Question Bank • QB_EM_ADV_014
MCQ • 1 Mark
Section 1: Engineering Mathematics
Linear Algebra
Subtopic: Matrices and determinants
Question Statement:
If $A$ is an involutory matrix such that $A^2 = I$, where $I$ is the identity matrix, then the matrix $P = \frac{1}{2}(I + A)$ is always:
Options:
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(A) Nilpotent
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(B) Idempotent
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(C) Skew-symmetric
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(D) Orthogonal
Step-by-Step Solution & Explanation
Compute $P^2$:
$P^2 = \left[ \frac{1}{2}(I + A) \right]^2 = \frac{1}{4}(I^2 + IA + AI + A^2)$
Since $I^2 = I$, $IA = AI = A$, and $A^2 = I$:
$P^2 = \frac{1}{4}(I + 2A + I) = \frac{1}{4}(2I + 2A) = \frac{1}{2}(I + A) = P$
Since $P^2 = P$, the matrix $P$ is idempotent.
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