JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Matrices

Expanding, $(I-A{)}^3=I-3A+3A^2-A^3$, so the condition $(I-A{)}^3=I-A^3$ reduces to $A^2=A$. For $A=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, $A^2=\left(\begin{array}{cc}{a}^2& b(a+d)\\ 0& d^2\end{array}\right)$, so $A^2=A$ needs $a^2=a$, $d^2=d$ and $b(a+d-1)=0$. Thus $a,d\in \{0,1\}$. When $a+d=1$, from the pairs $(0,1)$ and $(1,0)$, $b$ may be any of $-1,0,1$, giving $2\times 3=6$ matrices; when $a+d\ne 1$, from $(0,0)$ and $(1,1)$, $b=0$, giving $2$ matrices. The number of elements is $8$.