Matrices & Determinants
Properties of Matrices
Grade 12
Question:
<p>If <i>A</i> and <i>B</i> are square matrices of the same order and <i>A</i> is nonsingular, then for a positive integer <i>n</i>, \((A^{-1} B A)^n\) is equal to</p>
<p>\(A^{-n} B^n A^n\)</p>
<p>\(A^n B^n A^{-n}\)</p>
<p>\(A^{-1} B^n A\)</p>
<p>\(n(A^{-1} B A)\)</p>
Step-by-Step Solution
Key Concept: Use the associative property of matrix multiplication repeatedly: (A⁻¹BA)ⁿ = A⁻¹BA · A⁻¹BA · ... · A⁻¹BA, where consecutive terms A·A⁻¹ collapse to I, leaving A⁻¹BⁿA.
<p><strong>Step 1:</strong> Write out the product for general n:</p><p>(A⁻¹BA)ⁿ = (A⁻¹BA)(A⁻¹BA)(A⁻¹BA)...(A⁻¹BA) [n times]</p><p><strong>Step 2:</strong> Apply associativity by regrouping:</p><p>= A⁻¹B(AA⁻¹)B(AA⁻¹)B...(AA⁻¹)BA</p><p><strong>Step 3:</strong> Since AA⁻¹ = I, each middle term collapses:</p><p>= A⁻¹B·I·B·I·B...·I·BA</p><p><strong>Step 4:</strong> Simplify using I as identity:</p><p>= A⁻¹(BBB...B)A = A⁻¹BⁿA</p><p><strong>Verification:</strong> For n=1: (A⁻¹BA)¹ = A⁻¹B¹A ✓</p><p>∴ Answer: <strong>A⁻¹BⁿA</strong></p>
Correct Answer: C