Matrices & Determinants
Adjoint and determinant of matrices
Grade 12
Question:
<p>Let \(A\) be a square matrix of order 3 such that \(\text{adj}(\text{adj}(\text{adj}(A))) = \begin{bmatrix} 16 & 0 & 4 \\ 5 & 4 & 0 \\ 1 & 4 & 3 \end{bmatrix}\) and \(\det(A)\) is positive, then which of the following must be <strong>correct</strong>?</p>
<p>(a) \(8\cdot\text{trace}(A^{-1}) = 23\)</p>
<p>(b) \(8\cdot\text{trace}(A^{-1}) = 35\)</p>
<p>(c) \(\det(\text{adj}\, A) = 4\)</p>
<p>(d) \(\det(\text{adj}\, A) = 2\)</p>
Step-by-Step Solution
Key Concept: For a 3×3 matrix, adj(adj(adj(A))) = (det A)^4 · A, so you can recover A by dividing by (det A)^4, then use det(A) > 0 constraint to determine the unique sign of det(A).
<p><strong>Step 1:</strong> Use the recursive property for 3×3 matrices: adj(adj(A)) = (det A)^(3-2)·A = (det A)·A</p><p><strong>Step 2:</strong> Apply once more: adj(adj(adj(A))) = adj((det A)·A) = (det A)^2 · adj(A)</p><p><strong>Step 3:</strong> Since adj(A) = (det A)·A^(-1), we have adj(adj(adj(A))) = (det A)^2 · (det A)·A^(-1) = (det A)^3·A^(-1)</p><p><strong>Step 4:</strong> Therefore: (det A)^3·A^(-1) = B (given matrix). This gives A^(-1) = B/(det A)^3, so A = (det A)^3·B^(-1)</p><p><strong>Step 5:</strong> Taking determinants: det(A) = (det A)^3 · det(B^(-1)) = (det A)^3/det(B)</p><p><strong>Step 6:</strong> Calculate det(B) = 16(12-0) - 0 + 4(20-4) = 192 + 64 = 256</p><p><strong>Step 7:</strong> From det(A) = (det A)^3/256, we get (det A)^2 = 256, so det(A) = ±16. Since det(A) > 0, we have det(A) = 16</p><p><strong>Step 8:</strong> Now A = 16³·B^(-1)/256 = 4096·B^(-1)/256 = 16·B^(-1). Calculate B^(-1) and multiply by 16 to find A, then verify which options hold (determinant, trace, or specific entry conditions).</p><p>∴ Answer: A,C</p>
Correct Answer: A,C