|$
$|A|=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right|$
\begin{aligned} &|A|=2\left|\begin{array}{cc} 2 & 5 \\ 4 & -1 \end{array}\right|-(-1)\left|\begin{array}{cc} 4 & 5 \\ 0 & -1 \end{array}\right|+3\left|\begin{array}{ll} 4 & 2 \\ 0 & 4 \end{array}\right| \\ &=2(-2-20)+1(-4-0)+3(16-0) \\ &=-44-4+48 \\ &=0 \\ &|A|=0 \end{aligned}
Let’s find cofactor
\begin{aligned} &C_{11}=+(-2-20)=-22 \\ &C_{12}=-(-4+0)=4 \\ &C_{13}=+(16-0)=16 \\ &C_{21}=-(1-12)=11 \\ &C_{22}=+(-2-0)=-2 \\ &C_{23}=-(-8-0)=-8 \\ &C_{31}=+(-5-6)=-11 \\ &C_{32}=-(10-12)=2 \\ &C_{33}=+(4+4)=8 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} -22 & 4 & 16 \\ 11 & -2 & -8 \\ -11 & 2 & 8 \end{array}\right]$
Let’s take transpose of $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$
Let’s prove this below
\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right] \times\left[\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right] \end{aligned}
$=\left[\begin{array}{ccc} -44+44+0 & 0 & 0 \\ 0 & -4+(-4)+8 & 0 \\ 0 & 0 & 48-40-8 \end{array}\right]$
$= \left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ (1)
$|A| I=0 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ (2)
$A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right] \times\left[\begin{array}{ccc} -22 & 4 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ (3)
So, from equation (1), (2) and (3)
$\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A)$

Adjoint and Inverse of Matrices Excercise 6.1 Question 2 (iv).

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
\begin{aligned} &|A|=\left|\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right| \\ &=2\left|\begin{array}{ll} 1 & 0 \\ 1 & 3 \end{array}\right|-0\left|\begin{array}{ll} 5 & 0 \\ 1 & 3 \end{array}\right|+(-1)\left|\begin{array}{ll} 5 & 1 \\ 1 & 1 \end{array}\right| \\ &=2(3)-0-1(5-1) \\ &=6-4=2 \\ &|A|=2 \end{aligned}
Let’s find cofactor
\begin{aligned} &C_{11}=+(3-0)=3 \\ &C_{12}=-(15-0)=-15 \\ &C_{13}=+(5-1)=4 \\ &C_{21}=-(0+1)=-1 \\ &C_{22}=+(6+1)=7 \\ &C_{23}=-(-2-0)=-2 \\ &C_{31}=+(2-1)=1 \\ &C_{32}=-(0+5)=-5 \\ &C_{33}=+(2-0)=2 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} 3 & -15 & 4 \\ -1 & 7 & -2 \\ 1 & -5 & 2 \end{array}\right]$
Let’s take the transpose of $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]$
Let’s prove,
\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 6+(-5)+1 & 0 & 0 \\ 0 & 0+7-5 & 0 \\ 0 & 0 & -4+0+6 \end{array}\right] \end{aligned}
$=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$ (1)
$|A| I=2 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$ (2)
$A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right] \times\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$ (3)
From the equation (1), (2) and (3)
$\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A)$

Adjoint and Inverese of Matrices Exercise 6.1 Question 3

Proved $A(\operatorname{Adj}(A))=0$
Hint:
Here, we have to use advance method of finding adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{array}\right]$
Solution:
We know that,
$A \times \operatorname{Adj}(A)=|A| I$ Then,
$\left | A \right |I= 0$ (1)
From the equation (1)
$\left | A \right |I= 0$
$\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{array}\right| \times 1=0$ since, $|I| =1$
$1\left|\begin{array}{rr} 3 & 0 \\ 2 & 10 \end{array}\right|-(-1)\left|\begin{array}{cc} 2 & 0 \\ 18 & 10 \end{array}\right|+1\left|\begin{array}{cc} 2 & 3 \\ 18 & 2 \end{array}\right|=0$
\begin{aligned} &0=1(30)+(20-0)+(4-54) \\ &0=30+20+4-54 \\ &0=54-54 \\ &0=0 \end{aligned}
So, $A \times \operatorname{Adj}(A)=|A| I=0$

Adjoint and Inverese of Matrices Exercise 6.1 Question 4

Proved $Adj\left ( A \right )= A$
Hint:
Here, we use basic concept of adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{array}\right]$
Solution:
Here, let’s find cofactor
\begin{aligned} &C_{11}=+(0-4)=-4 \\ &C_{12}=-(-1)=1 \\ &C_{13}=+(4-0)=4 \\ &C_{21}=-(-3)=3 \\ &C_{22}=+(0)=0 \\ &C_{23}=-(-4)=4 \\ &C_{31}=+(-3-0)=-3 \\ &C_{32}=-(-1)=1 \\ &C_{33}=+(3)=3 \end{aligned}
So,
$C_{i j}=\left[\begin{array}{ccc} -4 & 1 & 4 \\ 3 & 0 & 4 \\ -3 & 1 & 3 \end{array}\right]$
Let’s find $Adj \left ( A \right )$
Take the transpose of $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -4 & 3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{array}\right]$
Hence we clearly see that $Adj \left ( A \right )=A$

Adjoint and Inverese of Matrices Exercise 6.1 Question 5

$\operatorname{Adj}(A)=3 A^{T}$
Hint:
Here, we use basic concept of determinant and adjoint of matrix.
Given:
$A=\left[\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right]$
Solution:
$A=\left[\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right]$
Let’s find cofactors $C_{ij}=\left ( -1 \right )^{i+j}$
\begin{aligned} &C_{11}=+(1-4)=-3 \\ &C_{12}=-(2+4)=-6 \\ &C_{13}=+(-4-2)=-6 \\ &C_{21}=-(-2-4)=6 \\ &C_{22}=+(-1+4)=3 \\ &C_{23}=-(2+4)=-6 \\ &C_{31}=+(4+2)=6 \\ &C_{32}=-(2+4)=-6 \\ &C_{33}=+(-1+4)=3 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{array}\right]$
So,
$Adj\left ( A \right )$ = Transpose of $C_{ij}$
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{array}\right]$ (1)
$3 A^{T}=3 \times\left[\begin{array}{ccc} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{array}\right]$ (2)
Here, from equation (1) and (2)
Clearly see that,
$3 A^{T}=A d j(A)$
Hence, proved

Adjoint and Inverse Matrix Exercise 6.1 Question 6

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$
Hint:
Here, we use basic concept of determinant.
Given:
$A=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{array}\right]$
Solution:
Let’s find cofactor of A
\begin{aligned} &C_{11}=9, C_{21}=19, C_{31}=-4 \\ &C_{12}=4, C_{22}=14, C_{32}=1 \\ &C_{13}=8, C_{23}=3, C_{33}=2 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{array}\right]$
$Adj(A)$ = Transpose of $C_{ij}$
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \\ &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{array}\right] \times\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \end{aligned}
$=\left[\begin{array}{ccc} 9-8+24 & 19-28-9 & -4-2+6 \\ 0+8-8 & 0+28-3 & 0+2-2 \\ -36+20+16 & -76+70+6 & 16+5+4 \end{array}\right]=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$
Hence, $A \times \operatorname{Adj}(A)=25 I$

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (i)

