Question
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I_2 = 0.$
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Answer
Given: $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
Now,
$ \text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}_2$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=0$
Hence proved.
Now,
$ \text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}_2$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=0$
Hence proved.
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