Question

If $A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, find $k$ so that $A^2=kA-2I$

Answer 1

$A^2=A.A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$

$=\left[\begin{array}{cc}3\left(3\right)+(-2)\left(4\right)& 3(-2)+(-2)(-2)\\ 4\left(3\right)+(-2)\left(4\right)& 4(-2)+(-2)(-2)\end{array}\right]=\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]$

Now, $A^2=kA-2I$

$A^2=kA-2I$

$\Rightarrow \left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]=k\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]-2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]=\left[\begin{array}{cc}3k& -2k\\ 4k& -2k\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]=\left[\begin{array}{cc}3k-2& 2k\\ 4k& -2k-2\end{array}\right]$

Comparing the corresponding element, we have:
$3k-2=1$

$\Rightarrow 3k=3$

$\Rightarrow k=1$