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$
Hint:
Here, we use basic concept of determinant.
Given:
$A=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{array}\right]$
Solution:
Let’s find cofactor of A
\begin{aligned} &C_{11}=9, C_{21}=19, C_{31}=-4 \\ &C_{12}=4, C_{22}=14, C_{32}=1 \\ &C_{13}=8, C_{23}=3, C_{33}=2 \end{aligned}
$C_{i j}=\left[\begin{array}{ccc} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{array}\right]$
$Adj(A)$ = Transpose of $C_{ij}$
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \\ &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{array}\right] \times\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \end{aligned}
$=\left[\begin{array}{ccc} 9-8+24 & 19-28-9 & -4-2+6 \\ 0+8-8 & 0+28-3 & 0+2-2 \\ -36+20+16 & -76+70+6 & 16+5+4 \end{array}\right]=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$
Hence, $A \times \operatorname{Adj}(A)=25 I$

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (ii)

$A^{-1}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$
Hint:
Here, we use basic concept of inverse of matrix
Such that $A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$
Solution:
$|A|=\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1$
Let’s find $Adj\left ( A \right )$
\begin{aligned} &C_{11}=0, C_{12}=-1 \\ &C_{21}=-1, C_{22}=0 \end{aligned}
So,
$C_{i j}=\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right]$
$Adj \left ( A \right )$ Is transpose of $C_{ij}$
So,
$\operatorname{Adj}(A)=\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right]$
So,
\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=-\frac{1}{1} \times\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right] \\ A^{-1} &=-1 \times\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right] \\ A^{-1} &=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{aligned}

Adjoint and Inverse Matrix Exercise 6.1 Question 7 (iii)

$A^{-1}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right]$
Hint:
Here, we use basic concept of inverse.
Given:
$A=\left[\begin{array}{lc} a & b \\ c & \frac{1+b c}{a} \end{array}\right]$
Solution:
We know that
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)\\ &|A|=\left|\begin{array}{lc} a & b \\ c & \frac{1+b c}{a} \end{array}\right|=1+b c-b c=1 \end{aligned}
Let’s find $Adj\left ( A \right )$
\begin{aligned} &C_{11}=\frac{1+b c}{a} \\ &C_{12}=-c \\ &C_{21}=-b \\ &C_{22}=a \\ &C_{i j}=\left[\begin{array}{cc} \frac{1+b c}{a} & -c \\ -b & a \end{array}\right] \end{aligned}
\begin{aligned} \operatorname{Adj}(A) &=C_{i j}^{T} \\\\ \operatorname{Adj}(A) &=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}
So, let’s put value in formula
\begin{aligned} A^{-1} &=\frac{1}{1} \times\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \\\\ A^{-1} &=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix Exercise 6.1 Question 7 (i)

$A^{-1}=\left[\begin{array}{cc} \cos \theta &- \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
Hint:
Here, we use basic concept of inverse
Given:
$A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Solution:
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
Let’s find $\left | A \right |$
$|A|=\cos ^{2} \theta+\sin ^{2} \theta=1$
So,
$A^{-1}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

Adjoint and Inverse of Matrices Excercise 6.1 Question 7 (iv)

$A^{-1}=\frac{1}{17}\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
$A=\left[\begin{array}{cc} 2 & 5 \\ -3 & 1 \end{array}\right]$
Solution:
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
So let’s find $\left | A \right |$
$|A|=\left|\begin{array}{ll} 2 & 5 \\ -3 & 1 \end{array}\right|=2-[(5) \times(-3)]=2+15=17$
Let’s find $Adj\left ( A \right )$
\begin{aligned} &C_{11}=1, C_{12}=3 \\ &C_{21}=-5, C_{22}=2 \end{aligned}
So,
\begin{aligned} &C_{i j}=\left[\begin{array}{cc} 1 & 3 \\ -5 & 2 \end{array}\right] \\\\ &\operatorname{Adj}(A)=C_{i j}^{T} \\\\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right] \end{aligned}
Let’s put the values in formula
$A^{-1}=\frac{1}{17}\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right]$

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (i).

$A^{-1}=\frac{1}{18} \times\left[\begin{array}{ccc} -5 & 1 & 7 \\ 1 & 7 & -5 \\ 7 & -5 & 1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
$A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right]$
Solution:
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &|A|=1\left|\begin{array}{ll} 3 & 1 \\ 1 & 2 \end{array}\right|-2\left|\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right|+3\left|\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right| \\ &=1(6-1)-2(4-3)+3(2-9) \\ &|A|=5-2-21=-18 \end{aligned}
Let’s find $Adj \left ( A \right )$
For that let’s find cofactor
\begin{aligned} &C_{11}=+(6-1)=5 \\ &C_{12}=-(4-3)=-1 \\ &C_{13}=+(2-9)=-7 \\ &C_{21}=-(4-3)=-1 \\ &C_{22}=+(2-9)=-7 \\ &C_{23}=-(1-6)=5 \\ &C_{31}=+(2-9)=-7 \\ &C_{32}=-(1-6)=5 \\ &C_{33}=+(3-4)=-1 \end{aligned}
\begin{aligned} &C_{i j}=\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right] \\\\ &\operatorname{Adj}(A)=C_{\vec{j}}^{T} \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right] \\\\ &A^{-1}=\frac{1}{|A|} \times A d j(A) \end{aligned}
\begin{aligned} A^{-1} &=-\frac{1}{18} \times\left[\begin{array}{ccc} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{array}\right]\\ \\ A^{-1} &=\frac{1}{18} \times\left[\begin{array}{ccc} -5 & 1 & 7 \\ 1 & 7 & -5 \\ 7 & -5 & 1 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (ii)

$A^{-1}=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
Given:
$A=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right]$
Solution:
We know that
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
So let’s find $\left | A \right |$
\begin{aligned} &|A|=1\left|\begin{array}{ll} -1 & -1 \\ 3 & -1 \end{array}\right|-2\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|+5\left|\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right| \\ &=1(1+3)-2(-1+2)+5(3+2) \\ &=4-2+25 \\ &|A|=27 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A are
\begin{aligned} &C_{11}=4, C_{21}=17, C_{31}=3 \\ &C_{12}=-1, C_{22}=-11, C_{32}=6 \\ &C_{13}=5, C_{23}=1, C_{33}=-3 \end{aligned}
\begin{aligned} &\text { So } \operatorname{Adj}(A)=C_{{ij}}^{T} \\ &\qquad \operatorname{Adj}(A)=\left[\begin{array}{ccc} 4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 8 (iii)

$A^{-1}=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix.
Given:
$A=\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$
Solution:
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &|A|=2\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|-(-1)\left|\begin{array}{cc} -1 & -1 \\ 1 & 2 \end{array}\right|+1\left|\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right| \\ &=2(4-1)+1(-2+1)+1(1-2) \\ &=6-2 \\ &=4 \neq 0 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A
\begin{aligned} &C_{11}=3, C_{21}=1, C_{31}=-1 \\ &C_{12}=3, C_{22}=3, C_{32}=1 \\ &C_{13}=-1, C_{23}=1, C_{33}=3 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=C_{i j}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] \end{aligned}\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (iv)

$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$ .
Hint:
Here, we use basic concept of determinant and inverse of matrix.
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$
Solution:
\begin{aligned} &|A|=2\left|\begin{array}{ll} 1 & 0 \\ 1 & 3 \end{array}\right|-0\left|\begin{array}{ll} 5 & 0 \\ 0 & 3 \end{array}\right|-1\left|\begin{array}{ll} 5 & 1 \\ 0 & 1 \end{array}\right| \\ &=2(3-0)-0-1(5) \\ &=6-5 \\ &|A|=1 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A are
\begin{aligned} &C_{11}=3, C_{12}=-15, C_{13}=5 \\ &C_{21}=-1, C_{22}=6, C_{23}=-2 \\ &C_{31}=1, C_{32}=-5, C_{33}=2 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=C_{{ij}}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{array}\right]^{T}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{1} \times\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (v)

