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
Answer & explanation
Correct answer: option 4
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 $5A$.