Given, A=[a0bd]:a,b,d∈−1,0,1
(I−A)3=I−A3
I−A3−3A+3A2=I−A3
3A2−3A=0
3A(A−I)=0
A2=A....(i)
A2=A.A=[a0bd][a0bd]
A2=[a20ab+bdd2]
From (i)
[a20ab+bdd2]=[a0bd]
Comparing on both sides
a2=a
a(a−1)=0
a=0,1
d2=d
d(d−1)=0
d=0,1
b(a+d)=b
Case I: b=0⇒(a,d)≡(0,1),(0,0),(1,1)→4 ways
Case II: a+d=1⇒(1,0),(0,1) and b=±1→4 ways
Total =8 ways