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Q. If $A=\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix},$ then $A^{2} -5A +6I =$

Matrices

Solution:

$A^{2} -5A +6I = \begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}$
$-\begin{bmatrix}10&0&5\\ 10&5&15\\ 5&-5&0\end{bmatrix} +\begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix}$
$= \begin{bmatrix}5&-1&2\\ 9&-2&5\\ 0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\ 10&5&15\\ 5&-5&0\end{bmatrix} + \begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix}$
$= \begin{bmatrix}-5&-1&-3\\ -1&-7&-10\\ -5&4&-2\end{bmatrix}+\begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix} =\begin{bmatrix}1&-1&-3\\ -1&-1&-10\\ -5&4&4\end{bmatrix}$