Question
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},$ find k such that A2 = kA - 2I2.

Answer

Given: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2=\text{kA}-2\text{I}_2$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\text{k}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3\text{k}&-2\text{k}\\4\text{k}&-2\text{k}\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3\text{k}-2&-2\text{k}-0\\4\text{k}-0&-2\text{k}-2\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ 1=3\text{k}-2$
$\Rightarrow1+2=3\text{k}$
$\Rightarrow3=3\text{k}$
$\Rightarrow\text{k}=1$

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