Question
If $\text{A}=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix},$ show that $A^2 = 0$

Answer

Given: $\text{A}=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{a}^2\text{b}^2-\text{a}^2\text{b}^2&\text{ab}^3-\text{ab}^3\\-\text{a}^3\text{b}+\text{a}^3\text{b}&-\text{a}^2\text{b}^2+\text{a}^2\text{b}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2=0.$
Hence proved.

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