Question

If $A$ is a square matrix such that $A^{2} = A$, then $(I + A)^{3} - 7A$ will be:

• A. $A$
• B. $I - A$
• C. $I$
• D. $3A$

Hint:

For idempotent matrices ($A^{2} = A$), higher powers reduce back to $A$ itself, i.e., $A^{n} = A$ for $n \geq 1$.

Answer 1

The Correct Option is C

Step 1: Expand $(I+A)^{3$.}
\[ (I + A)^{3} = I^{3} + 3I^{2}A + 3IA^{2} + A^{3} \] \[ = I + 3A + 3A^{2} + A^{3} \]

Step 2: Use the condition $A^{2 = A$.}
Since $A^{2} = A$, \[ A^{3} = A \cdot A^{2} = A \cdot A = A \] So, \[ (I + A)^{3} = I + 3A + 3A + A = I + 7A \]

Step 3: Simplify expression.
\[ (I + A)^{3} - 7A = (I + 7A) - 7A = I \]

Step 4: Conclusion.
The correct answer is (C) $I$.