If A is a square matrix such that \(A^2 = I\), where I is the identity matrix of the same order, then \((I - A)^2 + 2A\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2I
Since A and I commute (AI = IA = A), the square can be expanded like an ordinary binomial:
\((I - A)^2 = I - 2A + A^2\)
Using \(A^2 = I\):
\((I - A)^2 = 2I - 2A\)
Hence
\((I - A)^2 + 2A = 2I - 2A + 2A = 2I\)