Exercise 4.4 (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths
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Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations
Ex 4.4 Question 1:
Find adjoint of each of the matrices.
$
\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$.
We have,
$\begin{aligned}
& A_{11}=4, A_{12}=-3, A_{21}=-2, A_{22}=1 \\
& \therefore \text { adj } A=\left[\begin{array}{ll}
A_{11} & A_{21} \\
A_{23} & A_{22}
\end{array}\right]=\left[\begin{array}{lr}
4 & -2 \\
-3 & 1
\end{array}\right]
\end{aligned}$
Ex 4.4 Question 2:
Find adjoint of each of the matrices.
$
\left[\begin{array}{lrr}
1 & -1 & 2 \\
2 & 3 & 5 \\
-2 & 0 & 1
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{lrr}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.
We have,
$\begin{aligned}
& A_{11}=\left|\begin{array}{ll}
3 & 5 \\
0 & 1
\end{array}\right|=3-0=3 \\
& A_{12}=-\left|\begin{array}{ll}
2 & 5 \\
-2 & 1
\end{array}\right|=-(2+10)=-12
\end{aligned}$
$\begin{aligned}
& A_{13}=\left|\begin{array}{ll}
2 & 3 \\
-2 & 0
\end{array}\right|=0+6=6 \\
& A_{21}=-\left|\begin{array}{ll}
-1 & 2 \\
0 & 1
\end{array}\right|=-(-1-0)=1
\end{aligned}$
$\begin{aligned}
& A_{22}=\left|\begin{array}{ll}
1 & 2 \\
-2 & 1
\end{array}\right|=1+4=5 \\
& A_23=-\left|\begin{array}{cc}
1 & -1 \\
-2 & 0
\end{array}\right|=-(0-2)=2
\end{aligned}$
$\begin{aligned}
& A_{31}=\left|\begin{array}{ll}
-1 & 2 \\
3 & 5
\end{array}\right|=-5-6=-11 \\
& A_{32}=-\left|\begin{array}{cc}
1 & 2 \\
2 & 5
\end{array}\right|=-(5-4)=-1 \\
& A_{33}=\left|\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right|=3+2=5
\end{aligned}$
$\text { Hence, } \operatorname{adj} A=\left[\begin{array}{lll}
A_{14} & A_{21} & A_{31} \\
A_{13} & A_{23} & A_{12} \\
A_{33} & A_{23} & A_{33}
\end{array}\right]=\left[\begin{array}{lll}
3 & 1 & -11 \\
-12 & 5 & -1 \\
6 & 2 & 5
\end{array}\right] \text {. }$
Ex 4.4 Question 3:
$
\begin{aligned}
& \text { Verify } A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I \text {. } \\
& {\left[\begin{array}{rr}
2 & 3 \\
-4 & -6
\end{array}\right]}
\end{aligned}
$
Answer
$
A=\left[\begin{array}{rr}
2 & 3 \\
-4 & -6
\end{array}\right]
$
we have,
$
\begin{aligned}
& |A|=-12-(-12)=-12+12=0 \\
& \therefore|A|=0\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
& A_{11}=-6, A_{22}=4, A_{21}=-3, A_{22}=2 \\
& \therefore \text { adj } A=\left[\begin{array}{rr}
-6 & -3 \\
4 & 2
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
A(\operatorname{adj} A) & =\left[\begin{array}{rr}
2 & 3 \\
-4 & -6
\end{array}\right]\left[\begin{array}{rr}
-6 & -3 \\
4 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
-12+12 & -6+6 \\
24-24 & 12-12
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
Also, $(\operatorname{adj} A) A=\left[\begin{array}{rr}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{rr}2 & 3 \\ -4 & -6\end{array}\right]$
$
=\left[\begin{array}{cc}
-12+12 & -18+18 \\
8-8 & 12-12
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
$
Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.
