If A is a square matrix and \(I\) is an identity matrix of same order such that \(A^2 = A\), then \((I+A)^3 - 8I\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(7(A-I)\)
Since I and A commute (IA = AI = A), \((I + A)^3\) expands like a binomial:
\((I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3\)
\((I + A)^3 = I + 3A + 3A^2 + A^3\)
From \(A^2 = A\):
\(A^3 = A^2 \cdot A = A \cdot A = A^2 = A\)
So
\((I + A)^3 = I + 3A + 3A + A = I + 7A\)
\((I + A)^3 - 8I = 7A - 7I = 7(A - I)\)