CUET · MATHS · PYQ PAPER 2023
If \(A\) is a square matrix such that \(A^2=A\), then \((I+A)^3-8 A\) is equal to:
- A I
- B A
- C I-A
- D 3A
Answer & Solution
Correct Answer
(C) I-A
Step-by-step Solution
Detailed explanation
\((I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3\) \(= I + 3A + 3A^2 + A^3\) \(A^2=A \implies A^3 = A^2 \cdot A = A \cdot A = A^2 = A\) \((I+A)^3 = I + 3A + 3A + A = I + 7A\) \((I+A)^3 - 8A = (I+7A) - 8A\) \(= I - A\)
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