Ex 4.4 Question 4:
$
\begin{aligned}
& \text { Verify } A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I \text {. } \\
& {\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]} \\
&
\end{aligned}
$
Answer
$
\begin{aligned}
& A=\left[\begin{array}{ccr}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right] \\
& |A|=1(0-0)+1(9+2)+2(0-0)=11 \\
& \therefore|A| I=11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
& A_{11}=0, A_{12}=-(9+2)=-11, A_{13}=0 \\
& A_{21}=-(-3-0)=3, A_{22}=3-2=1, A_{23}=-(0+1)=-1 \\
& A_{31}=2-0=2, A_{32}=-(-2-6)=8, A_{33}=0+3=3 \\
\therefore \operatorname{adj} A & =\left[\begin{array}{llr}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
A(\operatorname{adj} A) & =\left[\begin{array}{llr}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\left[\begin{array}{lll}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right] \\
& =\left[\begin{array}{lll}
0+11+0 & 3-1-2 & 2-8+6 \\
0+0+0 & 9+0+2 & 6+0-6 \\
0+0+0 & 3+0-3 & 2+0+9
\end{array}\right] \\
& =\left[\begin{array}{lll}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]
\end{aligned}
$
Also,
$
\begin{aligned}
(\operatorname{adj} A) \cdot A & =\left[\begin{array}{lll}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right]\left[\begin{array}{llr}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right] \\
& =\left[\begin{array}{lll}
0+9+2 & 0+0+0 & 0-6+6 \\
-11+3+8 & 11+0+0 & -22-2+24 \\
0-3+3 & 0+0+0 & 0+2+9
\end{array}\right] \\
& =\left[\begin{array}{lll}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]
\end{aligned}
$
Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.
Ex 4.4 Question 5:
Find the inverse of each of the matrices (if it exists) $\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
Let $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
We know that
$
\begin{aligned}
& A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\
& \quad \begin{aligned}
|A| & =\left|\begin{array}{ll}
2 & 2 \\
4 & 3
\end{array}\right| \\
& =2 \times 3-4 \times(-2)=6+8=14
\end{aligned}
\end{aligned}
$
Since $|A| \neq 0, A^{-1}$ exists
Calculating adj $\mathrm{A}$
$
A=\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]
$
Interchange Change sign
$
\begin{aligned}
\operatorname{adj}(A) & = \\
& =\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]
\end{aligned}
$
Calculating A ${ }^{-1}$
$
\begin{aligned}
A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\
& =\frac{1}{14}\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]
\end{aligned}
$
Ex 4.4 Question 6:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$.
we have,
$
|A|=-2+15=13
$
Now,
$
\begin{aligned}
& A_{11}=2, A_{12}=3, A_{21}=-5, A_{22}=-1 \\
& \therefore \operatorname{adj} A=\left[\begin{array}{ll}
2 & -5 \\
3 & -1
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{13}\left[\begin{array}{ll}
2 & -5 \\
3 & -1
\end{array}\right]
\end{aligned}
$
Ex 4.4 Question 7:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$.
We have,
$
|A|=1(10-0)-2(0-0)+3(0-0)=10
$
Now,
$
\begin{aligned}
& A_{11}=10-0=10, A_{12}=-(0-0)=0, A_{13}=0-0=0 \\
& A_{21}=-(10-0)=-10, A_{22}=5-0=5, A_{23}=-(0-0)=0 \\
& A_{31}=8-6=2, A_{32}=-(4-0)=-4, A_{33}=2-0=2 \\
& \therefore \operatorname{adj} A=\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]
\end{aligned}
$
$
\therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{10}\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]
$
Ex 4.4 Question 8:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$.
We have,
$
|A|=1(-3-0)-0+0=-3
$
Now,
$
\begin{aligned}
& A_{11}=-3-0=-3, A_{12}=-(-3-0)=3, A_{13}=6-15=-9 \\
& A_{21}=-(0-0)=0, A_{22}=-1-0=-1, A_{23}=-(2-0)=-2 \\
& A_{31}=0-0=0, A_{32}=-(0-0)=0, A_{33}=3-0=3
\end{aligned}
$
$\begin{aligned}
& \therefore \text { adj } A=\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{3}\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right]
\end{aligned}$
Ex 4.4 Question 9:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{lll}
2 & 1 & 3 \\
4 & -1 & 0 \\
-7 & 2 & 1
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{lll}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]$.
