Matrices & Determinants
Adjoint and Inverse of a Matrix
Grade 12
Question:
<p>If \(A\) and \(B\) are square matrices of order 3 such that \(\det(A) = -2\) and \(\det(B) = 1\), then find the value of \(\det\left(A^{-1} \cdot \text{adj}(B^{-1}) \cdot \text{adj}(2A^{-1})\right)\).</p>
Step-by-Step Solution
Key Concept: Use the properties: det(adj(M)) = [det(M)]^(n-1) for order n, det(M^(-1)) = 1/det(M), and det(kM) = k^n·det(M) for scalar k and order n matrices. Chain these with the product rule det(XYZ) = det(X)·det(Y)·det(Z).
<p><strong>Step 1: Find det(A⁻¹)</strong></p><p>Since det(A⁻¹) = 1/det(A) = 1/(-2) = -1/2</p><p><strong>Step 2: Find det(adj(B⁻¹))</strong></p><p>First, det(B⁻¹) = 1/det(B) = 1/1 = 1</p><p>For order 3: det(adj(B⁻¹)) = [det(B⁻¹)]² = (1)² = 1</p><p><strong>Step 3: Find det(2A⁻¹)</strong></p><p>det(2A⁻¹) = 2³·det(A⁻¹) = 8·(-1/2) = -4</p><p><strong>Step 4: Find det(adj(2A⁻¹))</strong></p><p>det(adj(2A⁻¹)) = [det(2A⁻¹)]² = (-4)² = 16</p><p><strong>Step 5: Apply product rule</strong></p><p>det(A⁻¹·adj(B⁻¹)·adj(2A⁻¹)) = det(A⁻¹)·det(adj(B⁻¹))·det(adj(2A⁻¹))</p><p>= (-1/2)·(1)·(16)</p><p>= -8</p><p>∴ Answer: <strong>-8</strong></p>
Correct Answer: -8