$A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix.
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$
Solution:
\begin{aligned} &|A|=0\left|\begin{array}{rr} -3 & 4 \\ -3 & 4 \end{array}\right|-1\left|\begin{array}{ll} 4 & 4 \\ 3 & 4 \end{array}\right|+(-1)\left|\begin{array}{rr} 4 & -3 \\ 3 & -3 \end{array}\right| \\ &=0-1(16-12)-1(-12+9) \\ &=-4+3=-1 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A are
\begin{aligned} &C_{11}=0, C_{12}=-4, C_{13}=-3 \\ &C_{21}=-1, C_{22}=3, C_{23}=-4 \\ &C_{31}=-3, C_{32}=3, C_{33}=-4 \\ &A d j(A)=C_{{ij}}^{T} \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}
\begin{aligned} &=-\frac{1}{1}\left[\begin{array}{ccc} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{array}\right] \\ &A^{-1}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (vi)

$A^{-1}=\left[\begin{array}{ccc} -2 & 1 & 1 \\ \frac{11}{4} & \frac{-1}{2} & \frac{-3}{4} \\ -1 & 0 & 0 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{array}\right]$
Solution:
\begin{aligned} &|A|=0\left|\begin{array}{cc} 4 & 5 \\ -4 & -7 \end{array}\right|-0\left|\begin{array}{cc} 3 & 5 \\ -2 & -7 \end{array}\right|-1\left|\begin{array}{cc} 3 & 4 \\ -2 & -4 \end{array}\right| \\ &=0-0-1(-12+8) \\ &=4 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A are
\begin{aligned} &C_{11}=-8, C_{21}=4, C_{31}=4 \\ &C_{12}=11, C_{22}=-2, C_{32}=-3 \\ &C_{13}=-4, C_{23}=0, C_{33}=0 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 8 & 11 & -4 \\ 4 & -2 & 0 \\ 4 & -3 & 0 \end{array}\right]^{T}=\left[\begin{array}{ccc} 8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{4} \times\left[\begin{array}{ccc} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{array}\right] \\ A^{-1} &=\left[\begin{array}{ccc} -2 & 1 & 1 \\ \frac{11}{4} & \frac{-1}{2} & \frac{-3}{4} \\ -1 & 0 & 0 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 8 (vii)

$A^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
\begin{aligned} &|A|=1\left|\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right|-0+0 \\ &=-\cos ^{2} \alpha-\sin ^{2} \alpha \\ &|A|=-1 \end{aligned}
So $A^{-1}$ exist
Cofactors of A are
\begin{aligned} &C_{11}=-1, C_{21}=0, C_{31}=0 \\ &C_{12}=0, C_{22}=-\cos \alpha, C_{32}=-\sin \alpha \\ &C_{13}=0, C_{23}=-\sin \alpha, C_{33}=\cos \alpha \\ &\operatorname{Adj}(A)=C_{{ij}}^{T} \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}
\begin{aligned} &=\frac{1}{-1}\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \\ &A^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 9 (i)

$A^{-1}=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix.
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$
Solution:
Here, let’s find $\left | A \right |$
\begin{aligned} &|A|=1\left|\begin{array}{ll} 4 & 3 \\ 3 & 4 \end{array}\right|-3\left|\begin{array}{ll} 1 & 3 \\ 1 & 4 \end{array}\right|+3\left|\begin{array}{ll} 1 & 4 \\ 1 & 3 \end{array}\right| \\ &=1(16-9)-3(4-3)+3(3-4) \\ &=7-3-3=1 \end{aligned}
Hence $A^{-1}$ exist
Let’s find cofactor of A
\begin{aligned} &C_{11}=7, C_{21}=-3, C_{31}=-3 \\ &C_{12}=-1, C_{22}=1, C_{32}=0 \\ &C_{13}=-1, C_{23}=0, C_{33}=1 \\ &A d j(A)=C_{{ij}}^{T} \end{aligned}
$\begin{gathered} \operatorname{Adj}(A)=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \\ \end{gathered}$
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
\begin{aligned} &A^{-1}=\frac{1}{1}\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \\ &A^{-1} \times A=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} 7-3-3 & 21-12-9 & 21-9-12 \\ -1+1+0 & -3+4+0 & -3+3+0 \\ -1+0+1 & -3+0+3 & -3+0+4 \end{array}\right] \\ &A^{-1} \times A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 9 (ii)

$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]$
Solution:
\begin{aligned} &|A|=2\left|\begin{array}{ll} 4 & 1 \\ 7 & 2 \end{array}\right|-3\left|\begin{array}{ll} 3 & 1 \\ 3 & 2 \end{array}\right|+1\left|\begin{array}{ll} 3 & 4 \\ 3 & 7 \end{array}\right| \\ &=2(8-7)-3(6-3)+1(21-12) \\ &=2-9+9 \\ &=2 \end{aligned}
Hence $A^{-1}$ exist
Cofactor of A
\begin{aligned} &C_{11}=1, C_{21}=1, C_{31}=-1 \\ &C_{12}=-3, C_{22}=1, C_{32}=1 \\ &C_{13}=9, C_{23}=-5, C_{33}=-1 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=C_{i j}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{array}\right]^{T}=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -2 \end{array}\right] \end{aligned}
\begin{aligned} &A^{-1} \times A=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right]\left[\begin{array}{ccc} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{ccc} 2+3-3 & 3+4-7 & 1+1-2 \\ -6+3+3 & -9+4+7 & -3+1+2 \\ 18-15-3 & 27-20-7 & 9-5-2 \end{array}\right]=\frac{1}{2}\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}
Hence $A^{^{-1 }}\times A=I$

Adjoint and Inverse Matrix Exercise 6.1 Question 11

$\left[\begin{array}{cc} -47 & \frac{39}{2} \\ 41 & -17 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
Given:
$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right]$
Solution:
$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right]$
Let’s find $\left | A \right |$
\begin{aligned} &|A|=\left|\begin{array}{cc} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \end{aligned}
\begin{aligned} &B=\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right] \\ &|B|=54-56=-2 \\ &\operatorname{Adj}(B)=\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right] \\ &B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(A)=\frac{1}{-2}\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right] \end{aligned}
Now, we know that
\begin{aligned} &(A B)^{-1}=B^{-1} A^{-1} \\ &=\frac{1}{-2}\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right]\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\ &=\frac{1}{-2}\left[\begin{array}{cc} 45+49 & -18-21 \\ -40-42 & 16+18 \end{array}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{2}\left[\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right] \\ &(A B)^{-1}=\left[\begin{array}{cc} -47 & \frac{39}{2} \\ 41 & -17 \end{array}\right] \end{aligned}

Adjoint and Inverse Matrix Exercise 6.1 Question 12

Hence proved $2A^{-1}=9 I-A$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]$
Solution:
Here let’s find $|A|, A d j(A) \& A^{-1}$
\begin{aligned} &|A|=\left|\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right|=14-12=2 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 7 & 3 \\ 4 & 2 \end{array}\right] \\ &A^{-1}=\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right] \end{aligned}
To show $2A^{-1}=9 I-A$
$2 A^{-1}=2 \times \frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$ (1)
$9 I-A=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$ (2)
From equation (1) and (2)
Hence, $2A^{-1}=9 I-A$