We have,
$
\begin{aligned}
|A| & =2(-1-0)-1(4-0)+3(8-7) \\
& =2(-1)-1(4)+3(1) \\
& =-2-4+3 \\
& =-3
\end{aligned}
$
Now,
$
\begin{aligned}
& A_{11}=-1-0=-1, A_{12}=-(4-0)=-4, A_{13}=8-7=1 \\
& A_{21}=-(1-6)=5, A_{22}=2+21=23, A_{23}=-(4+7)=-11 \\
& A_{31}=0+3=3, A_{32}=-(0-12)=12, A_{33}=-2-4=-6 \\
& \therefore \text { adj } A=\left[\begin{array}{lll}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \text { adj } A=-\frac{1}{3}\left[\begin{array}{lll}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]
\end{aligned}
$
Ex 4.4 Question 10:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \text {. }
$
Answer
Let $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$.
By expanding along $\mathrm{C}_1$, we have:
$
|A|=1(8-6)-0+3(3-4)=2-3=-1
$
Now,
$
\begin{aligned}
& A_{11}=8-6=2, A_{12}=-(0+9)=-9, A_{13}=0-6=-6 \\
& A_{21}=-(-4+4)=0, A_{22}=4-6=-2, A_{23}=-(-2+3)=-1 \\
& A_{31}=3-4=-1, A_{32}=-(-3-0)=3, A_{33}=2-0=2
\end{aligned}
$
$
\begin{aligned}
& \therefore \text { adj } A=\left[\begin{array}{lll}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \text { adj } A=-\left[\begin{array}{lll}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]=\left[\begin{array}{lll}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]
\end{aligned}
$
Ex 4.4 Question 11:
Find the inverse of each of the matrices (if it exists).
$
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]
$
Answer
Let $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$.
We have,
$
|A|=1\left(-\cos ^2 \alpha-\sin ^2 \alpha\right)=-\left(\cos ^2 \alpha+\sin ^2 \alpha\right)=-1
$
Now,
$
\begin{aligned}
& A_{11}=-\cos ^2 \alpha-\sin ^2 \alpha=-1, A_{12}=0, A_{13}=0 \\
& A_{21}=0, A_{22}=-\cos \alpha, A_{23}=-\sin \alpha \\
& A_{31}=0, A_{32}=-\sin \alpha, A_{33}=\cos \alpha
\end{aligned}
$
$\begin{aligned}
& \therefore \operatorname{adj} A=\left[\begin{array}{lll}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A=-\left[\begin{array}{lll}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]
\end{aligned}$
Ex 4.4 Question 12:
Let $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]$. Verify that $(A B)^{-1}=B^{-1} A^{-1}$
Answer
Let $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$.
We have,
$
|A|=15-14=1
$
Now,
$
\begin{aligned}
& A_{11}=5, A_{12}=-2, A_{21}=-7, A_{22}=3 \\
& \therefore \text { adj } A=\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \cdot \text { adj } A=\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right]
\end{aligned}
$
Now, let $B=\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]$.
We have,
$
\begin{aligned}
& |B|=54-56=-2 \\
& \therefore \operatorname{adj} B=\left[\begin{array}{rr}
9 & -8 \\
-7 & 6
\end{array}\right]
\end{aligned}
$
$
\therefore B^{-1}=\frac{1}{|B|} \operatorname{adj} B=-\frac{1}{2}\left[\begin{array}{rr}
9 & -8 \\
-7 & 6
\end{array}\right]=\left[\begin{array}{cc}
-\frac{9}{2} & 4 \\
\frac{7}{2} & -3
\end{array}\right]
$
Now,
$
\begin{aligned}
B^{-1} A^{-1} & =\left[\begin{array}{cc}
-\frac{9}{2} & 4 \\
\frac{7}{2} & -3
\end{array}\right]\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
-\frac{45}{2}-8 & \frac{63}{2}+12 \\
\frac{35}{2}+6 & -\frac{49}{2}-9
\end{array}\right]=\left[\begin{array}{ll}
-\frac{61}{2} & \frac{87}{2} \\
\frac{47}{2} & -\frac{67}{2}
\end{array}\right]
\end{aligned}
$
Then,
$
\begin{aligned}
A B & =\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right] \\
& =\left[\begin{array}{ll}
18+49 & 24+63 \\
12+35 & 16+45
\end{array}\right] \\
& =\left[\begin{array}{ll}
67 & 87 \\
47 & 61
\end{array}\right]
\end{aligned}
$
Therefore, we have $|A B|=67 \times 61-87 \times 47=4087-4089=-2$.
