Question
If $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}, f(x) = x^2 - 2x - 3, $ show that $f(A) = 0$

Answer

Given : $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $f(x) = x^2 - 2x - 3$
$\text{f(A)}=\text{A}^2-2\text{A}-3\text{I}$
$=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}-2\begin{bmatrix}1&2\\2&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}1+4&2+2\\2+2&4+1\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ =\begin{bmatrix}5&4\\4&5\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$=\begin{bmatrix}5-2-3&4-4-0\\4-4-0&5-2-3\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
So $, \text{f(A)}=0$

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