The number of elements in the set A = [ cases a & b 0 & d cases ]:...

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

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

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

$3{A}^{2} - 3A = 0$

$3A ( A - I ) = 0$

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

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

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

From \left(i\right)</p><p>\left[\begin{cases}{a}^{2} & ab + bd \\ 0 & {d}^{2}\end{cases}\right] = \left[\begin{cases}a & b \\ 0 & d\end{cases}\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 $I:$ $b = 0 ⇒ ( a, d ) ≡ ( 0, 1 ) , ( 0, 0 ) , ( 1, 1 ) \to 4$ ways

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

Total $= 8$ ways