Also,
$
\begin{aligned}
& \operatorname{adj}(A B)=\left[\begin{array}{rr}
61 & -87 \\
-47 & 67
\end{array}\right] \\
& \begin{aligned}
\therefore(A B)^{-1}=\frac{1}{|A B|} \operatorname{adj}(A B) & =-\frac{1}{2}\left[\begin{array}{ll}
61 & -87 \\
-47 & 67
\end{array}\right] \\
& =\left[\begin{array}{ll}
-\frac{61}{2} & \frac{87}{2} \\
\frac{47}{2} & -\frac{67}{2}
\end{array}\right] .
\end{aligned}
\end{aligned}
$
From (1) and (2), we have:
$
(A B)^{-1}=B^{-1} A^{-1}
$
Hence, the given result is proved.
Ex 4.4 Question 13:
If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^2-5 A+7 I=O$. Hence find $A^{-1}$.
Answer
$
\begin{aligned}
& A=\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right] \\
& A^2=A \cdot A=\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{ll}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right] \\
& \therefore A^2-5 A+7 I
\end{aligned}
$
$=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]$
$=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]$
$
=\left[\begin{array}{ll}
-7 & 0 \\
0 & -7
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
$
Hence, $A^2-5 A+7 I=O$.
$\begin{aligned}
& \therefore A \cdot A-5 A=-7 I \\
& \Rightarrow A \cdot A\left(A^{-1}\right)-5 A A^{-1}=-7 I A^{-1} \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \\
& \Rightarrow A\left(A A^{-1}\right)-5 I=-7 A^{-1} \\
& \Rightarrow A I-5 I=-7 A^{-1} \\
& \Rightarrow A^{-1}=-\frac{1}{7}(A-5 I) \\
& \Rightarrow A^{-1}=\frac{1}{7}(5 I-A)
\end{aligned}$
$\begin{aligned}
& =\frac{1}{7}\left(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]-\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\right)=\frac{1}{7}\left[\begin{array}{rr}
2 & -1 \\
1 & 3
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{7}\left[\begin{array}{rr}
2 & -1 \\
1 & 3
\end{array}\right]
\end{aligned}$
Ex 4.4 Question 14:
$
A=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right] \text {, find the numbers } a \text { and } b \text { such that } A^2+a A+b I=O .
$
Answer
$
\begin{aligned}
& A=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right] \\
& \therefore A^2=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
9+2 & 6+2 \\
3+1 & 2+1
\end{array}\right]=\left[\begin{array}{ll}
11 & 8 \\
4 & 3
\end{array}\right]
\end{aligned}
$
Now,
$
\begin{aligned}
& A^2+a A+b I=O \\
& \Rightarrow(A A) A^{-1}+a A A^{-1}+b I A^{-1}=O \quad \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \\
& \Rightarrow A\left(A A^{-1}\right)+a I+b\left(I A^{-1}\right)=O \\
& \Rightarrow A I+a I+b A^{-1}=O \\
& \Rightarrow A+a I=-b A^{-1} \\
& \Rightarrow A^{-1}=-\frac{1}{b}(A+a I)
\end{aligned}
$
Now,
$
A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{1}\left[\begin{array}{rr}
1 & -2 \\
-1 & 3
\end{array}\right]=\left[\begin{array}{rc}
1 & -2 \\
-1 & 3
\end{array}\right]
$
We have:
$
\left[\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right]=-\frac{1}{b}\left(\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]+\left[\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right]\right)=-\frac{1}{b}\left[\begin{array}{ll}
3+a & 2 \\
1 & 1+a
\end{array}\right]=\left[\begin{array}{cc}
\frac{-3-a}{b} & -\frac{2}{b} \\
-\frac{1}{b} & \frac{-1-a}{b}
\end{array}\right]
$
Comparing the corresponding elements of the two matrices, we have:
$
\begin{aligned}
& -\frac{1}{b}=-1 \Rightarrow b=1 \\
& \frac{-3-a}{b}=1 \Rightarrow-3-a=1 \Rightarrow a=-4
\end{aligned}
$
Hence, -4 and 1 are the required values of $a$ and $b$ respectively.