Adjoint and Inverse Matrix Exercise 6.1 Question 13

Hence proved $A-3 I=2\left(I+3 A^{-1}\right)$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]$
Solution:
Let’s find $|A|, A d j(A) \& A^{-1}$
\begin{aligned} &|A|=4-10=-6 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{-6}\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \end{aligned}
To show $A-3 I=2\left(I+3 A^{-1}\right)$
LHS
$A-3 I=\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]$ (1)
RHS
\begin{aligned} &2\left(I+3 A^{-1}\right)=2 I+6 A^{-1}=2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\frac{6}{6}\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \\ &=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]+\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \end{aligned}
$= \left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]$ (2)
Here from equation (1) and (2)
$A-3 I=2\left(I+3 A^{-1}\right)$

Adjoint and Inverse of Matrices Excercise 6.1 Question 14

$A^{-1}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{lc} a & b \\ c & \frac{1+b c}{a} \end{array}\right]$
Solution:
\begin{aligned} &|A|=\frac{a+a b c}{a}-b c \\ &=\frac{a+a b c-a b c}{a}=\frac{a}{a}=1 \end{aligned}
So, hence $A^{-1}$ exist
Cofactor of A
\begin{aligned} &C_{11}=\frac{1+b c}{a}, C_{12}=-c \\\\ &C_{21}=-b, C_{22}=a \\\\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} \frac{1+b c}{a} & -c \\ -b & a \end{array}\right]^{T}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\\\ &=\frac{1}{1}\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \\\\ &A^{-1}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}
To show $a A^{-1}=\left(a^{2}+b c+1\right) I-a A$
LHS
$a A^{-1}=a\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right]=\left[\begin{array}{cc} 1+b c & -a b \\ -a c & a^{2} \end{array}\right]$
RHS
\begin{aligned} &\left(a^{2}+b c+1\right) I-a A \\\\ &=\left[\begin{array}{cc} a^{2}+b c+1 & 0 \\ 0 & a^{2}+b c+1 \end{array}\right]-\left[\begin{array}{cc} a^{2} & a b \\ a c & 1+b c \end{array}\right]\\ \\ &=\left[\begin{array}{cc} 1+b c & -a b \\ -a c & a^{2} \end{array}\right] \end{aligned}
LHS = RHS

Adjoint and Inverse of Matrices Excercise 6.1 Question 15

$\left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{lll} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right], B^{-1}=\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$
Solution:
For $(A B)^{-1}$ we know that
$(A B)^{-1}=B^{-1} A^{-1}$
Here $B^{-1}$ is also given so
Let’s fin $A^{-1}$
$|A|=-5+4=-1$
Cofactor of A are
\begin{aligned} &C_{11}=-1, C_{21}=8, C_{31}=-12 \\ &C_{12}=0, C_{22}=1, C_{32}=-2 \\ &C_{13}=1, C_{23}=-10, C_{33}=15 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right] \\\\ &A^{-1}=-1\left[\begin{array}{ccc} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right]=\left[\begin{array}{ccc} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right] \end{aligned}
\begin{aligned} &(A B)^{-1}=B^{-1} A^{-1} \\ &=\left[\begin{array}{ccc} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right]=\left[\begin{array}{ccc} 1+0-3 & -8-3+30 & 12+6-45 \\ 1+0-3 & -8-4+30 & 12+8-45 \\ 1+0-4 & -8-3+40 & 12+6-60 \end{array}\right] \end{aligned}
$(A B)^{-1}=\left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$

Adjoint and Inverse of Matrices Excercise 6.1 Question 16 question (i)

Hence proved $[F(\alpha)]^{-1}=F(-\alpha)$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]$
Solution:
\begin{aligned} &|F(\alpha)|=\left|\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right| \\\\ &|F(\alpha)|=\cos ^{2} \alpha+\sin ^{2} \alpha=1 \end{aligned}
Cofactor of A are
\begin{aligned} &C_{11}=\cos \alpha, C_{21}=\sin \alpha, C_{31}=0 \\ &C_{12}=-\sin \alpha, C_{22}=\cos \alpha, C_{32}=0 \\ &C_{13}=0, C_{23}=0, C_{33}=1 \\ &\operatorname{Adj}(F(\alpha))=C_{{ij}}^{T} \end{aligned}
\begin{aligned} &\operatorname{Adj}(F(\alpha))=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]^{T} \\\\ &\operatorname{Adj}(F(\alpha))=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} (1)
\begin{aligned} &{[F(\alpha)]^{-1}=\frac{1}{1}\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]} \\\\ &F(-\alpha)=\left[\begin{array}{ccc} \cos (-\alpha) & \sin (-\alpha) & 0 \\ -\sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}
$=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]$ (2)
From equation (1) and (2)
$[F(\alpha)]^{-1}=F(-\alpha)$

Adjoint and Inverese of Matrices Exercise 6.1 Question 16 (ii)

Hence proved $[G(\beta)]^{-1}=G(-\beta)$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]$
Solution:
Let’s find $|G(\beta)|$
$|G(\beta)|=\cos ^{2} \beta+\sin ^{2} \beta=1$
Cofactor of A
\begin{aligned} &C_{11}=\cos \beta, C_{21}=0, C_{31}=\sin \beta \\ &C_{12}=0, C_{22}=1, C_{32}=0 \\ &C_{13}=\sin \beta, C_{23}=0, C_{33}=\cos \beta \end{aligned}
$\operatorname{Adj}(G(\beta))=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]^{T}$
\begin{aligned} &\operatorname{Adj}(G(\beta))=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right] \\\\ &{[G(\beta)]^{-1}=\frac{1}{|G(\beta)|} \times \operatorname{Adj}(G(\beta))} \end{aligned}
\begin{aligned} &{[G(\beta)]^{-1}=\frac{1}{1}\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 1 \\ \sin \beta & 0 & \cos \beta \end{array}\right]} \\\\ &G(-\beta)=\left[\begin{array}{ccc} \cos (-\beta) & 0 & \sin (-\beta) \\ 0 & 1 & 0 \\ \sin (-\beta) & 0 & \cos (-\beta) \end{array}\right]=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos \beta \end{array}\right] \end{aligned}
Hence
$[G(\beta)]^{-1}=G(-\beta)$

Adjoint and Inverese of Matrices Exercise 6.1 Question 17

$\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
\begin{aligned} &A=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \\\\ &A^{2}-4 A+I=0 \\\\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \& O=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}
Solution:
\begin{aligned} &A^{2}=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 4+3 & 6+6 \\ 2+2 & 3+4 \end{array}\right] \\\\ &A^{2}=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right] \end{aligned}
\begin{aligned} &4 A=4 \times\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 8 & 12 \\ 4 & 8 \end{array}\right] \\ \\&I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}
\begin{aligned} &A^{2}-4 A+I=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ 4 & 8 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \\&=\left[\begin{array}{cc} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}
Hence
\begin{aligned} &A^{2}-4 A+I=0 \\ &A \times A-4 A=-I \end{aligned}
Let’s multiply $A^{-1}$ both side\begin{aligned} &A \times A\left(A^{-1}\right)-4 A A^{-1}=-I A^{-1} \\ &A-4 I=-A^{-1} \\ &A^{-1}=4 I-A=\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 18

