Q: Find the adjoint of the matrix $\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$

Let $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$ Then, $$ \begin{array}{ll} A_{11}=4 & A_{12}=-3 \\ A_{21}=-2 & A_{22}=1 \end{array} $$ Therefore, $$ \begin{aligned} \operatorname{adj} A & =\left(\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right) \\ & =\left(\begin{array}{cc} 4 & -2 \\ -3 & 1 \end{array}\right) \end{aligned} $$

Q: Find the adjoint of the matrix $\left(\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right)$

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right)$ Then, $$ \begin{array}{lll} A_{11}=\left|\begin{array}{ll} 3 & 5 \\ 0 & 1 \end{array}\right|=3 & A_{12}=-\left|\begin{array}{cc} 2 & 5 \\ -2 & 1 \end{array}\right|=-12 & A_{13}=\left|\begin{array}{cc} 2 & 3 \\ -2 & 0 \end{array}\right|=6 \\ A_{21}=-\left|\begin{array}{cc} -1 & 2 \\ 0 & 1 \end{array}\right|=1 & A_{22}=\left|\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right|=5 & A_{23}=-\left|\begin{array}{cc} 1 & -1 \\ -2 & 0 \end{array}\right|=2 \\ A_{31}=\left|\begin{array}{cc} -1 & 2 \\ 3 & 5 \end{array}\right|=-11 & A_{32}=-\left|\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right|=-1 & A_{33}=\left|\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right|=5 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ccc} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{array}\right) $$

Q: Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$ for $\left(\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right)$

Let $A=\left(\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right)$ Then, $$ \begin{aligned} |A| & =-12-(-12) \\ & =0 \end{aligned} $$ Also, $$ \begin{aligned} |A| I & =0\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ & =\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned} $$ Now, $$ \begin{array}{ll} A_{11}=-6 & A_{12}=4 \\ A_{21}=-3 & A_{22}=2 \end{array} $$ Hence, $$ \operatorname{adj} A=\left(\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right) $$ Now, $$ \begin{aligned} A(\operatorname{adj} A) & =\left(\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right)\left(\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right) \\ & =\left(\begin{array}{cc} -12+12 & -6+6 \\ 24-24 & 12-12 \end{array}\right) \\ & =\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned} $$ Also, $$ \begin{aligned} (\operatorname{adj} A) A & =\left(\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right)\left(\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right) \\ & =\left(\begin{array}{cc} -12+12 & -18+18 \\ 8-8 & 12-12 \end{array}\right) \\ & =\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned} $$ Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.

Q: Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$ for $\left(\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right)$

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right)$ Then, $$ \begin{aligned} |A| & =1(0-0)+1(9+2)+2(0-0) \\ & =11 \end{aligned} $$ Also, $$ \begin{aligned} |A| I & =11\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ & =\left(\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right) \end{aligned} $$ Now, $$ \begin{array}{lll} A_{11}=0 & A_{12}=-11 & A_{13}=0 \\ A_{21}=3 & A_{22}=1 & A_{23}=-1 \\ A_{31}=2 & A_{32}=8 & A_{33}=3 \end{array} $$ Hence, $$ \operatorname{adj} A=\left(\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right) $$ Now, $$ \begin{aligned} A(\operatorname{adj} A) & =\left(\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right)\left(\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right) \\ & =\left(\begin{array}{ccc} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{array}\right)=\left(\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right) \end{aligned} $$ Also, $$ \begin{aligned} (\operatorname{adj} A) A & =\left(\begin{array}{ccc} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{array}\right)\left(\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right) \\ & =\left(\begin{array}{ccc} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 2+0+9 \end{array}\right) \\ & =\left(\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right) \end{aligned} $$ Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.

Q: Find the inverse of each of the matrix $\left(\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right)$ (if it exists).

Let $A=\left(\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right)$ Then, $$ \begin{aligned} |A| & =6+8 \\ & =14 \end{aligned} $$ Now, $$ \begin{array}{ll} A_{11}=3 & A_{12}=-4 \\ A_{21}=2 & A_{22}=2 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =\frac{1}{14}\left(\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right) \end{aligned} $$

Q: Find the inverse of each of the matrix $\left(\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right)$ (if it exists)

Let $A=\left(\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right)$ Then, $$ |A|=-2+15=13 $$ Now, $$ \begin{array}{ll} A_{11}=2 & A_{12}=3 \\ A_{21}=-5 & A_{22}=-1 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =\frac{1}{13}\left(\begin{array}{ll} 2 & -5 \\ 3 & -1 \end{array}\right) \end{aligned} $$

Q: Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right)$ (if it exists)

