<p>Match the following lists:</p><p><b>List I</b></p><p>a. \((I - A)^n\) is if \(A\) is idempotent</p><p>b. \((I - A)^n\) is if \(A\) is involuntary</p><p>c. \((I - A)^n\) is if \(A\) is nilpotent of index 2</p><p>d. If \(A\) is orthogonal, then \((A^T)^{-1}\)</p><p><b>List II</b></p><p>p. \(2^{n-1}(I - A)\)</p><p>q. \(I - nA\)</p><p>r. \(A\)</p><p>s. \(I - A\)</p>
Step-by-Step Solution
Key Concept: Recognize special matrix properties (idempotent A²=A, involuntary A²=I, nilpotent A²=0, orthogonal A^T·A=I) and use them to simplify powers of (I-A) or matrix expressions through binomial expansion and telescoping series.
<p><strong>a) A is idempotent (A² = A):</strong></p><p>For (I-A)ⁿ: Note that I-A commutes with A. Since A² = A, we have (I-A)² = I - 2A + A² = I - 2A + A = I - A.</p><p>By induction: (I-A)ⁿ = I - A for all n ≥ 1. ✓ <strong>Answer: s</strong></p><p><strong>b) A is involuntary (A² = I):</strong></p><p>For (I-A)ⁿ: We have (I-A)² = I - 2A + A² = I - 2A + I = 2(I-A).</p><p>By induction: (I-A)ⁿ = 2ⁿ⁻¹(I-A). ✓ <strong>Answer: p</strong></p><p><strong>c) A is nilpotent of index 2 (A² = 0):</strong></p><p>For (I-A)ⁿ: Using binomial expansion, (I-A)ⁿ = I - nA + C(n,2)A² - ... = I - nA (since A² = 0, all higher terms vanish). ✓ <strong>Answer: q</strong></p><p><strong>d) A is orthogonal (A·Aᵀ = I):</strong></p><p>If A is orthogonal, then A⁻¹ = Aᵀ, so (Aᵀ)⁻¹ = (A⁻¹)ᵀ = (Aᵀ)ᵀ = A. ✓ <strong>Answer: r</strong></p><p>∴ Answer: a-s, b-p, c-q, d-r</p>
Correct Answer: a-p, b-s, c-q, d-r