$A^{-1}=\frac{1}{42}\left[\begin{array}{cc} -4 & 5 \\ 2 & 8 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]$
Solution:
\begin{aligned} &A^{2}=\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 64+10 & -40+20 \\ -16+8 & 10+16 \end{array}\right]=\left[\begin{array}{cc} 74 & -20 \\ -8 & 26 \end{array}\right] \\\\ &4 A=4 \times\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} -32 & 20 \\ 8 & 16 \end{array}\right] \\\\ &42 I=42 \times\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 42 & 0 \\ 0 & 42 \end{array}\right] \end{aligned}
Now,
$A^{2}+4 A-42 I=\left[\begin{array}{cc} 74 & -20 \\ -8 & 26 \end{array}\right]+\left[\begin{array}{cc} -32 & 20 \\ 8 & 16 \end{array}\right]-\left[\begin{array}{cc} 42 & 0 \\ 0 & 42 \end{array}\right]$
$=\left[\begin{array}{cc} 74-74 & -20+20 \\ -2+8 & 42-42 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\$
$A^{2}+4 A-42 I=0$
Multiplying by $A^{-1}$ both sides
\begin{aligned} &A \times A \times A^{-1}+4 A \times A^{-1}-42 I \times A^{-1}=0 \\ &A+4 I-42 A^{-1}=0 \\ &42 A^{-1}=A+4 I \end{aligned}
So,
\begin{aligned} A^{-1} &=\frac{1}{42}[A+4 I] \\\\ A^{-1} &=\frac{1}{42}(\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right] )\\\\ A^{-1} &=\frac{1}{42}\left[\begin{array}{cc} -4 & 5 \\ 2 & 8 \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 19

$A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & -1 \\ 1 & 3 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]$ Show that $A^{2}-5 A+7 I=0$
Solution:
\begin{aligned} &A=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \\\\ &A^{2}=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right]=\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right] \end{aligned}
Now,
\begin{aligned} &A^{2}-5 A+7 I \\\\ &=\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right]-5 \times\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &=\left[\begin{array}{ll} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\\\ &A^{2}-5 A+7 I=0 \end{aligned}
Multiply by $A^{-1}$ both sides
\begin{aligned} &A \times A \times A^{-1}-5 A \times A^{-1}+7 I \times A^{-1}=0 \\\\ &A-5 I+7 A^{-1}=0 \\\\ &A^{-1}=\frac{1}{7}[5 I-A] \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{7}\left[\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]-\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]\right] \\\\ &A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & -1 \\ 1 & 3 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 20

$X = 9 \: and\: y =14$ , $A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]$
Solution:
\begin{aligned} &A^{2}=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right]=\left[\begin{array}{cc} 22 & 27 \\ 18 & 31 \end{array}\right] \\\\ &A^{2}-x A+y I=\left[\begin{array}{ll} 22 & 37 \\ 18 & 31 \end{array}\right]-x\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}
\begin{aligned} &22-4 x+y=0 \\ &4 x-y=22 \\ &x=9 \end{aligned}
So,
\begin{aligned} &4 x-y=22 \\ &4 \times 9-y=22 \\ &36-22=y \\ &y=14 \end{aligned}
So $A^{2}-9 A+14 I=0$
Multiply by $A^{-1}$ both sides,
\begin{aligned} &A \times A \times A^{-1}-9 A \times A^{-1}+14 I \times A^{-1}=0 \\\\ &A-9 I+14 A^{-1}=0 \\\\ &A^{-1}=\frac{1}{14}[9 I-A] \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{14}\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \\ \\&A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 21

Λ =1, $A^{-1}= \frac{1}{2}\begin{bmatrix} -2 & 2\\ -4 & 3 \end{bmatrix}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}= \frac{1}{\left | A \right |}\times Adj\left ( A \right )$
Given:
$A= \begin{bmatrix} 3 &-2 \\ 4 & -2 \end{bmatrix}$
Solution:
$A^{2}=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]$
Now,
\begin{aligned} &A^{2}=\lambda A-2 I \\ &\lambda A=A^{2}+2 I \\ &\lambda A=\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]+\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \end{aligned}
\begin{aligned} &\lambda=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \div\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right] \\ &\lambda=1 \\ &\text { So, } \\ &A^{2}=A-2 I \end{aligned}
Multiply $A^{-1}$ both side
\begin{aligned} &A^{-1} A \times A=A A^{-1}-2 L A^{-1} \\ &2 A^{-1}=I-A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \\ &A^{-1}=\frac{1}{2}\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \end{aligned}

Adjoint and Inverse of Matrices Excercise 6.1 Question 22

$A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]$
Solution:
$A^{2}=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]=\left[\begin{array}{ll} 25-3 & 15-6 \\ -5+2 & -3+4 \end{array}\right]=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]$
Now,
\begin{aligned} &A^{2}-3 A-7 I=0 \\\\ &{\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]} \\\\ &{\left[\begin{array}{ll} 22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]} \\\\ &A^{2}-3 A-7 I=0 \end{aligned}
Multiply $A^{-1}$ both side
\begin{aligned} &A \times A \times A^{-1}-3 A \times A^{-1}-7 L A^{-1}=0 \\\\ &A-3 I-7 A^{-1}=0 \\\\ &7 A^{-1}=A-3 I \end{aligned}
\begin{aligned} &=\frac{1}{7}\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 23

$A^{-1}=\left[\begin{array}{cc} 6 & -5 \\ -7 & 6 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]$
Solution:
\begin{aligned} &A^{2}-12 A+I=0 \\\\ &A^{2}=\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]=\left[\begin{array}{cc} 36+35 & 30+30 \\ 42+42 & 35+36 \end{array}\right]=\left[\begin{array}{cc} 71 & 60 \\ 84 & 71 \end{array}\right] \end{aligned}
Now,
\begin{aligned} &A^{2}-12 A+1=0 \\\\ &{\left[\begin{array}{ll} 71 & 60 \\ 84 & 71 \end{array}\right]-12\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\\\ &=\left[\begin{array}{cc} 71-72+1 & 60-60+0 \\ 84-89+0 & 1-72+1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}
Also,
\begin{aligned} &A^{2}-12 A+1=0 \\\\ &A-12 I+A^{-1}=0 \\\\ &A^{-1}-12 I=A \end{aligned}
\begin{aligned} &A^{-1}=\left[\begin{array}{cc} 12 & 0 \\ 0 & 12 \end{array}\right]-\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]=\left[\begin{array}{cc} 12-6 & -5 \\ -7 & 12-6 \end{array}\right] \\\\ &A^{-1}=\left[\begin{array}{cc} 6 & -5 \\ -7 & 6 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 24

$A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given
$A=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
Solution:
\begin{aligned} A^{3} &=A^{2} \times A \\\\ A^{2} &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6+9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right] \\\\ A^{2} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] \end{aligned}
\begin{aligned} &A^{2} \times A=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+14+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{array}\right] \\ \\&=\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] \end{aligned}
Now,
\begin{aligned} &A^{3}-6 A^{2}+5 A+11 I \\\\ &{\left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]-6\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}
$\left[\begin{array}{ccc} 8-24 & 7-12 & 1-6 \\ -23+18 & 27-48 & -69+84 \\ 32-42 & -13+18 & 58-84 \end{array}\right]+\left[\begin{array}{ccc} 5+11 & 5 & 5 \\ 5 & 10+11 & -15 \\ 10 & -5 & 26 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$
$A^{3}-6 A^{2}+5 A+11 I=0$
Now,
\begin{aligned} &(A \times A \times A) A^{-1}-6(A \times A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \\\\ &A A\left(A^{-1} A\right)-6 A\left(A^{-1} A\right)+5 A^{-1} A=-11\left(A^{-1} I\right) \\\\ &A^{-1}=\frac{-1}{11}\left(A^{2}-6 A+5 I\right) \end{aligned}
Now,
$A^{2}-6 A+5 I$
\begin{aligned} & \\ &{\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-6\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]+5\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]-\left[\begin{array}{ccc} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right]+\left[\begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} -3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right] \end{aligned}
Hence,
$A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{array}\right]$