Ex 4.4 Question 15:
For the matrix $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]_{\text {show that }} A^3-6 A^2+5 A+11 I=0$. Hence, $A^{-1}$.
Answer
$\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \\
A^2 & =\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
1+1+2 & 1+2-1 & 1-3+3 \\
1+2-6 & 1+4+3 & 1-6-9 \\
2-1+6 & 2-2-3 & 2+3+9
\end{array}\right]=\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]
\end{aligned}$
$\begin{aligned}
& A^3=A^2 \cdot A=\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \\
&=\left[\begin{array}{lll}
4+2+2 & 4+4-1 & 4-6+3 \\
-3+8-28 & -3+16+14 & -3-24-42 \\
7-3+28 & 7-6-14 & 7+9+42
\end{array}\right] \\
&=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right] \\
& \therefore A^3-6 A^2+5 A+11 I
\end{aligned}$
$\begin{aligned}
& =\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+11\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right]+\left[\begin{array}{ccc}
5 & 5 & 5 \\
5 & 10 & -15 \\
10 & -5 & 15
\end{array}\right]+\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right] \\
& =\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right]-\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right] \\
& =\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=O
\end{aligned}$
Thus, $A^3-6 A^2+5 A+11 I=O$.
Now,
$
A^3-6 A^2+5 A+11 I=O
$
$
\begin{aligned}
& \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+5 A A^{-1}+11 L A^{-1}=0 \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \\
& \Rightarrow A A\left(A A^{-1}\right)-6 A\left(A A^{-1}\right)+5\left(A A^{-1}\right)=-11\left(I A^{-1}\right) \\
& \Rightarrow A^2-6 A+5 I=-11 A^{-1} \\
& \Rightarrow 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] \\
& =\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}
9 & 2 & 1 \\
-3 & 13 & -14 \\
7 & -3 & 19
\end{array}\right]-\left[\begin{array}{ccc}
6 & 6 & 6 \\
6 & 12 & -18 \\
12 & -6 & 18
\end{array}\right] \\
& =\left[\begin{array}{ccc}
3 & -4 & -5 \\
-9 & 1 & 4 \\
-5 & 3 & 1
\end{array}\right]
\end{aligned}
$
From equation (1), we have:
$A^{-1}=-\frac{1}{11}\left[\begin{array}{lll}
3 & -4 & -5 \\
-9 & 1 & 4 \\
-5 & 3 & 1
\end{array}\right]=\frac{1}{11}\left[\begin{array}{lll}
-3 & 4 & 5 \\
9 & -1 & -4 \\
5 & -3 & -1
\end{array}\right]$
Ex 4.4 Question 16:
$
A=\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]_{\text {verify that }} A^3-6 A^2+9 A-4 I=O \text { and hence find } A^{-1}
$
Answer
$
\begin{aligned}
A & =\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
A^2 & =\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
& =\left[\begin{array}{lll}
4+1+1 & -2-2-1 & 2+1+2 \\
-2-2-1 & 1+4+1 & -1-2-2 \\
2+1+2 & -1-2-2 & 1+1+4
\end{array}\right] \\
& =\left[\begin{array}{lll}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]
\end{aligned}
$
$\begin{aligned}
A^3=A^2 A & =\left[\begin{array}{lll}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
& =\left[\begin{array}{lll}
12+5+5 & -6-10-5 & 6+5+10 \\
-10-6-5 & 5+12+5 & -5-6-10 \\
10+5+6 & -5-10-6 & 5+5+12
\end{array}\right] \\
& =\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]
\end{aligned}$
Now,
$
\begin{aligned}
& 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 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-\left[\begin{array}{ccc}
