Question

If \( A \) is a square matrix such that \( A^2 = A \), then \( (A - I)^3 - A \) is equal to :

• A. \( I \)
• B. \( -I \)
• C. \( A \)
• D. \( A^2 \)

Hint:

For an idempotent matrix, any power \( A^n = A \).
Remember that \( (I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A \), which is a useful shortcut in similar problems.

Answer 1

The Correct Option is B

Concept:
A matrix \(A\) is idempotent if: \[ A^2=A \] This also implies: \[ A^3=A \] The identity matrix \(I\) commutes with every square matrix, so the binomial expansion can be applied to \((A-I)^3\). 
Step 1: Expand \((A-I)^3\)
Using the binomial expansion: \[ (A-I)^3=A^3-3A^2I+3AI^2-I^3 \] Since \(AI=A\) and \(I^2=I^3=I\): \[ (A-I)^3=A^3-3A^2+3A-I \] 
Step 2: Use the idempotent property
Given: \[ A^2=A \] Therefore: \[ A^3=A^2A=A^2=A \] Substituting \(A^2=A\) and \(A^3=A\): \[ (A-I)^3=A-3A+3A-I \] \[ (A-I)^3=A-I \] 
Step 3: Evaluate \((A-I)^3-A\)
Using the above result: \[ (A-I)^3-A=(A-I)-A \] \[ =A-I-A \] \[ =-I \] 
Final Answer:
Therefore, \[ \boxed{-I} \]