The number of elements in the set $\{A = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0,…
The number of elements in the set $\{A = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\}\}$ and $\{(I - A)^3 = I - A^3\}$, where $I$ is $2 \times 2$ identity matrix, is $\underline{\hspace{3cm}}$.
Solution
$A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$, $a, b, d \in \{-1,0,1\}$
$(I-A)^3=I-A^3$
$I-A^3-3A+3A^2=I-A^3$
$3A^2-3A=0$
$3A(A-I)=0$
$A^2=A...i$
$A^2=A.A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix}\begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$
$A^2=\begin{bmatrix} a^2 & ab+bd \\ 0 & d^2 \end{bmatrix}$
From $i$
$\begin{bmatrix} a^2 & ab+bd \\ 0 & d^2 \end{bmatrix}=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$
Comparing on both sides
$a^2=a$
$a(a-1)=0$
$a=0, 1$
$d^2=d$
$d(d-1)=0$
$d=0, 1$
$b(a+d)=b$
Case I: $b=0 \Rightarrow (a, d) \equiv (0,1), (0,0), (1,1) \rightarrow 4$ ways
Case II: $a+d=1 \Rightarrow (1,0), (0,1)$ and $b=\pm1 \rightarrow 4$ ways
Total $=8$ ways