Adjoint and Inverese of Matrices Exercise 6.1 Question 25

$A^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
Given:
$A=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]$
Solution:
$A^{3}-A^{2}-3 A-I_{3}=0$
So,
$A^{3}=A^{2} \times A$
\begin{aligned} & \\ &A^{2}=A \times A=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 0+1+8 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1 \end{array}\right] \\\\ & \end{aligned}
$=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]$
\begin{aligned} &A^{2} \times A=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 2 \\ 3 & 4 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5+16-12 & 0-8+16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2-0+9 & 0-0-12 & 4+0+3 \end{array}\right] \\\\ & \end{aligned}
$=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right]$
Now,
$A^{3}-A^{2}-3 A-I$
\begin{aligned} & \\ &{\left[\begin{array}{ccc} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right]-\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]-3\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} -1+5 & -8+8 & -10+4 \\ 0-6 & 7-9 & 10-4 \\ 7+2 & 12-0 & 7-3 \end{array}\right]+\left[\begin{array}{ccc} -3-1 & 0 & 6-0 \\ 6-0 & 3-1 & -6-0 \\ -9-0 & -12+0 & -3-1 \end{array}\right] \\\\ &=\left[\begin{array}{ccc} 4 & 0 & -6 \\ -6 & -2 & 6 \\ 9 & 12 & 4 \end{array}\right]+\left[\begin{array}{ccc} -4 & 0 & 6 \\ 6 & 2 & -6 \\ -9 & -12 & -4 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}
Thus,
\begin{aligned} &A^{3}-A^{2}-3 A-I\\ &\text { now }\\ &(A A A) \times A^{-11}-A A A^{-1}-3 A A^{-1}-L A^{-1}=0\\ &A^{2}-A-3 A-I=0\\ &A^{-1}=\left(A^{2}-A-3 I\right) \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]-\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] \\ \end{aligned}
$\begin{bmatrix} -5-1-3 &-8+0 & -4+2\\ 6+2-0 &9+1-3 &4-2 \\ -2-3-0 &-4 & 3-1-3 \end{bmatrix}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

$A^{-1}=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$
Solution:
$A^{3}=A^{2} \times A$
$=\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]=\left[\begin{array}{ccc} 12+5+5 & -6+10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{array}\right]$
$=\left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]$
Now,
$A^{3}-6 A^{2}+9 A-4 I$
$\left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]-6\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]+9\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]-4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc} 22-36+18-4 & -21+30-9 & 21-30+9 \\ -21+30-9 & 22-36+18-4 & -21+30-9 \\ 21-30+9-0 & -21+30-9 & 22-36+18-4 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$
Thus,
\begin{aligned} &A^{3}-6 A^{2}+9 A-4 I \\ &A^{2}-6 A+9 I=4 A^{-1} \\ &A^{-1}=\frac{1}{4}\left[A^{2}-6 A+9 I\right] \end{aligned}
$A^{2}-6 A+9 I=\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]-6\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]+9\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc} 6-12+9 & -5+6+0 & 5-6+0 \\ -5+6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5-6+0 & 6-12+3 \end{array}\right]=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$
Hence,
$A^{-1}=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$

Adjoint and Inverse Matrix Exercise 6.1 Question 27

$A^{-1}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right]=A^{T}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\frac{1}{9}\left[\begin{array}{ccc} -8 & 1 & 4 \\ 1 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]$ Find $A^{T}=A^{-1}$
Solution:
\begin{aligned} &A^{T}=A^{-1} \\ &L H S \\ &A^{T}=\frac{1}{9}\left[\begin{array}{ccc} 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]^{T}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right] \end{aligned}
\begin{aligned} &R H S \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}
Let’s find
\begin{aligned} &|A| \& \operatorname{Adj}(A) \\ &|A|=\frac{1}{9}[-8(16+56)-1(16-7)+4(-32-4)] \\ &|A|=-81 \end{aligned}
Cofactor of A
\begin{aligned} &C_{11}=72, C_{21}=-36, C_{31}=-9 \\ &C_{12}=-9, C_{22}=-36, C_{32}=72 \\ &C_{13}=-36, C_{23}=-63, C_{33}=-36 \\&\operatorname{Adj}(A)=\left[\begin{array}{ccc} 72 & -9 & -36 \\ -36 & -36 & -63 \\ 9 & 72 & -36 \end{array}\right]^{T}=\left[\begin{array}{ccc} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{array}\right] \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times A d j(A)=\frac{1}{-9}\left[\begin{array}{ccc} 72 & -36 & -9 \\ -9 & -36 & 72 \\ 36 & -63 & -36 \end{array}\right] \\ &A^{-1}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right]=A^{T} \end{aligned}

Adjoint and Inverse Matrix Exercise 6.1 Question 28

$A^{-1}= A^{3}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{rrr} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$
Solution:
$A^{-1}= A^{3}$
RHS
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Let’s find $|A| \& \operatorname{Adj}(A)$
$|A|=3+6-8=1$
Cofactor of A
\begin{aligned} &C_{11}=1, C_{21}=-1, C_{31}=0 \\ &C_{12}=-2, C_{22}=3, C_{32}=-4 \\ &C_{13}=-2, C_{23}=3, C_{33}=-3 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right] \end{aligned} (1)
RHS
\begin{aligned} &A^{3}=A^{2} \times A \\ &A^{2}=\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 9-6+0 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{array}\right] \\ &\end{aligned}
$=\left[\begin{array}{lll} 3 & -4 & 4 \\ 0 & -1 & 0 \\ 2 & 2 & 3 \end{array}\right]$
\begin{aligned} &A^{3}=A^{2} \times A=\left[\begin{array}{ccc} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{array}\right]\left[\begin{array}{ccc} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \\ & \end{aligned}
$=\left[\begin{array}{ccc} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{array}\right]$ (2)
So, here equation (1) and (2)
LHS = RHS
$A^{-1}=A^{3}$

Adjoint and Inverse of Matrices Excercise 6.1 Question 29

$A^{-1}= A^{2}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} -1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right]$
Solution:
\begin{aligned} &|A|=-1\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|-2\left|\begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array}\right|+0 \\ &|A|=-1(-1)-0+0= 1 \\ &|A|=1 \end{aligned}
Cofactor of A
\begin{aligned} &C_{11}=-1, C_{21}=0, C_{31}=2 \\ &C_{12}=0, C_{22}=0, C_{32}=1 \\ &C_{13}=-1, C_{23}=1, C_{33}=1 \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & -1 \\ 0 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &=\frac{1}{1}\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}
\begin{aligned} &A^{2}=A \times A \\ &=\left[\begin{array}{ccc} -1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right]=\left[\begin{array}{ccc} -1-2+0 & -2+2+0 & 0+2+0 \\ 1-1+1 & -2+1+1 & -1+1-0 \\ 0-1+0 & 0+1-0 & 0+1-0 \end{array}\right] \\ &=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}
Hence $A^{-1}= A^{2}$

