Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>Let \(A\) and \(B\) be two invertible matrices of order \(3 \times 3\). If \(\det(ABA^T) = 8\) and \(\det(AB^{-1}) = 8\), then \(\det(BA^{-1}B^T)\) is equal to:</p>
<p>\(\dfrac{1}{4}\)</p>
<p>1</p>
<p>\(\dfrac{1}{16}\)</p>
<p>16</p>
Step-by-Step Solution
Key Concept: Use the property that det(XYZ) = det(X)·det(Y)·det(Z) and det(M^T) = det(M) to express unknowns in terms of det(A) and det(B), then solve the system of equations from the two given conditions.
<p><strong>Step 1:</strong> Simplify det(ABA^T) using determinant properties.</p><p>det(ABA^T) = det(A)·det(B)·det(A^T) = det(A)·det(B)·det(A) = [det(A)]²·det(B) = 8</p><p><strong>Step 2:</strong> Simplify det(AB^{-1}) using determinant properties.</p><p>det(AB^{-1}) = det(A)·det(B^{-1}) = det(A)·(1/det(B)) = 8</p><p>Therefore: det(A)/det(B) = 8 ... (equation 2)</p><p><strong>Step 3:</strong> From equation 2: det(A) = 8·det(B)</p><p>Substitute into equation 1: [8·det(B)]²·det(B) = 8</p><p>64·[det(B)]²·det(B) = 8</p><p>64·[det(B)]³ = 8</p><p>[det(B)]³ = 1/8</p><p>det(B) = 1/2</p><p><strong>Step 4:</strong> Find det(A): det(A) = 8·(1/2) = 4</p><p><strong>Step 5:</strong> Calculate det(BA^{-1}B^T).</p><p>det(BA^{-1}B^T) = det(B)·det(A^{-1})·det(B^T)</p><p>= det(B)·(1/det(A))·det(B)</p><p>= [det(B)]²/det(A)</p><p>= (1/2)²/4 = (1/4)/4 = 1/16</p><p>∴ Answer: A (or 1/16)</p>
Correct Answer: A