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
Since A and I commute (AI = IA = A), the binomial expansions hold as for numbers:
\((A - I)^3 = A^3 - 3A^2 + 3A - I\)
\((A + I)^3 = A^3 + 3A^2 + 3A + I\)
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:
\((A - I)^3 + (A + I)^3 - 3A = 2A + 6A - 3A = 5A\)