Question
Given $A=\left[\begin{array}{ll}p & 0 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -q \\ 1 & 0\end{array}\right], C=\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]$ and if $B A=C^2$, find $p$ and $q$.

Answer

$\begin{aligned} & BA =\left[\begin{array}{cc}0 & - q \\ 1 & 0\end{array}\right]\left[\begin{array}{ll} p & 0 \\ 0 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}0-0 & 0-2 q \\ p -0 & 0+0\end{array}\right] \\ & =\left[\begin{array}{cc}0 & -2 q \\ p & 0\end{array}\right] \\ & C ^2=\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ 2 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}4-4 & -4-4 \\ 4+4 & -4+4\end{array}\right] \\ & =\left[\begin{array}{cc}0 & -8 \\ 8 & 0\end{array}\right] \\ & \text { But BA }= C ^2 \\ & {\left[\begin{array}{cc}0 & -2 q \\ p & 0\end{array}\right]=\left[\begin{array}{cc}0 & -8 \\ 8 & 0\end{array}\right]} \\ & -2 q =-8\end{aligned}$
$
\begin{aligned}
& q=\frac{8}{2}=4 \\
& p=8
\end{aligned}
$
$\therefore$ The value of $p=8$ and $q=4$

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