Given A2=A and I is the identity matrix of the same order as A.
Since A2=A:
A3=A2⋅A=A⋅A=A2=A
So every power of A equals A itself.
Expanding (2I+A)3 using the binomial formula:
(2I+A)3=(2I)3+3(2I)2(A)+3(2I)(A2)+A3
(2I)3=8I
3(2I)2(A)=3⋅4I⋅A=12A
3(2I)(A2)=6I⋅A2=6A (since A2=A)
A3=A (since A3=A)
So:
(2I+A)3=8I+12A+6A+A
=8I+19A
(2I+A)3−19A−3I
=(8I+19A)−19A−3I
=8I+19A−19A−3I
=5I