Question

If a square matrix A is such that \( A^2 = A \) and \( (I - A)^3 = xA + I \), then value of x must be :

• A. \( 7 \)
• B. \( 5 \)
• C. \( -7 \)
• D. \( -1 \)

Hint:

For an idempotent matrix where \( A^2 = A \), any positive integer power satisfies \( A^n = A \). This allows immediate simplification of any polynomial expression containing \( A \).

Answer 1

The Correct Option is C

Concept: A matrix \( A \) that satisfies \( A^2 = A \) is called an idempotent matrix. Since the identity matrix \( I \) commutes with any square matrix \( A \) (\( IA = AI = A \)), we can safely expand binomial expressions like \( (I - A)^3 \) using standard algebraic binomial expansion formulas.

Step 1: Expand the binomial expression \( (I - A)^3 \).
Using the identity \( (X - Y)^3 = X^3 - 3X^2Y + 3XY^2 - Y^3 \) where \( X = I \) and \( Y = A \): \[ (I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3 \]

Step 2: Simplify high powers of identity matrix \( I \) and matrix \( A \).
We know that:
• \( I^3 = I \) and \( I^2 = I \)
• \( IA = A \) and \( IA^2 = A^2 \) Substituting these into the expanded formula yields: \[ (I - A)^3 = I - 3A + 3A^2 - A^3 \]

Step 3: Apply the idempotent property \( A^2 = A \).
Since \( A^2 = A \), we can also determine higher powers like \( A^3 \): \[ A^3 = A^2 \cdot A = A \cdot A = A^2 = A \] Now replace both \( A^2 \) and \( A^3 \) with \( A \) in our simplified expression: \[ (I - A)^3 = I - 3A + 3(A) - (A) \] Combining the like terms of \( A \): \[ (I - A)^3 = I - 3A + 3A - A = I - A \]

Step 4: Equate with the given equation to solve for x.
We are given: \[ (I - A)^3 = xA + I \] Substituting our derived result: \[ I - A = xA + I \] Subtracting \( I \) from both sides: \[ -A = xA \quad \Rightarrow \quad x = -1 \] Let us re-verify carefully. If \( (I-A)^3 = I - A \), then matching it to \( xA + I \) gives \( x = -1 \). Thus, option (D) is the correct choice.