Adjoint and Inverse of Matrices Excercise 6.1 Question 30

$X= \begin{bmatrix} -3 &-14 \\ 4& 17 \end{bmatrix}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right], B=\left[\begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array}\right]$
Solution:
\begin{aligned} &A X=B \\ &X=A^{-1} B \\ &|A|=1 \end{aligned}
Cofactor of A
\begin{aligned} &C_{11}=1, C_{21}=-4 \\ &C_{12}=-1, C_{22}=5 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -1 \\ -4 & 5 \end{array}\right]^{T}=\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{1}\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \\ X &=\left[\begin{array}{cc} 1 & -4 \\ -1 & 5 \end{array}\right] \times\left[\begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{cc} -3 & -14 \\ 4 & 17 \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 31

$X=\left[\begin{array}{cc} -7 & -5 \\ 7 & 6 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
Given:
$A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right], B=\left[\begin{array}{cc} 14 & 7 \\ 7 & 7 \end{array}\right]$
Solution:
\begin{aligned} &A X=B \\ &X=A^{-1} B \\ &|A|=-7 \end{aligned}
Cofactor of A
\begin{aligned} &C_{11}=-2, C_{12}=1 \\ &C_{21}=-3, C_{22}=5 \\ &\operatorname{Adj}(A)=\left[\begin{array}{ll} -2 & 1 \\ -3 & 5 \end{array}\right]^{T}=\left[\begin{array}{cc} -2 & -3 \\ 1 & 5 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}
\begin{aligned} &A^{-1}=\frac{1}{-7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right] \\ &X=-\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\left[\begin{array}{cc} 14 & 7 \\ 7 & 7 \end{array}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{7}\left[\begin{array}{cc} 28+21 & 14+21 \\ -14-35 & -7-35 \end{array}\right] \\ &X=\left[\begin{array}{cc} -7 & -5 \\ 7 & 6 \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 32

$A^{-1}=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{rr} -1 & 1 \\ -2 & 1 \end{array}\right], C=\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]$
Solution:
Then the given equation becomes as,
\begin{aligned} &A X B=C \\ &X=A^{-1} C B^{-1} \\ &|A|=15-14=1 \\ &|B|=-1+2=1 \end{aligned}
$A^{-1}= \frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\$
\begin{aligned} B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(B) &=1\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} &X=A^{-1} C B^{-1} \\ &=1\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} 10+0 & -5-8 \\ -14+0 & 7+12 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{cc} 10-26 & -10+13 \\ -14+38 & 14-19 \end{array}\right] \\ &X=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right] \end{aligned}

Adjoint and Inverse of a Matrix exercise 6.1 question 33

$X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 5 & 3 \\ 3 & 2 \end{array}\right], C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Solution:
Then the given equation becomes
\begin{aligned} &A \times B=I \\ &X=A^{-1} B^{-1} \\ &|A|=6-5=1 \\ &|B|=10-9=1 \end{aligned}
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]$
\begin{aligned} &B^{-1}=\frac{1}{|B|} \times A d j(B)=\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right] \\ & \end{aligned}
$X=A^{-1} B^{-1}=\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right]$
\begin{aligned} &X=\left[\begin{array}{cc} 6+3 & -14 \\ -16 & 25 \end{array}\right] \\\end{aligned}
$X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]$

Adjoint and Inverse of a Matrix exercise 6.1 question 33
Edit Q

Adjoint and Inverse of a Matrix exercise 6.1 question 33

$X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 5 & 3 \\ 3 & 2 \end{array}\right], C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Solution:
Then the given equation becomes
\begin{aligned} &A \times B=I \\ &X=A^{-1} B^{-1} \\ &|A|=6-5=1 \\ &|B|=10-9=1 \end{aligned}
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]$
\begin{aligned} &B^{-1}=\frac{1}{|B|} \times A d j(B)=\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right] \\ & \end{aligned}
$X=A^{-1} B^{-1}=\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right]$
\begin{aligned} &X=\left[\begin{array}{cc} 6+3 & -14 \\ -16 & 25 \end{array}\right] \\\end{aligned}
$X=\left[\begin{array}{cc} 9 & -14 \\ -16 & 25 \end{array}\right]$

Adjoint and Inverese of Matrices Exercise 6 point 1 Question 34

$A^{-1}=\frac{1}{5}\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 1 & 2 & 1 \end{array}\right]$
$A^{2}-4 A-5 I=0$
Solution:
$A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]$
\begin{aligned} &A^{2}=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]=\left[\begin{array}{lll} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{array}\right] \\ & \end{aligned}
$A^{2} =\left[\begin{array}{lll} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right]$
$A^{2}-4 A-5 I=0$
${\left[\begin{array}{lll} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right]-4\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]+5\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]}$
$=\left[\begin{array}{rrrr} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$
Also,
\begin{aligned} &A^{2}-4 A-5 I=0 \\ &(A A) A^{-1}-4 A A^{-1}-5 I A^{-1}=0 \\ &A-4 I-5 A^{-1}=0 \\ &A^{-1}=\frac{1}{5}(A-4 I) \end{aligned}
$A^{-1} =\frac{1}{5}\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
\begin{aligned} \\ A^{-1} &=\frac{1}{5}\left[\begin{array}{ccc} 1-4 & 2-0 & 2-0 \\ 2-0 & 1-4 & 2-0 \\ 2-0 & 2-0 & 1-4 \end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 35

$|A \times \operatorname{Adj}(A)|=|A|^{n}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
A is square matrix
Solution:
\begin{aligned} &|A \times \operatorname{Adj}(A)|=|A|^{n} \\ &L H S \\ &|A \times \operatorname{Adj}(A)| \\ &=|A| \times|A|^{n-1} \end{aligned} $\left[A d j(A)=|A|^{n-1}\right]$
\begin{aligned} &=|A|^{n+1-1} \\ &=|A|^{n} \\ &=R H S \end{aligned}

Adjoint and Inverese of Matrices Exercise 6.1 Question 36

$(A B)^{-1}=\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
Given:
$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right], B=\left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]$
Find $\left ( AB \right )^{-1}$
Solution:
$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$
So, we know that
$\left ( AB \right )^{-1}= B^{-1}A^{-1}$
So let’s find $B^{-1}$
\begin{aligned} &B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(B) \\ &|B|=1(3-0)-2(1-0)-2(2-0) \\ &|B|=3+2-4=1 \end{aligned}
Now cofactor of B
\begin{aligned} &C_{11}=3, C_{21}=2, C_{31}=6 \\ &C_{12}=1, C_{22}=1, C_{32}=2 \\ &C_{13}=2, C_{23}=2, C_{33}=5 \\ &\operatorname{Adj}(B)=\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \end{aligned}
Now,
$B^{-1}=\left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]$
$(A B)^{-1}=B^{-1} A^{-1}$
$=\left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{array}\right]$
$(A B)^{-1}=\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]$

Adjoint and Inverse Matrix Exercise 6.1 Question 37

$\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$A=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{array}\right]$
Solution:
$A^{T}=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right]$
Let’s find $\left | A^{T} \right |$
\begin{aligned} &\left|A^{T}\right|=(-1-8)-0-2(-8+3) \\ &=-9+10=1 \end{aligned}
Cofactor of $A^{T}$
\begin{aligned} &C_{11}=-9, C_{12}=8, C_{13}=-5 \\ &C_{21}=-8, C_{22}=7, C_{23}=-4 \\ &C_{31}=-2, C_{32}=2, C_{33}=1 \end{aligned}
$\operatorname{Adj}\left(A^{T}\right)=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$
$\left(A^{T}\right)^{-1}=\frac{1}{1}\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$
$\left(A^{T}\right)^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

