Question
If $A=\left[\begin{array}{cc}4 & 8 \\ -2 & -4\end{array}\right]$, prove that $A^2=0$
$\begin{aligned} & =\left[\begin{array}{cc}4 & 8 \\ -2 & -4\end{array}\right]\left[\begin{array}{cc}4 & 8 \\ -2 & -4\end{array}\right] \\ & =\left[\begin{array}{cc}16-16 & 32-32 \\ -8+8 & -16+16\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0\end{aligned}$
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$\frac{x}{2}+\frac{y}{4}=1$