If $A$ is a square matrix and $I$ is the identity matrix of same order such that $A^2 = I$, then $(A-I)^3 + (A + I)^3 - 3A$ is equal to

5A

The correct answer is Option (4) → 5A

Given $A^2=I$

Expand

$(A-I)^3=A^3-3A^2I+3AI^2-I^3$

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

$(A+I)^3=A^3+3A^2I+3AI^2+I^3$

$=A^3+3A^2+3A+I$

Add

$(A-I)^3+(A+I)^3=2A^3+6A$

Subtract $3A$

$=2A^3+3A$

Using $A^2=I$

$A^3=A$

$2A^3+3A=2A+3A$

$=5A$

The value of $(A-I)^3+(A+I)^3-3A$ is $5A$.