Matrices & Determinants
Adjoint of a Matrix
Grade 12
Question:
<p>Given that matrix \(A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}\). If \(xyz = 60\) and \(8x + 4y + 3z = 20\), then \(A(\text{adj }A)\) is equal to</p>
<p>\(\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 88 & 0 & 0 \\ 0 & 88 & 0 \\ 0 & 0 & 88 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 34 & 0 & 0 \\ 0 & 34 & 0 \\ 0 & 0 & 34 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: For any square matrix A, the product A(adj A) = (det A)·I, where det A is a scalar multiple of the identity matrix. The constraint equations allow us to find det A, which is the only information needed to determine A(adj A).
<p><strong>Step 1:</strong> Recall the fundamental property: For any square matrix A of order n, <br/>A(adj A) = (adj A)·A = (det A)·I<br/>where I is the identity matrix.</p><p><strong>Step 2:</strong> Calculate det A using cofactor expansion along row 1:<br/>det A = x(yz - 8) - 3(z - 8) + 2(2 - 2y)<br/>det A = xyz - 8x - 3z + 24 + 4 - 4y<br/>det A = xyz - 8x - 4y - 3z + 28</p><p><strong>Step 3:</strong> Substitute the given constraints xyz = 60 and 8x + 4y + 3z = 20:<br/>det A = 60 - (8x + 4y + 3z) + 28<br/>det A = 60 - 20 + 28<br/>det A = 68</p><p><strong>Step 4:</strong> Therefore, A(adj A) = 68·I = <br/>∴ Answer: C (which is 68I, a 3×3 matrix with 68 on diagonal and 0 elsewhere)</p>
Correct Answer: C