The number of elements in the set {A=[ … ]:a,b,d∈ -1,0,1 and (I-A)^3=I-A^3}, where I is 2× 2 identity matrix, is .

Given, $A=[\begin{matrix}a & b \\ 0 & d\end{matrix}]:a,b,d\in {-1,0,1}$A=[a0​bd​]:a,b,d∈−1,0,1

$${(I-A)}^{3}=I-{A}^{3}$$(I−A)3=I−A3

$$I-{A}^{3}-3A+3{A}^{2}=I-{A}^{3}$$I−A3−3A+3A2=I−A3

$$3{A}^{2}-3A=0$$3A2−3A=0

$$3A(A-I)=0$$3A(A−I)=0

$${A}^{2}=A....(i)$$A2=A....(i)

$${A}^{2}=A.A=[\begin{matrix}a & b \\ 0 & d\end{matrix}][\begin{matrix}a & b \\ 0 & d\end{matrix}]$$A2=A.A=[a0​bd​][a0​bd​]

$${A}^{2}=[\begin{matrix}{a}^{2} & ab+bd \\ 0 & {d}^{2}\end{matrix}]$$A2=[a20​ab+bdd2​]

From $(i)$(i)

$$[\begin{matrix}{a}^{2} & ab+bd \\ 0 & {d}^{2}\end{matrix}]=[\begin{matrix}a & b \\ 0 & d\end{matrix}]$$[a20​ab+bdd2​]=[a0​bd​]

Comparing on both sides

$${a}^{2}=a$$a2=a

$$a(a-1)=0$$a(a−1)=0

$$a=0,1$$a=0,1

$${d}^{2}=d$$d2=d

$$d(d-1)=0$$d(d−1)=0

$$d=0,1$$d=0,1

$$b(a+d)=b$$b(a+d)=b

Case $I:$I: $b=0\Rightarrow (a,d)\equiv (0,1),(0,0),(1,1)\rightarrow 4$b=0⇒(a,d)≡(0,1),(0,0),(1,1)→4 ways

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

Total $=8$=8 ways