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: Use the property that A² = A for idempotent matrices to expand (I - 0.4A)⁻¹ and match coefficients with I - αA. The key is recognizing that when you multiply (I - 0.4A)(I - αA), the A² term collapses to A due to idempotency.
<p><strong>Step 1:</strong> Since (I - 0.4A)⁻¹ = I - αA, we have:</p><p>(I - 0.4A)(I - αA) = I</p><p><strong>Step 2:</strong> Expand the left side:</p><p>I - αA - 0.4A + 0.4αA² = I</p><p><strong>Step 3:</strong> Use the idempotent property A² = A:</p><p>I - αA - 0.4A + 0.4αA = I</p><p><strong>Step 4:</strong> Collect A terms:</p><p>I + A(-α - 0.4 + 0.4α) = I</p><p><strong>Step 5:</strong> For this to equal I, the coefficient of A must be zero:</p><p>-α - 0.4 + 0.4α = 0</p><p>-0.6α - 0.4 = 0</p><p>-0.6α = 0.4</p><p>α = -2/3</p><p><strong>Step 6:</strong> Verify: -(-2/3) - 0.4 + 0.4(-2/3) = 2/3 - 0.4 - 4/15 = 10/15 - 6/15 - 4/15 = 0 ✓</p><p>∴ Answer: D (α = -2/3)</p>
Correct Answer: D