Matrices & Determinants
Adjoint and Inverse
Grade 12
Question:
<p>If \(A\) and \(B\) are two invertible matrices of the same order, then \(\text{adj}(AB)\) is equal to</p>
<p>(1) \(\text{adj}(B)\,\text{adj}(A)\)</p>
<p>(2) \(|B|\,|A|\,B^{-1}A^{-1}\)</p>
<p>(3) \(|B|\,|A|\,A^{-1}B^{-1}\)</p>
<p>(4) \(|A|\,|B|\,(AB)^{-1}\)</p>
Step-by-Step Solution
Key Concept: The adjugate function reverses the order of multiplication: adj(AB) = adj(B)·adj(A), analogous to how (AB)⁻¹ = B⁻¹A⁻¹. This reversal occurs because the adjugate involves cofactors which transform multiplicatively in reverse order.
<p><strong>Step 1:</strong> Recall that for invertible matrices: <strong>A·adj(A) = det(A)·I</strong></p><p><strong>Step 2:</strong> Consider the product AB. We know: <strong>(AB)·adj(AB) = det(AB)·I</strong></p><p><strong>Step 3:</strong> Also, det(AB) = det(A)·det(B), so: <strong>AB·adj(AB) = det(A)·det(B)·I</strong></p><p><strong>Step 4:</strong> Now compute: <strong>AB·adj(B)·adj(A) = A·[B·adj(B)]·adj(A) = A·det(B)·I·adj(A) = det(B)·A·adj(A)</strong></p><p><strong>Step 5:</strong> Continue: <strong>det(B)·A·adj(A) = det(B)·det(A)·I = det(AB)·I</strong></p><p><strong>Step 6:</strong> Since AB·adj(AB) = det(AB)·I and AB·adj(B)·adj(A) = det(AB)·I, and AB is invertible, we conclude: <strong>adj(AB) = adj(B)·adj(A)</strong></p><p>∴ Answer: <strong>adj(B)·adj(A)</strong> (option D)</p>
Correct Answer: D