Let $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right)$ Then, $$ \begin{aligned} |A| & =1(10-0)-2(0-0)+3(0-0) \\ & =10 \end{aligned} $$ Now, $$ \begin{array}{lll} A_{11}=10 & A_{12}=0 & A_{13}=0 \\ A_{21}=-10 & A_{22}=5 & A_{23}=0 \\ A_{31}=2 & A_{32}=-4 & A_{33}=2 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =\frac{1}{10}\left(\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right) \end{aligned} $$

Q: Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right)$ (if it exists)

Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right)$ Then, $$ |A|=1(-3-0)-0+0=-3 $$ Now, $$ \begin{array}{lll} A_{11}=-3 & A_{12}=3 & A_{13}=-9 \\ A_{21}=0 & A_{22}=-1 & A_{23}=-2 \\ A_{31}=0 & A_{32}=0 & A_{33}=3 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ccc} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =\frac{-1}{3}\left(\begin{array}{ccc} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right) \end{aligned} $$

Q: Find the inverse of each of the matrix $\left(\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right)$ (if it exists)

Let $A=\left(\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right)$ Then, $$ \begin{aligned} |A| & =2(-1-0)-1(4-0)+3(8-7) \\ & =2(-1)-1(4)+3(1) \\ & =-3 \end{aligned} $$ Now, $$ \begin{array}{lll} A_{11}=-1 & A_{12}=-4 & A_{13}=1 \\ A_{21}=5 & A_{22}=23 & A_{23}=-11 \\ A_{31}=3 & A_{32}=12 & A_{33}=-6 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ccc} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =-\frac{1}{3}\left(\begin{array}{ccc} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right) \end{aligned} $$

Q: Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right)$ (if it exists)

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right)$ Then, expanding along $C_{1}$, $$ \begin{aligned} |A| & =1(8-6)-0+3(3-4)=2-3 \\ & =-1 \end{aligned} $$ Now, $$ \begin{array}{llr} A_{11}=2 & A_{12}=-9 & A_{13}=-6 \\ A_{21}=0 & A_{22}=-2 & A_{23}=-1 \\ A_{31}=-1 & A_{32}=3 & A_{33}=2 \end{array} $$ Therefore, $$ \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right) $$ Hence, $$ \begin{aligned} A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\ & =-1\left(\begin{array}{ccc} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right) \\ & =\left(\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right) \end{aligned} $$

Question 1

Find the adjoint of the matrix $\left(\begin{array}{ll}1 & 2 \ 3 & 4\end{array}ight)$

Accepted Answer

Let $A=\left(\begin{array}{ll}1 & 2 \ 3 & 4\end{array}ight)$

Then,

$$

\begin{array}{ll}

A_{11}=4 & A_{12}=-3 \

A_{21}=-2 & A_{22}=1

\end{array}

$$

Therefore,

$$

\begin{aligned}

\operatorname{adj} A & =\left(\begin{array}{ll}

A_{11} & A_{12} \

A_{21} & A_{22}

\end{array}ight) \

& =\left(\begin{array}{cc}

4 & -2 \

-3 & 1

\end{array}ight)

\end{aligned}

$$

Question 2

Find the adjoint of the matrix $\left(\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}ight)$

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}ight)$

Then,

$$

\begin{array}{lll}

A_{11}=\left|\begin{array}{ll}

3 & 5 \

0 & 1

\end{array}ight|=3 & A_{12}=-\left|\begin{array}{cc}

2 & 5 \

-2 & 1

\end{array}ight|=-12 & A_{13}=\left|\begin{array}{cc}

2 & 3 \

-2 & 0

\end{array}ight|=6 \

A_{21}=-\left|\begin{array}{cc}

-1 & 2 \

0 & 1

\end{array}ight|=1 & A_{22}=\left|\begin{array}{cc}

1 & 2 \

-2 & 1

\end{array}ight|=5 & A_{23}=-\left|\begin{array}{cc}

1 & -1 \

-2 & 0

\end{array}ight|=2 \

A_{31}=\left|\begin{array}{cc}

-1 & 2 \

3 & 5

\end{array}ight|=-11 & A_{32}=-\left|\begin{array}{ll}

1 & 2 \

2 & 5

\end{array}ight|=-1 & A_{33}=\left|\begin{array}{cc}

1 & -1 \

2 & 3

\end{array}ight|=5

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

3 & 1 & -11 \

-12 & 5 & -1 \

6 & 2 & 5

\end{array}ight)

$$

Question 3

Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$ for $\left(\begin{array}{cc}2 & 3 \ -4 & -6\end{array}ight)$

Accepted Answer

Let $A=\left(\begin{array}{cc}2 & 3 \ -4 & -6\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =-12-(-12) \

