If A is a square matrix and I is the identify matrix of same order such that $A^2 = I$, then $3(A-I)^3 + 3(A + I)^3 - 15A$ is equal to

9A

The correct answer is Option (2) → 9A

Given: $A^2 = I$ and $I$ is the identity matrix

We are to evaluate: $3(A - I)^3 + 3(A + I)^3 - 15A$

Use the identity: $(X)^3 = X^3$ and expand using binomial expansion:

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

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

Now compute:

$3[(A - I)^3 + (A + I)^3] - 15A$

$= 3[(A^3 - 3A^2 + 3A - I) + (A^3 + 3A^2 + 3A + I)] - 15A$

$= 3[2A^3 + 6A] - 15A$

$= 6A^3 + 18A - 15A = 6A^3 + 3A$

Now use the given: $A^2 = I \Rightarrow A^3 = A \cdot A^2 = A \cdot I = A$

So, $6A^3 + 3A = 6A + 3A = 9A$