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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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