Question
If $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix},$ find k such that $A^2 - 8A + kI = 0.$

Answer

Given: $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}1-0&0+0\\-1-7&0+49\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-8&49\end{bmatrix}$
$ \text{A}^2-8\text{A}+\text{kI}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-8\begin{bmatrix}1&0\\-1&7\end{bmatrix}+\text{k}\begin{bmatrix}1&0\\0&1\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-\begin{bmatrix}8&0\\-8&56\end{bmatrix}+\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}1-8+\text{k}&0-0+0\\-8+8+0&49-56+\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}-7+\text{k}&0\\0&-7+\text{k}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore-7+\text{k}=0$
$\Rightarrow\text{k}=7$

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