If $A$ is a square matrix such that $A^2 = I$, then $(A - I)^3 + (A + I)^3 - 7A$ is equal to

$A$

The correct answer is Option (1) → $A$ ##

We have, $A^2 = I$

$∴(A - I)^3 + (A + I)^3 - 7A = [((A - I) + (A + I))((A - I)^2 + (A + I)^2 - (A - I)(A + I))] - 7A$

$\left[ ∵a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \right]$

$= [(2A)(A^2 + I^2 - 2AI + A^2 + I^2 + 2AI - (A^2 - I^2))] - 7A$

$= 2A[I + I + I + I - (I - I)] - 7A \quad [∵A^2 = I]$

$= 2A[5I - I] - 7A \quad [∵A = AI]$

$= 8AI - 7AI$

$= AI = A$