Question

If $A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$, find $K$ such that $A^2=KA-2I$, where $I$ is the identity element.

Answer 1

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

$A^2=KA-2I$

$A^2=A\times A$

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

$=\left[\begin{array}{cc}9-8& -6+4\\ 12-8& -8+4\end{array}\right]=\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]$

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

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

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

$3K-2=1$

$\Rightarrow 3K=3$

$\Rightarrow K=1$

$-2K=-2$

$\Rightarrow K=1$

$4K=4$

$\Rightarrow K=1$

$-2K-2=-4$

$\Rightarrow -2K=-2$

$\Rightarrow K=1$

$\therefore K=1$