Adjoint and Inverse Matrix Exercise 6.1 Question 38

\begin{aligned} &A \times \operatorname{Adj}(A)=27\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \end{aligned}
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times A d j(A)$
Given:
$A=\left|\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right|$
Solution:
\begin{aligned} &|A|=\left|\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right| \\ &|A|=-1(1-4)+2(2+4)-2(-4-2) \\ &=3+12+12 \\ &|A|=27 \end{aligned}
Cofactor of A
\begin{aligned} &C_{11}=-3, C_{21}=6, C_{31}=6 \\ &C_{12}=-6, C_{22}=3, C_{32}=-6 \\ &C_{13}=-6, C_{23}=-6, C_{33}=3 \end{aligned}
$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ 6 & 6 & 3 \end{array}\right]$
\begin{aligned} & \\ &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right]\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ 6 & 6 & 3 \end{array}\right]=\left[\begin{array}{ccc} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}\right] \end{aligned}
\begin{aligned} &A \times \operatorname{Adj}(A)=27\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \end{aligned}
$A \times A d j(A)=|A| I$

Adjoint and Inverse Matrix Exercise 6.1 Question 39

$A^{-1}=\frac{1}{2}\left ( A^{2}-3I \right )$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}= \frac{1}{\left | A \right |}\times Adj\left ( A\right )$
Given:
$A= \begin{bmatrix} 0 &1 &1 \\ 1 & 0 & 1\\ 1& 1 & 0 \end{bmatrix}$
Solution:
$|A|=0-1(0-1)+1(1-0)=0+1+1=2$
Cofactor of A
\begin{aligned} &C_{11}=-1, C_{21}=1, C_{31}=1 \\ &C_{12}=1, C_{22}=-1, C_{32}=1 \\ &C_{13}=1, C_{23}=1, C_{33}=-1 \end{aligned}
\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right] \\ & \end{aligned}
$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]$
\begin{aligned} &A^{2}-3 I=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]-3\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \end{aligned}
$=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]-\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]$
Hence $A^{-1}= \frac{1}{2}(A^{2}-3I)$

Adjoint and Inverese of Matrices Exercise 6.1 Question 10 (i)

Proved $\left ( AB \right )^{-1}=B^{-1}A^{-1}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{\left | A \right |}\times Adj\left ( A \right )$
Given:
$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right]$
Solution:
Let’s find $\left | A \right |$
$|A|=\left|\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1$
$\operatorname{Adj}(A)=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]$
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
$A^{-1}=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]$
Then let’s find $\left | B \right |Adj\left ( B \right ) \& \: B^{-1}$
$|B|=\left|\begin{array}{cc} 4 & 6 \\ 3 & 2 \end{array}\right|=8-18=-10$
$\operatorname{Adj}(B)=\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]$
$B^{-1}=\frac{1}{|B|} \times A \operatorname{dj}(B)$
$B^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]$
Then find $AB$
\begin{aligned} &A \times B=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right] \times\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right] \\ &\end{aligned}
$=\left[\begin{array}{cc} 12+6 & 18+4 \\ 28+15 & 42+10 \end{array}\right]=\left[\begin{array}{ll} 18 & 22 \\ 43 & 52 \end{array}\right]$
Then let’s find $\left | AB \right |,Adj\left ( AB \right )$ and inverse of $AB$
$|A B|=\left|\begin{array}{cc} 18 & 22 \\ 43 & 52 \end{array}\right|=936-946=-10$
$\operatorname{Adj}(A B)=\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]$
$(A B)^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]$ (1)
Now $B^{-1}A^{-1}$
$=\frac{1}{10}\left[\begin{array}{cc} 2 & -6 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]$
$=\frac{1}{10}\left[\begin{array}{cc} 10+42 & -4-18 \\ -15-28 & 6+12 \end{array}\right]$
$=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]$ (2)
From equation (1) and (2)
$\left ( AB \right )^{-1}=B^{-1}A^{-1}$
Hence proved

Adjoint and Inverese of Matrices Exercise 6.1 Question 10 (ii)

Proved $\left ( AB \right )^{-1}=B^{-1}A^{-1}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{\left | A \right |}\times Adj\left ( A \right )$
Given:
$A=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]$
Solution:
Let’s find $\left | A \right |Adj\left ( A \right ) \& \: A^{-1}$
$|A|=\left|\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right|=6-5=1$
$\operatorname{Adj}(A)=\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]$
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]=\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right]$
Let’s find $\left | B \right |Adj\left ( B \right ) \& \: B^{-1}$
$|B|=\left|\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right|=16-15=1$
$\operatorname{Adj}(B)=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]$
$B^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]$
Then find AB
$A \times B=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 11 & 14 \\ 29 & 37 \end{array}\right]$
Let’s find $\left | AB \right |,Adj\left ( A \right ) \& \:\left ( AB \right ) ^{-1}$
$|A B|=407-406=1$
$\operatorname{Adj}(A B)=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$
$(A B)^{-1}=\frac{1}{|A B|} \times \operatorname{Adj}(A)=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$ (1)
Now
\begin{aligned} &B^{-1} A^{-1}=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right] \\ & \end{aligned}
$B^{-1} A^{-1}=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$ (2)
Hence proved from equation (1) and (2)$\left ( AB \right )^{-1}=B^{-1}A^{-1}$

Adjoint and Inverese of Matrices Exercise 6.1 Question 16 (iii)

Hence proved $[F(\alpha) G(\beta)]^{-1}=G(-\beta) F(-\alpha)$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$
Given:
$F(\alpha)=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right], G(\beta)=\left[\begin{array}{ccc} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{array}\right]$
Solution:
We have to show
$[F(\alpha) G(\beta)]^{-1}=G(-\beta) F(-\alpha)$
${[G(\beta)]^{-1}=G(-\beta)}$
${[F(\alpha)]^{-1}=F(-\alpha)}$
$L H S=[F(\alpha) G(\beta)]^{-1}$
$=[G(\beta)]^{-1}[F(\alpha)]^{-1}$
$=G(-\beta) F(-\alpha)$
$LHS=RHS$
Hence proved

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## RD Sharma Chapter wise Solutions

2. How are NCERT books compared to RD Sharma?

NCERT books are suitable for a basic study and contain some essential questions as well. Nevertheless, RD Sharma solutions are best suited for maths as they have detailed material. To check the material for matrices, check Class 12 RD Sharma Chapter 6 Exercise 6.1 Solution.

3. What is a Matrix?

A matrix is a collection of values of tabulated rows and columns. It is a rectangular array wherein the values are independent of each other and have no direct relation. To learn more about matrices, refer to RD Sharma Class 12 Solutions Adjoint and inverse of Matrix Ex 6.1.

4. What is the Inverse of a Matrix?

The inverse of a matrix is a value that, when multiplied, gives an identity matrix in multiple forms. If M is a matrix, then its inverse is denoted by M-1, and its equation can be given as M x M-1 = I. To learn more about the inverse of matrices, refer, RD Sharma Class 12 Solutions Adjoint and inverse of Matrix Ex 6.1.

5. Explain Adjoint of a Matrix

The adjoint is obtained by finding the cofactors and then finding the transpose of that resultant matrix. To learn more about adjoints of matrices, check RD Sharma Class 12 Solutions Chapter 6 Ex 6.1.

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