& =0

\end{aligned}

$$

Also,

$$

\begin{aligned}

|A| I & =0\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) \

& =\left(\begin{array}{ll}

0 & 0 \

0 & 0

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{array}{ll}

A_{11}=-6 & A_{12}=4 \

A_{21}=-3 & A_{22}=2

\end{array}

$$

Hence,

$$

\operatorname{adj} A=\left(\begin{array}{cc}

-6 & -3 \

4 & 2

\end{array}ight)

$$

Now,

$$

\begin{aligned}

A(\operatorname{adj} A) & =\left(\begin{array}{cc}

2 & 3 \

-4 & -6

\end{array}ight)\left(\begin{array}{cc}

-6 & -3 \

4 & 2

\end{array}ight) \

& =\left(\begin{array}{cc}

-12+12 & -6+6 \

24-24 & 12-12

\end{array}ight) \

& =\left(\begin{array}{ll}

0 & 0 \

0 & 0

\end{array}ight)

\end{aligned}

$$

Also,

$$

\begin{aligned}

(\operatorname{adj} A) A & =\left(\begin{array}{cc}

-6 & -3 \

4 & 2

\end{array}ight)\left(\begin{array}{cc}

2 & 3 \

-4 & -6

\end{array}ight) \

& =\left(\begin{array}{cc}

-12+12 & -18+18 \

8-8 & 12-12

\end{array}ight) \

& =\left(\begin{array}{ll}

0 & 0 \

0 & 0

\end{array}ight)

\end{aligned}

$$

Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.

Question 4

Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$ for $\left(\begin{array}{ccc}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}ight)$

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =1(0-0)+1(9+2)+2(0-0) \

& =11

\end{aligned}

$$

Also,

$$

\begin{aligned}

|A| I & =11\left(\begin{array}{lll}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

11 & 0 & 0 \

0 & 11 & 0 \

0 & 0 & 11

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

A_{11}=0 & A_{12}=-11 & A_{13}=0 \

A_{21}=3 & A_{22}=1 & A_{23}=-1 \

A_{31}=2 & A_{32}=8 & A_{33}=3

\end{array}

$$

Hence,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

0 & 3 & 2 \

-11 & 1 & 8 \

0 & -1 & 3

\end{array}ight)

$$

Now,

$$

\begin{aligned}

A(\operatorname{adj} A) & =\left(\begin{array}{ccc}

1 & -1 & 2 \

3 & 0 & -2 \

1 & 0 & 3

\end{array}ight)\left(\begin{array}{ccc}

0 & 3 & 2 \

-11 & 1 & 8 \

0 & -1 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

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}ight)=\left(\begin{array}{ccc}

11 & 0 & 0 \

0 & 11 & 0 \

0 & 0 & 11

\end{array}ight)

\end{aligned}

$$

Also,

$$

\begin{aligned}

(\operatorname{adj} A) A & =\left(\begin{array}{ccc}

0 & 3 & 2 \

-11 & 1 & 8 \

0 & -1 & 3

\end{array}ight)\left(\begin{array}{ccc}

1 & -1 & 2 \

3 & 0 & -2 \

1 & 0 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

0+9+2 & 0+0+0 & 0-6+6 \

-11+3+8 & 11+0+0 & -22-2+24 \

0-3+3 & 0+0+0 & 2+0+9

\end{array}ight) \

& =\left(\begin{array}{ccc}

11 & 0 & 0 \

0 & 11 & 0 \

0 & 0 & 11

\end{array}ight)

\end{aligned}

$$

Hence, $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$.

Question 5

Find the inverse of each of the matrix $\left(\begin{array}{cc}2 & -2 \ 4 & 3\end{array}ight)$ (if it exists).

Accepted Answer

Let $A=\left(\begin{array}{cc}2 & -2 \ 4 & 3\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =6+8 \

& =14

\end{aligned}

$$

Now,

$$

\begin{array}{ll}

A_{11}=3 & A_{12}=-4 \

A_{21}=2 & A_{22}=2

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{cc}

3 & 2 \

-4 & 2

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =\frac{1}{14}\left(\begin{array}{cc}

3 & 2 \

-4 & 2

\end{array}ight)

\end{aligned}

$$

Question 6

Find the inverse of each of the matrix $\left(\begin{array}{ll}-1 & 5 \ -3 & 2\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{ll}-1 & 5 \ -3 & 2\end{array}ight)$

Then,

$$

|A|=-2+15=13

$$

Now,

$$

\begin{array}{ll}

A_{11}=2 & A_{12}=3 \

A_{21}=-5 & A_{22}=-1

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ll}

2 & -5 \

3 & -1

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =\frac{1}{13}\left(\begin{array}{ll}

2 & -5 \

3 & -1

\end{array}ight)

\end{aligned}

$$

Question 7

Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =1(10-0)-2(0-0)+3(0-0) \

& =10

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

A_{11}=10 & A_{12}=0 & A_{13}=0 \

A_{21}=-10 & A_{22}=5 & A_{23}=0 \

A_{31}=2 & A_{32}=-4 & A_{33}=2

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

10 & -10 & 2 \

0 & 5 & -4 \

0 & 0 & 2

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

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}ight)

