Question

If matrix $A=\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]$ and $A^2=pA$, then write the value of $p$.

Answer 1

If $A=\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$ then $A^2$ can be computed as shown below:

$A^2=\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)=\left(\begin{array}{cc}2\times 2+-2\times -2& 2\times -2+(-2)\times 2\\ -2\times 2+2\times -2& -2\times -2+2\times 2\end{array}\right)$

implies $A^2=\left(\begin{array}{cc}8& -8\\ -8& 8\end{array}\right)$

Since $A^2=pA$, therefore,

$\left(\begin{array}{cc}8& -8\\ -8& 8\end{array}\right)=p\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$

$\left(\begin{array}{cc}8& -8\\ -8& 8\end{array}\right)=\left(\begin{array}{cc}2p& -2p\\ -2p& 2p\end{array}\right)$

Now equating the matrices we get $2p=8⟹p=4$.

Hence, $p=4$.