For the matrixshow that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
Question:

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

Solution:

$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$

$A^{2}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9\end{array}\right]=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]$

$A^{3}=A^{2} \cdot A=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$

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

$=\left[\begin{array}{ccc}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{array}\right]$

$\therefore A^{3}-6 A^{2}+5 A+11 I$

From equation (1), we have:

$A^{-1}=-\frac{1}{11}\left[\begin{array}{lll}3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1\end{array}\right]=\frac{1}{11}\left[\begin{array}{lll}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{array}\right]$

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