\end{aligned}

$$

Question 8

Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}ight)$

Then,

$$

|A|=1(-3-0)-0+0=-3

$$

Now,

$$

\begin{array}{lll}

A_{11}=-3 & A_{12}=3 & A_{13}=-9 \

A_{21}=0 & A_{22}=-1 & A_{23}=-2 \

A_{31}=0 & A_{32}=0 & A_{33}=3

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

-3 & 0 & 0 \

3 & -1 & 0 \

-9 & -2 & 3

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

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}ight)

\end{aligned}

$$

Question 9

Find the inverse of each of the matrix $\left(\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =2(-1-0)-1(4-0)+3(8-7) \

& =2(-1)-1(4)+3(1) \

& =-3

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

A_{11}=-1 & A_{12}=-4 & A_{13}=1 \

A_{21}=5 & A_{22}=23 & A_{23}=-11 \

A_{31}=3 & A_{32}=12 & A_{33}=-6

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

-1 & 5 & 3 \

-4 & 23 & 12 \

1 & -11 & -6

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =-\frac{1}{3}\left(\begin{array}{ccc}

-1 & 5 & 3 \

-4 & 23 & 12 \

1 & -11 & -6

\end{array}ight)

\end{aligned}

$$

Question 10

Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight)$

Then, expanding along $C_{1}$,

$$

\begin{aligned}

|A| & =1(8-6)-0+3(3-4)=2-3 \

& =-1

\end{aligned}

$$

Now,

$$

\begin{array}{llr}

A_{11}=2 & A_{12}=-9 & A_{13}=-6 \

A_{21}=0 & A_{22}=-2 & A_{23}=-1 \

A_{31}=-1 & A_{32}=3 & A_{33}=2

\end{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

2 & 0 & -1 \

-9 & -2 & 3 \

-6 & -1 & 2

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =-1\left(\begin{array}{ccc}

2 & 0 & -1 \

-9 & -2 & 3 \

-6 & -1 & 2

\end{array}ight) \

& =\left(\begin{array}{ccc}

-2 & 0 & 1 \

9 & 2 & -3 \

6 & 1 & -2

\end{array}ight)

\end{aligned}

$$

Question 11

Find the inverse of each of the matrix $\left(\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}ight)$ (if it exists)

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =1\left(-\cos ^{2} \alpha-\sin ^{2} \alphaight)=-\left(\cos ^{2} \alpha+\sin ^{2} \alphaight) \

& =-1

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

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{array}

$$

Therefore,

$$

\operatorname{adj} A=\left(\begin{array}{ccc}

-1 & 0 & 0 \

0 & -\cos \alpha & -\sin \alpha \

0 & -\sin \alpha & \cos \alpha

\end{array}ight)

$$

Hence,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =-1\left(\begin{array}{ccc}

-1 & 0 & 0 \

0 & -\cos \alpha & -\sin \alpha \

0 & -\sin \alpha & \cos \alpha

\end{array}ight) \

& =\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & \cos \alpha & \sin \alpha \

0 & \sin \alpha & -\cos \alpha

\end{array}ight)

\end{aligned}

$$

Question 12

Let $A=\left(\begin{array}{ll}3 & 7 \ 2 & 5\end{array}ight)$ and $B=\left(\begin{array}{ll}6 & 8 \ 7 & 9\end{array}ight)$. Verify that $(A B)^{-1}=B^{-1} A^{-1}$.

Accepted Answer

Let $A=\left(\begin{array}{ll}3 & 7 \ 2 & 5\end{array}ight)$

Then,

$$

\begin{aligned}

|A| & =15-14 \

& =1

\end{aligned}

$$

Now,

$$

\begin{array}{lr}

A_{11}=5 & A_{12}=-2 \

A_{21}=-7 & A_{22}=3

\end{array}

$$

Then,

$$

\operatorname{adj} A=\left(\begin{array}{cc}

5 & -7 \

-2 & 3

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =\left(\begin{array}{cc}

5 & -7 \

-2 & 3

\end{array}ight)

