Show that the matrix,

Solution:

We have, $A=\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\begin{array}{lll}3 & 4 & 1]\end{array}$

$\Rightarrow|A|=\mid \begin{array}{lll}1 & 0 & -2\end{array}$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$3 \quad 4 \quad 1 \mid=1(-9)+0-2(-8)=-9+16=7$

Since, $|A| \neq 0$

Hence, $A^{-1}$ exists.

Now,

$A^{2}=\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\left.\begin{array}{lll}3 & 4 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\begin{array}{lll}3 & 4 & 1\end{array}=\left[\begin{array}{lllllll}1+0-6 & 0+0-8 & -2+0-2-2+2+6 & 0+1+8 & 4-2+2 & 3-8+3 & 0-4+4 & -6+8+1\end{array}=\right.$

$\left[\begin{array}{lll}-5 & -8 & -4\end{array}\right.$

$\begin{array}{lll}6 & 9 & 4\end{array}$

$\left.\begin{array}{lll}-2 & 0 & 3\end{array}\right]$

$A^{3}=A^{2} \cdot A=\left[\begin{array}{lll}-5 & -8 & -4\end{array}\right.$

$\begin{array}{lll}6 & 9 & 4\end{array}$

$\left.\begin{array}{lll}-2 & 0 & 3\end{array}\right]\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\left.\begin{array}{lll}3 & 4 & 1\end{array}\right]=\left[\begin{array}{llllllll}-5+16-12 & 0+8-16 & 10-16-4 & 6-18+12 & 0-9+16 & -12+18+4 & -2+0+9 & 0+0+12 & 4+0+3\end{array}\right]$

$=\left[\begin{array}{lll}-1 & -8 & -10\end{array}\right.$

$\begin{array}{ccc}0 & 7 & 10 \\ 7 & 12 & 7]\end{array}$

Now, $A^{3}-A^{2}-3 A-I_{3}=\left[\begin{array}{lll}-1 & -8 & -10\end{array}\right.$

$\begin{array}{lll}0 & 7 & 10\end{array}$

$\left.\begin{array}{lll}7 & 12 & 7\end{array}\right]-\left[\begin{array}{lll}-5 & -8 & -4\end{array}\right.$

$\begin{array}{lll}6 & 9 & 4\end{array}$

$\left.\begin{array}{lll}-2 & 0 & 3\end{array}\right]-3\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\left.\begin{array}{lll}3 & 4 & 1\end{array}\right]-\left[\begin{array}{lll}1 & 0 & 0\end{array}\right.$

$\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1]\end{array}$

$=\left[\begin{array}{lllll}-1+5-3-1 & -8+8+0+0 & -10+4+6-00-6+6-0 & 7-9+3-1 & 10-4-6-0 & 7+2-9-0 & 12+0-12-0 & 7-3-3-1\end{array}\right]$

$=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right.$

$\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0]=O\end{array}$

Hence proved.

Now, $A^{3}-A^{2}-3 A-I_{3}=O$ (: matrix)

$\Rightarrow A^{-1}\left(A^{3}-A^{2}-3 A-I_{3}\right)=A^{-1} O \quad\left(\right.$ Pre $-$ multiplying by $\left.A^{-1}\right)$

$\Rightarrow A^{2}-A^{1}-3 I_{3}=A^{-1}$

$\Rightarrow\left[\begin{array}{lll}-5 & -8 & -4\end{array}\right.$

$\begin{array}{lll}6 & 9 & 4\end{array}$

$\left.\begin{array}{lll}-2 & 0 & 3\end{array}\right]-\left[\begin{array}{lll}1 & 0 & -2\end{array}\right.$

$\begin{array}{lll}-2 & -1 & 2\end{array}$

$\left.\begin{array}{lll}3 & 4 & 1\end{array}\right]-3\left[\begin{array}{lll}1 & 0 & 0\end{array}\right.$

$\begin{array}{lll}0 & 1 & 0\end{array}$

$\begin{array}{lll}0 & 0 & 1] & =A^{-1}\end{array}$

$\Rightarrow\left[\begin{array}{llllllll}-5-1-3 & -8-0-0 & -4+2+0 & 6+2+0 & 9+1-3 & 4-2 & -2-3-0 & 0-4-0 & 3-1-3\end{array}\right]$

$=\left[\begin{array}{lll}-9 & -8 & -2\end{array}\right.$

$\begin{array}{lll}8 & 7 & 2\end{array}$

$\begin{array}{lll}-5 & -4 & -1]=A^{-1}\end{array}$

$\Rightarrow A^{-1}=\left[\begin{array}{lll}-9 & -8 & -2\end{array}\right.$

$\begin{array}{lll}8 & 7 & 2\end{array}$

$\begin{array}{lll}-5 & -4 & -1]\end{array}$