Matrices & Determinants
Determinants of Products
Grade 12
Question:
<p>Let <i>A</i> and <i>B</i> be two invertible matrices of order 3 × 3. If det(<i>ABA</i><sup><i>T</i></sup>) = 8 and det(<i>AB</i><sup>−1</sup>) = 8, then det(<i>BA</i><sup>−1</sup><i>B</i><sup><i>T</i></sup>) is equal to</p>
<p>(a) 1</p>
<p>(b) \(\frac{1}{4}\)</p>
<p>(c) \(\frac{1}{16}\)</p>
<p>(d) 16</p>
Step-by-Step Solution
Key Concept: Use the properties det(XY) = det(X)det(Y), det(X^T) = det(X), and det(X^{-1}) = 1/det(X) to convert the given conditions into equations involving det(A) and det(B), then solve the system to find det(BA^{-1}B^T).
<p><strong>Step 1:</strong> Express det(AB A^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 ... (i)</p><p><strong>Step 2:</strong> Express det(AB^{-1}) using determinant properties.</p><p>det(AB^{-1}) = det(A)·det(B^{-1}) = det(A)· rac{1}{det(B)} = 8 ... (ii)</p><p><strong>Step 3:</strong> From equation (ii), find det(A) in terms of det(B).</p><p>\frac{det(A)}{det(B)} = 8</p><p>det(A) = 8·det(B) ... (iii)</p><p><strong>Step 4:</strong> Substitute (iii) into equation (i).</p><p>[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 = \frac{8}{64} = \frac{1}{8}</p><p>det(B) = \frac{1}{2}</p><p><strong>Step 5:</strong> Find det(A) using equation (iii).</p><p>det(A) = 8·\frac{1}{2} = 4</p><p><strong>Step 6:</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)·\frac{1}{det(A)}·det(B)</p><p>= [det(B)]^2·\frac{1}{det(A)}</p><p>= \left(\frac{1}{2}\right)^2·\frac{1}{4}</p><p>= \frac{1}{4}·\frac{1}{4}</p><p>= \frac{1}{16}</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c