\end{aligned}

$$

Now,

Let $B=\left(\begin{array}{ll}6 & 8 \ 7 & 9\end{array}ight)$

Then,

$$

\begin{aligned}

|B| & =54-56 \

& =-2

\end{aligned}

$$

Now,

$$

\begin{array}{ll}

A_{11}=9 & A_{12}=-7 \

A_{21}=-8 & A_{22}=6

\end{array}

$$

Then,

$$

\operatorname{adjB}=\left(\begin{array}{cc}

9 & -8 \

-7 & 6

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

B^{-1} & =\frac{1}{|B|} \operatorname{adj} B \

& =-\frac{1}{2}\left(\begin{array}{cc}

9 & -8 \

-7 & 6

\end{array}ight) \

& =\left(\begin{array}{cc}

-\frac{9}{2} & 4 \

\frac{7}{2} & -3

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{align*}

B^{-1} A^{-1} & =\left(\begin{array}{cc}

-\frac{9}{2} & 4 \

\frac{7}{2} & -3

\end{array}ight)\left(\begin{array}{cc}

5 & -7 \

-2 & 3

\end{array}ight) \

& =\left(\begin{array}{cc}

-\frac{45}{2}-8 & \frac{63}{2}+12 \

\frac{35}{2}+6 & -\frac{49}{2}-9

\end{array}ight) \

& =\left(\begin{array}{cc}

-\frac{61}{2} & \frac{87}{2} \

\frac{47}{2} & \frac{-67}{2}

\end{array}ight) 	ag{1}

\end{align*}

$$

Also,

$$

\begin{aligned}

A B & =\left(\begin{array}{ll}

3 & 7 \

2 & 5

\end{array}ight)\left(\begin{array}{ll}

6 & 8 \

7 & 9

\end{array}ight) \

& =\left(\begin{array}{ll}

18+49 & 24+63 \

12+35 & 16+45

\end{array}ight) \

& =\left(\begin{array}{ll}

67 & 87 \

47 & 61

\end{array}ight)

\end{aligned}

$$

Then, we have

$$

\begin{aligned}

|A B| & =67(61)-87(47) \

& =4087-4089 \

& =-2

\end{aligned}

$$

Therefore,

$$

\operatorname{adj}(A B)=\left(\begin{array}{cc}

61 & -87 \

-47 & 67

\end{array}ight)

$$

Thus,

$$

\begin{align*}

(A B)^{-1} & =\frac{1}{|A B|} \operatorname{adj}(A B) \

& =-\frac{1}{2}\left(\begin{array}{cc}

61 & -87 \

-47 & 67

\end{array}ight) \

& =\left(\begin{array}{cc}

-\frac{61}{2} & \frac{87}{2} \

\frac{47}{2} & -\frac{67}{2}

\end{array}ight) 	ag{2}

\end{align*}

$$

From (1) and (2),

$$

(A B)^{-1}=B^{-1} A^{-1}

$$

Hence, proved.

Question 13

If $A=\left(\begin{array}{cc}3 & 1 \ -1 & 2\end{array}ight)$, show that $A^{2}-5 A+7 I=0$. Hence find $A^{-1}$.

Accepted Answer

Let $A=\left(\begin{array}{cc}3 & 1 \ -1 & 2\end{array}ight)$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A=\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight)\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight) \

& =\left(\begin{array}{cc}

9-1 & 3+2 \

-3-2 & -1+4

\end{array}ight) \

& =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

A^{2}-5 A+7 I & =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)-5\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight)+7\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) \

& =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)-\left(\begin{array}{cc}

15 & 5 \

-5 & 10

\end{array}ight)+\left(\begin{array}{ll}

7 & 0 \

0 & 7

\end{array}ight) \

& =\left(\begin{array}{cc}

-7 & 0 \

0 & -7

\end{array}ight)+\left(\begin{array}{cc}

7 & 0 \

0 & 7

\end{array}ight) \

& =\left(\begin{array}{ll}

0 & 0 \

0 & 0

\end{array}ight)

\end{aligned}

$$

Hence, $A^{2}-5 A+7 I=0$.

Now,

$$

\begin{aligned}

& \Rightarrow A \cdot A-5 A=-7 I \

& \Rightarrow A \cdot A\left(A^{-1}ight)-5 A \cdot A^{-1}=-7 I A^{-1} \quad\left[	ext { post-multiplying by } A^{-1} 	ext { as }|A| 
eq 0ight] \

& \Rightarrow A\left(A A^{-1}ight)-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) \

& \Rightarrow A^{-1}=\frac{1}{7}\left[\left(\begin{array}{cc}

5 & 0 \

0 & 5

\end{array}ight)-\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight)ight] \

& \Rightarrow A^{-1}=\frac{1}{7}\left(\begin{array}{cc}

2 & -1 \

1 & 3

\end{array}ight)

\end{aligned}

$$

Thus,

$$

A^{-1}=\frac{1}{7}\left(\begin{array}{cc}

2 & -1 \

1 & 3

\end{array}ight)

$$

Question 14

For the matrix $A=\left(\begin{array}{ll}3 & 2 \ 1 & 1\end{array}ight)$, find the numbers $a$ and $b$ such that $A^{2}+a A+b I=0$.

Accepted Answer

Let $A=\left(\begin{array}{ll}3 & 2 \ 1 & 1\end{array}ight)$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A=\left(\begin{array}{ll}

3 & 2 \

1 & 1

\end{array}ight)\left(\begin{array}{ll}

3 & 2 \

1 & 1

\end{array}ight) \

& =\left(\begin{array}{ll}

9+2 & 6+2 \

3+1 & 2+1

\end{array}ight)=\left(\begin{array}{ll}

11 & 8 \

4 & 3

\end{array}ight)

\end{aligned}

$$

Now, $A^{2}+a A+b I=0$.

