Mathematics - Matrices Question with Solution | TestHub

Given, $A=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right] :a,b,d\in \left\{-1,0,1\right\}$

${\left(I-A\right)}^3=I-A^3$

$I-A^3-3A+3A^2=I-A^3$

$3A^2-3A=0$

$3A(A-I)=0$

$A^2=A....\left(i\right)$

$A^2=A.A=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]$

$A^2=\left[\begin{array}{cc}{a}^2& ab+bd\\ 0& d^2\end{array}\right]$

From $\left(i\right)$

$\left[\begin{array}{cc}{a}^2& ab+bd\\ 0& d^2\end{array}\right]=\left[\begin{array}{ll}a& b\\ 0& d\end{array}\right]$

Comparing on both sides
$a^2=a$

$a\left(a-1\right)=0$

$a=0, 1$

$d^2=d$

$d\left(d-1\right)=0$

$d=0, 1$

$b(a+d)=b$

Case $\mathrm{I}:$ $b=0\Rightarrow (a, d)\equiv (0, 1), (0, 0), (1, 1)\to 4$ ways

Case $\mathrm{II} :$ $a+d=1\Rightarrow (1, 0), (0, 1)$ and $b=\pm 1\to 4$ ways

Total $=8$ ways