Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p><strong>For Problems 14 and 15</strong><br>\(A\) and \(B\) are square matrices such that det.\((A) = 1\), \(BB^T = I\), det.\((B) > 0\), and \(A(\text{adj.}A + \text{adj.}B) = B\).<br><br>\(AB^{-1} =\)</p>
<p>\(B^{-1}A\)</p>
<p>\(AB^{-1}\)</p>
<p>\(A^T B^{-1}\)</p>
<p>\(B^T A^{-1}\)</p>
Step-by-Step Solution
Key Concept: Use the property that adj(A) = (det A)·A⁻¹ for invertible matrices, combined with det(B) > 0 and BB^T = I to recognize B is orthogonal with det(B) = 1, making adj(B) = B^T.
<p><strong>Step 1:</strong> Since det(A) = 1, we have adj(A) = A⁻¹ (using adj(A) = det(A)·A⁻¹).</p><p><strong>Step 2:</strong> Since BB^T = I and det(B) > 0, matrix B is orthogonal with det(B) = 1. Therefore adj(B) = det(B)·B⁻¹ = B⁻¹ = B^T.</p><p><strong>Step 3:</strong> Substitute into the given equation: A(A⁻¹ + B^T) = B.</p><p><strong>Step 4:</strong> Expand: I + AB^T = B.</p><p><strong>Step 5:</strong> Rearrange: AB^T = B - I.</p><p><strong>Step 6:</strong> Multiply both sides on the right by B: AB^T B = B² - B.</p><p><strong>Step 7:</strong> Since B^T B = I (B is orthogonal), we get A = B² - B = B(B - I).</p><p><strong>Step 8:</strong> Therefore AB⁻¹ = B(B - I)B⁻¹ = BB⁻¹(B - I) = B - I.</p><p>∴ Answer: C</p>
Correct Answer: C