Hence,

$$

\begin{align*}

& \Rightarrow(A \cdot A) A^{-1}+a A \cdot A^{-1}+b I A^{-1}=0 \quad\left[	ext { post-multiplying by } A^{-1} 	ext { as }|A| 
eq 0ight] \

& \Rightarrow A\left(A A^{-1}ight)+a I+b\left(I A^{-1}ight)=0 \

& \Rightarrow A I+a I+b A^{-1}=0 \

& \Rightarrow A+a I=-b A^{-1} \

& \Rightarrow A^{-1}=-\frac{1}{b}(A+a I) \quad \ldots(1) 	ag{1}

\end{align*}

$$

Now,

$$

\begin{align*}

A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \

& =\frac{1}{1}\left(\begin{array}{cc}

1 & -2 \

-1 & 3

\end{array}ight) \

& =\left(\begin{array}{cc}

1 & -2 \

-1 & 3

\end{array}ight) 	ag{2}

\end{align*}

$$

From (1) and (2), we have,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{cc}

1 & -2 \

-1 & 3

\end{array}ight)=\frac{1}{b}\left[\left(\begin{array}{ll}

3 & 2 \

1 & 1

\end{array}ight)+\left(\begin{array}{cc}

a & 0 \

0 & a

\end{array}ight)ight] \

& \Rightarrow\left(\begin{array}{cc}

1 & -2 \

-1 & 3

\end{array}ight)=-\frac{1}{b}\left(\begin{array}{cc}

3+a & 2 \

1 & a

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

1 & -2 \

-1 & 3

\end{array}ight)=\left(\begin{array}{cc}

\frac{-3-a}{b} & -\frac{2}{b} \

-\frac{1}{b} & \frac{-1-a}{b}

\end{array}ight)

\end{aligned}

$$

Comparing the corresponding elements of the two matrices, we have:

$$

\begin{aligned}

& \Rightarrow-\frac{1}{b}=-1 \

& \Rightarrow h=1

\end{aligned}

$$

Also,

$$

\begin{aligned}

& \Rightarrow \frac{-3-a}{b}=1 \

& \Rightarrow-3-a=1 \

& \Rightarrow a=-4

\end{aligned}

$$

Thus, $a=-4$ and $b=1$.

Question 15

For the matrix $A=\left(\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}ight)$, show that $A^{3}-6 A^{2}+5 A+11 I=0$. Hence, find $A^{-1}$.

Accepted Answer

Let $A=\left(\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}ight)$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A=\left(\begin{array}{ccc}

1 & 1 & 1 \

1 & 2 & -3 \

2 & -1 & 3

\end{array}ight)\left(\begin{array}{ccc}

1 & 1 & 1 \

1 & 2 & -3 \

2 & -1 & 3

\end{array}ight) \

& =\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}ight)=\left(\begin{array}{ccc}

4 & 2 & 1 \

-3 & 8 & -14 \

7 & -3 & 14

\end{array}ight)

\end{aligned}

$$

And,

$$

\begin{aligned}

A^{3} & =A^{2} \cdot A=\left(\begin{array}{ccc}

4 & 2 & 1 \

-3 & 8 & -14 \

7 & -3 & 14

\end{array}ight)\left(\begin{array}{ccc}

1 & 1 & 1 \

1 & 2 & -3 \

2 & -1 & 3

\end{array}ight) \

& =\left(\begin{array}{cc}

4+2+2 & 4+4-1 \

-3+8-28 & 4-6+3 \

7-3+28 & 7-6-14

\end{array}ight)=\left(\begin{array}{ccc}

8 & 7 & 1 \

-23 & 27 & -69 \

32 & -13 & 58

\end{array}ight)

\end{aligned}

$$

Hence,

$$

\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}ight)-6\left(\begin{array}{ccc}

