If \(A\) is a square matrix such that \(A^2=A\) and \(I\) is the identity matrix of the same order as \(A\), then \((I+2A)^2 - 5A\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(I+3A\)
Since I commutes with every matrix (\(IA = AI = A\)), the square expands like an ordinary binomial:
\((I + 2A)^2 = I^2 + 4A + 4A^2\)
Using \(I^2 = I\) and the given \(A^2 = A\):
\((I + 2A)^2 = I + 4A + 4A = I + 8A\)
So
\((I + 2A)^2 - 5A = I + 8A - 5A = I + 3A\)