36 & -30 & 30 \\
-30 & 36 & -30 \\
30 & -30 & 36
\end{array}\right]+\left[\begin{array}{ccc}
18 & -9 & 9 \\
-9 & 18 & -9 \\
9 & -9 & 18
\end{array}\right]-\left[\begin{array}{ccc}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right] \\
& =\left[\begin{array}{ccc}
40 & -30 & 30 \\
-30 & 40 & -30 \\
30 & -30 & 40
\end{array}\right]-\left[\begin{array}{ccc}
40 & -30 & 30 \\
-30 & 40 & -30 \\
30 & -30 & 40
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& \therefore A^3-6 A^2+9 A-4 I=O
\end{aligned}
$
Now,
$
\begin{aligned}
& A^3-6 A^2+9 A-4 I=O \\
& \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=O \quad \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \\
& \Rightarrow A A\left(A A^{-1}\right)-6 A\left(A A^{-1}\right)+9\left(A A^{-1}\right)=4\left(I A^{-1}\right) \\
& \Rightarrow A A I-6 A I+9 I=4 A^{-1} \\
& \Rightarrow A^2-6 A+9 I=4 A^{-1} \\
& \Rightarrow A^{-1}=\frac{1}{4}\left(A^2-6 A+9 I\right) \\
& A^2-6 A+9 I
\end{aligned}
$
$
\begin{aligned}
& =\left[\begin{array}{lll}
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}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& =\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-\left[\begin{array}{ccc}
12 & -6 & 6 \\
-6 & 12 & -6 \\
6 & -6 & 12
\end{array}\right]+\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right] \\
& =\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]
\end{aligned}
$
From equation (1), we have:
$
A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]
$
Ex 4.4 Question 17:
Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|\operatorname{adj} A|$ is equal to
A. $|A|$
B. $|A|^2$
c. $|A|$
D. $3|A|$
We know that,
$
\begin{aligned}
& (\operatorname{adj} A) A=|A| I=\left[\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right] \\
& \Rightarrow|(\operatorname{adj} A) A|=\left|\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right| \\
& \Rightarrow|\operatorname{adj} A||A|=|A|^3\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|=|A|^3(I) \\
& \therefore|\operatorname{adj} A|=|A|^2 \\
&
\end{aligned}
$
Hence, the correct answer is B.
Ex 4.4 Question 18:
If $A$ is an invertible matrix of order 2 , then $\operatorname{det}\left(A^{-1}\right)$ is equal to
A. $\operatorname{det}(A)$
B. $\frac{1}{\operatorname{det}(A)}$
C. 1
D. 0
Answer
$
A^{-1} \text { exists and } A^{-1}=\frac{1}{|A|} \text { adj } A \text {. }
$
Since $A$ is an invertible matrix,
As matrix $A$ is of order 2 , let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$.
Then, $|A|=a d-b c$ and $a d j A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.
Now,
$A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left[\begin{array}{cc}
\frac{d}{|A|} & \frac{-b}{|A|} \\
\frac{-c}{|A|} & \frac{a}{|A|}
\end{array}\right]$
$
\begin{aligned}
& \therefore\left|A^{-1}\right|=\left|\begin{array}{ll}
\frac{d}{|A|} & \frac{-b}{|A|} \\
\frac{-c}{|A|} & \frac{a}{|A|}
\end{array}\right|=\frac{1}{|A|^2}\left|\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right|=\frac{1}{|A|^2}(a d-b c)=\frac{1}{\mid A^2} \cdot|A|=\frac{1}{|A|} \\
& \therefore \operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}
\end{aligned}
$
Hence, the correct answer is B.