4 & 2 & 1 \

-3 & 8 & -14 \

7 & -3 & 14

\end{array}ight)+5\left(\begin{array}{ccc}

1 & 1 & 1 \

1 & 2 & -3 \

2 & -1 & 3

\end{array}ight)+11\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

8 & 7 & 1 \

-23 & 27 & -69 \

32 & -13 & 58

\end{array}ight)-\left(\begin{array}{ccc}

24 & 12 & 6 \

-18 & 48 & -84 \

42 & -18 & 84

\end{array}ight)+\left(\begin{array}{ccc}

5 & 5 & 5 \

5 & 10 & -15 \

10 & -5 & 15

\end{array}ight)+\left(\begin{array}{ccc}

11 & 0 & 0 \

0 & 11 & 0 \

0 & 0 & 11

\end{array}ight) \

& =\left(\begin{array}{ccc}

24 & 12 & 6 \

-18 & 48 & -84 \

42 & -18 & 84

\end{array}ight)-\left(\begin{array}{ccc}

24 & 12 & 6 \

-18 & 48 & -84 \

42 & -18 & 84

\end{array}ight) \

& =\left(\begin{array}{ccc}

0 & 0 & 0 \

0 & 0 & 0

\end{array}ight) \

& =0

\end{aligned}

$$

Thus, $A^{3}-6 A^{2}+5 A+11 I=0$

Now,

$$

\begin{align*}

& \Rightarrow A^{3}-6 A^{2}+5 A+11 I=0 \

& \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+5 A A^{-1}+11 I A^{-1}=0 \quad\left[	ext { post-multiplying by } A^{-1} 	ext { as }|A| 
eq 0ight] \

& \Rightarrow A A\left(A A^{-1}ight)-6 A\left(A A^{-1}ight)+5\left(A A^{-1}ight)=-11\left(I A^{-1}ight) \

& \Rightarrow A^{2}-6 A+5 I=-11 A^{-1} \

& \Rightarrow A^{-1}=-\frac{1}{11}\left(A^{2}-6 A+5 Iight) 	ag{1}

\end{align*}

$$

Now,

$$

\begin{align*}

A^{2}-6 A+5 I & =\left(\begin{array}{ccc}

4 & 2 & 1 \

-3 & 8 & -14 \

7 & -3 & 14

\end{array}ight)-6\left(\begin{array}{ccc}

1 & 1 & 1 \

1 & 2 & -3 \

2 & -1 & 3

\end{array}ight)+5\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

4 & 2 & 1 \

-3 & 8 & -14 \

7 & -3 & 14

\end{array}ight)-\left(\begin{array}{ccc}

6 & 6 & 6 \

6 & 12 & -18 \

12 & -6 & 18

\end{array}ight)+\left(\begin{array}{ccc}

5 & 0 & 0 \

0 & 5 & 0 \

0 & 0 & 5

\end{array}ight) \

& =\left(\begin{array}{ccc}

9 & 2 & 1 \

-3 & 13 & -14 \

7 & -3 & 19

\end{array}ight)-\left(\begin{array}{ccc}

6 & 6 & 6 \

6 & 12 & -18 \

12 & -6 & 18

\end{array}ight) \

& =\left(\begin{array}{ccc}

3 & -4 & -5 \

-9 & 1 & 4 \

-5 & 3 & 1

\end{array}ight) 	ag{2}

\end{align*}

$$

From equation (1) and (2)

$$

\begin{aligned}

A^{-1} & =-\frac{1}{11}\left(\begin{array}{ccc}

3 & -4 & -5 \

-9 & 1 & 4 \

-5 & 3 & 1

\end{array}ight) \

& =\frac{1}{11}\left(\begin{array}{ccc}

-3 & 4 & 5 \

9 & -1 & -4 \

5 & -3 & -1

\end{array}ight)

\end{aligned}

$$

Question 16

$A=\left(\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}ight)$, verify that $A^{3}-6 A^{2}+9 A-4 I=0$. Hence, find $A^{-1}$.

Accepted Answer

Let

$$

A=\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A \

& =\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight)\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight) \

& =\left(\begin{array}{ccc}

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}ight) \

& =\left(\begin{array}{ccc}

6 & -5 & 5 \

-5 & 6 & -5 \

5 & -5 & 6

\end{array}ight)

\end{aligned}

$$

And

$$

\begin{aligned}

A^{3} & =A^{2} \cdot A \

& =\left(\begin{array}{ccc}

6 & -5 & 5 \

-5 & 6 & -5 \

5 & -5 & 6

\end{array}ight)\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight) \

& =\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}ight) \

& =\left(\begin{array}{ccc}

22 & -21 & 21 \

-21 & 22 & -21 \

21 & -21 & 22

\end{array}ight)

\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}ight)-6\left(\begin{array}{ccc}

6 & -5 & 5 \

-5 & 6 & -5 \

5 & -5 & 6

\end{array}ight)+9\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight)-4\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

