# For the matrix

Question:

For the matrix $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$. Show that $A^{-3}-6 A^{2}+5 A+11 I_{3}=O$. Hence, find $A^{-1}$.

Solution:

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

$\begin{array}{lll}1 & 2 & -3\end{array}$

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

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

$1 \quad 2-3$

$2-1 \quad 3 \mid=(1 \times 3)-(1 \times 9)+(1 \times-5)=3-9-5=-11$

Since, $|A| \neq 0$

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

Now,

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

$1 \quad 2-3$

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

$1 \quad 2-3$

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

$\begin{array}{lll}1+2-6 & 1+4+3 & 1-6-9\end{array}$

$2-1+6 \quad 2-2-3 \quad 2+3+9]=\left[\begin{array}{lll}4 & 2 & 1\end{array}\right.$

$\begin{array}{crr}-3 & 8 & -14 \\ 7 & -3 & 14]\end{array}$

and $A^{3}=A^{2} . A=\left[\begin{array}{lll}4 & 2 & 1\end{array}\right.$

$-3 \quad 8 \quad-14$

$\left.\begin{array}{lll}7 & -3 & 14\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1\end{array}\right.$

and $A^{3}=A^{2} \cdot A=\left[\begin{array}{lll}4 & 2 & 1\end{array}\right.$

$\begin{array}{lll}-3 & 8 & -14\end{array}$

$\left.\begin{array}{lll}7 & -3 & 14\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1\end{array}\right.$

$\begin{array}{lll}1 & 2 & -3\end{array}$

$\left.\begin{array}{lll}2 & -1 & 3\end{array}\right]=\left[\begin{array}{llllll}4+2+2 & 4+4-1 & 4-6+3-3+8-28 & -3+16+14 & -3-24-42 & 7-3+28 & 7-6-14 & 7+9+42\end{array}\right]$

$=\left[\begin{array}{lll}8 & 7 & 1\end{array}\right.$

$\begin{array}{ccc}-23 & 27 & -69 \\ 32 & -13 & 58]\end{array}$

Now, $A^{3}-6 A^{2}+5 A+11 I_{3}=\left[\begin{array}{lll}8 & 7 & 1\end{array}\right.$

$\begin{array}{lll}-23 & 27 & -69\end{array}$

$32 \quad-13 \quad 58]-6\left[\begin{array}{lll}4 & 2 & 1\end{array}\right.$

$\begin{array}{lll}-3 & 8 & -14\end{array}$

$7-3 \quad 14]+5\left[\begin{array}{lll}1 & 1 & 1\end{array}\right.$

$1 \quad 2-3$

$\left.\begin{array}{lll}2 & -1 & 3\end{array}\right]+11\left[\begin{array}{lll}1 & 0 & 0\end{array}\right.$

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

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

$=\left[\begin{array}{llllllll}8-24+5+11 & & 7-12+5+0 & 1-6+5+0-23+18+5+0 & 27-48+10+11 & -69+84-15+0 & 32-42+10+0 & -13+18-5+0 & 58-84+15+11\end{array}\right]$

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

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

$\left.\begin{array}{lll}0 & 0 & 0\end{array}\right]=O$ (: matrix)

Again, $A^{3}-6 A^{2}+5 A+11 I_{3}=O$

$\Rightarrow A^{-1} \times\left(\mathrm{A}^{3}-6 \mathrm{~A}^{2}+5 \mathrm{~A}+11 \mathrm{I}_{3}\right)=A^{-1} \times O \quad$ (Pre $-$ multiplying both sides because $A^{-1}$ exists)

$\Rightarrow\left(A^{2}-6 A+5 I_{3}+11 A^{-1}\right)=0$

$\Rightarrow\left(A^{2}-6 A+5 I_{3}+11 A^{-1}\right)=0$

$\Rightarrow\left[\begin{array}{lll}4 & 2 & 1\end{array}\right.$

$\begin{array}{lll}-3 & 8 & -14\end{array}$

$7 \quad-3 \quad 14]-6\left[\begin{array}{lll}1 & 1 & 1\end{array}\right.$

$\begin{array}{lll}1 & 2 & -3\end{array}$

$2-1 \quad 3]+5\left[\begin{array}{lll}1 & 0 & 0\end{array}\right.$

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

$0 \quad 0 \quad 1]=-11 A^{-1}$

$\Rightarrow\left[\begin{array}{llllllll}4-6+5 & 2-6+0 & 1-6+0-3-6+0 & 8-12+5 & -14+18+0 & 7-12+0 & -3+6+0 & 14-18+5\end{array}\right]=-11 A^{-1}$

$\Rightarrow A^{-1}=-\frac{1}{11}\left[\begin{array}{lll}3 & -4 & -5\end{array}\right.$

$\begin{array}{lll}-9 & 1 & 4\end{array}$

$\left.\begin{array}{lll}-5 & 3 & 1\end{array}\right]$