If A is a square matrix and I is the identity matrix of same order such that \(A^2 = I\), then \(3(A - I)^3 + 3(A + I)^3 - 15A\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9A
Since \(AI = IA = A\), the matrices A and I commute, so the binomial expansions hold:
\((A - I)^3 = A^3 - 3A^2 + 3A - I\)
\((A + I)^3 = A^3 + 3A^2 + 3A + I\)
Adding:
\((A - I)^3 + (A + I)^3 = 2A^3 + 6A\)
From \(A^2 = I\):
\(A^3 = A \cdot A^2 = A \cdot I = A\)
So:
\(3(A - I)^3 + 3(A + I)^3 - 15A = 3(2A + 6A) - 15A\)
\(= 24A - 15A = 9A\)