22 & -21 & 21 \

-21 & 22 & -21 \

21 & -21 & 22

\end{array}ight)-\left(\begin{array}{ccc}

36 & -30 & 30 \

-30 & 36 & -30 \

30 & -30 & 36

\end{array}ight)+\left(\begin{array}{ccc}

18 & -9 & 9 \

-9 & 18 & -9 \

9 & -9 & 18

\end{array}ight)-\left(\begin{array}{ccc}

4 & 0 & 0 \

0 & 4 & 0 \

0 & 0 & 4

\end{array}ight) \

& =\left(\begin{array}{ccc}

40 & -30 & 30 \

-30 & 40 & -30 \

30 & -30 & 40

\end{array}ight)-\left(\begin{array}{ccc}

40 & -30 & 30 \

-30 & 40 & -30 \

30 & -30 & 40

\end{array}ight) \

& =\left(\begin{array}{ccc}

0 & 0 & 0 \

0 & 0 & 0

\end{array}ight) \

& =0

\end{aligned}

$$

Thus,

$$

A^{3}-6 A^{2}+9 A-4 I=0

$$

Now,

$$

\begin{align*}

& \Rightarrow A^{3}-6 A^{2}+9 A-4 I=0 \

& \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A A^{-1}-4 I A^{-1}=0 \quad\left[	ext { post-multiplying by } A^{-1} 	ext { as }|A| 
eq 0ight] \

& \Rightarrow A A\left(A A^{-1}ight)-6 A\left(A A^{-1}ight)+9\left(A A^{-1}ight)=4\left(I A^{-1}ight) \

& \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ight) 	ag{1}

\end{align*}

$$

Now,

$$

\begin{align*}

A^{2}-6 A+9 I & =\left(\begin{array}{ccc}

6 & -5 & 5 \

-5 & 6 & -5 \

5 & -5 & 6

\end{array}ight)-6\left(\begin{array}{ccc}

2 & -1 & 1 \

-1 & 2 & -1 \

1 & -1 & 2

\end{array}ight)+9\left(\begin{array}{lll}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

6 & -5 & 5 \

-5 & 6 & -5 \

5 & -5 & 6

\end{array}ight)-\left(\begin{array}{ccc}

12 & -6 & 6 \

-6 & 12 & -6 \

6 & -6 & 12

\end{array}ight)+\left(\begin{array}{lll}

9 & 0 & 0 \

0 & 9 & 0 \

0 & 0 & 9

\end{array}ight) \

& =\left(\begin{array}{ccc}

3 & 1 & -1 \

1 & 3 & 1 \

-1 & 1 & 3

\end{array}ight) 	ag{2}

\end{align*}

$$

From equations (1) and (2),

$$

A^{-1}=\frac{1}{4}\left(\begin{array}{ccc}

3 & 1 & -1 \

1 & 3 & 1 \

-1 & 1 & 3

\end{array}ight)

$$

Question 17

Let $A$ be a non-singular square matrix of order $3 	imes 3$. Then $|\operatorname{adj} A|$ is equal to:

(A) $|A|$

(B) $|A|^{2}$

(C) $|A|^{3}$

(D) $3|A|$

Accepted Answer

Since $A$ be a non-singular square matrix of order $3 	imes 3$

$$

\begin{aligned}

(\operatorname{adj} A) A & =|A| I \

& =\left(\begin{array}{ccc}

|A| & 0 & 0 \

0 & |A| & 0 \

0 & 0 & |A|

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

|(\operatorname{adj} A) A| & =\left|\begin{array}{ccc}

|A| & 0 & 0 \

0 & |A| & 0 \

0 & 0 & |A|

\end{array}ight| \

|\operatorname{adj} A||A| & =|A|^{3}\left|\begin{array}{lll}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight| \

& =|A|^{3} I \

|\operatorname{adj} A| & =|A|^{2}

\end{aligned}

$$

Thus, the correct option is B.

Question 18

If $A$ is an invertible matrix of order 2 , the $\operatorname{det}\left(A^{-1}\right)$ is equal to:

(A) $\operatorname{det}(A)$

(B) $\frac{1}{\operatorname{det}(A)}$

(C) 1

(D) 0

Accepted Answer

Since $A$ is an invertible matrix, $A^{-1}$ exists and $A^{-1}=\frac{1}{|A|}$ adj $A$.

As matrix $A$ is of order 2, let $A=\left(\begin{array}{ll}a & b \ c & d\end{array}ight)$

Then,

$$

|A|=a d-b c

$$

And

$$

\operatorname{adj} A=\left(\begin{array}{cc}

d & -b \

-c & a

\end{array}ight)

$$

Now,

$$

\begin{aligned}

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}ight)

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\left|A^{-1}ight| & =\left|\frac{\frac{d}{|A|}}{\left\lvert\, \frac{-c}{|A|}ight.} \frac{\frac{-b}{|A|}}{|A|}ight| \

\left|A^{-1}ight| & =\frac{1}{|A|^{2}}\left|\begin{array}{cc}

d & -b \

-c & a

\end{array}ight| \

& =\frac{1}{|A|^{2}}(a d-b c) \

& =\frac{1}{|A|^{2}} \cdot|A| \

& =\frac{1}{|A|}

\end{aligned}

$$

Hence,

$$

\operatorname{det}\left(A^{-1}ight)=\frac{1}{\operatorname{det}(A)}

$$

Thus, the correct option is B.

\section*{EXERCISE 4.6}