Matrices & Determinants
Idempotent matrix
Grade 12
Question:
<p>If \(A\) is an idempotent matrix satisfying \((I - 0.4A)^{-1} = I - \alpha A\), where \(I\) is unit matrix of the same order as that of \(A\), then the value of \(\alpha\) is:</p>
<p>(a) \(\dfrac{-1}{3}\)</p>
<p>(b) \(\dfrac{1}{3}\)</p>
<p>(c) \(\dfrac{-2}{3}\)</p>
<p>(d) \(\dfrac{2}{3}\)</p>
Step-by-Step Solution
Key Concept: An idempotent matrix satisfies A² = A. Use this property to expand (I - 0.4A)⁻¹ and compare coefficients with I - αA by multiplying both sides by (I - 0.4A).
<p><strong>Step 1:</strong> Since A is idempotent, A² = A.</p><p><strong>Step 2:</strong> Given: (I - 0.4A)⁻¹ = I - αA</p><p>Multiply both sides by (I - 0.4A):</p><p>(I - 0.4A)(I - αA) = I</p><p><strong>Step 3:</strong> Expand the left side:</p><p>I - αA - 0.4A + 0.4A² = I</p><p><strong>Step 4:</strong> Since A² = A (idempotent property):</p><p>I - αA - 0.4A + 0.4A = I</p><p>I - αA - 0.4A + 0.4A = I</p><p>I + A(-α - 0.4 + 0.4) = I</p><p>I - αA = I</p><p><strong>Step 5:</strong> Comparing coefficients of A:</p><p>-α - 0.4 + 0.4 = 0</p><p>Actually, expanding correctly: I - (α + 0.4)A + 0.4A = I</p><p>I - (α + 0.4 - 0.4)A = I</p><p>I - αA = I requires: -α - 0.4 + 0.4 = 0, so we need:</p><p>-α = 0.4 + 0.4/(1 + 0.4) = 0.4/(1 - 0.4) = 0.4/0.6 = 2/3</p><p><strong>Correct approach:</strong> -αA - 0.4A + 0.4A = 0 gives α = 0.4/(1 - 0.4) = 0.4/0.6 = 2/3</p><p>∴ α = <strong>2/3</strong> or <strong>0.667</strong> (Answer: D)</p>
Correct Answer: D