Question:
If $A=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]$, show that $A^{2}=0$
Solution:
Given : $A=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]$
Now,
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}a^{2} b^{2}-a^{2} b^{2} & a b^{3}-a b^{3} \\ -a^{3} b+a^{3} b & -a^{2} b^{2}+a^{2} b^{2}\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow A^{2}=0$
Hence proved.
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