The number of elements in the set
$\left\{A=\left(\begin{array}{ll}a & b \\ 0 & d\end{array}\right): a, b, d \in\{-1,0,1\}\right.$ and $\left.(I-A)^{3}=I-A^{3}\right\}$
where I is $2 \times 2$ identity matrix, is :
$(\mathrm{I}-\mathrm{A})^{3}=\mathrm{I}^{3}-\mathrm{A}^{3}-3 \mathrm{~A}(\mathrm{I}-\mathrm{A})=\mathrm{I}-\mathrm{A}^{3}$
$\Rightarrow 3 \mathrm{~A}(\mathrm{I}-\mathrm{A})=0$ or $\mathrm{A}^{2}=\mathrm{A}$
$\Rightarrow\left[\begin{array}{cc}\mathrm{a}^{2} & \mathrm{ab}+\mathrm{bd} \\ 0 & \mathrm{~d}^{2}\end{array}\right]=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ 0 & \mathrm{~d}\end{array}\right]$
$\Rightarrow \mathrm{a}^{2}=\mathrm{a}, \mathrm{b}(\mathrm{a}+\mathrm{d}-1)=0, \mathrm{~d}^{2}=\mathrm{d}$
If $b \neq 0, a+d=1 \Rightarrow 4$ ways
If $b=0, a=0,1 \& d=0,1 \Rightarrow 4$ ways
$\Rightarrow$ Total 8 matrices
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