Matrices & Determinants
Adjoint and Determinant
Grade 12
Question:
<p>We have \(p = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}\). If \(|\text{adj}\, A| = |A|^2 \Rightarrow |\text{adj}\, A| = 16\), find \(\alpha\).</p>
<p>(1) \(\alpha = 11\)</p>
<p>(2) \(\alpha = 5\)</p>
<p>(3) \(\alpha = 4\)</p>
<p>(4) \(\alpha = 3\)</p>
Step-by-Step Solution
Key Concept: For a 3×3 matrix, |adj A| = |A|² always holds by the adjugate property. Use this with |adj A| = 16 to find |A| = ±4, then compute the determinant of P in terms of α and solve.
<p><strong>Step 1:</strong> Recall that for an n×n matrix, |adj A| = |A|^(n-1). For n=3: |adj P| = |P|²</p><p><strong>Step 2:</strong> Given |adj P| = 16, so |P|² = 16, which means |P| = ±4</p><p><strong>Step 3:</strong> Calculate |P| using cofactor expansion along row 1:</p><p>|P| = 1(3·4 - 3·4) - α(1·4 - 3·2) + 3(1·4 - 3·2)</p><p>|P| = 1(0) - α(-2) + 3(-2)</p><p>|P| = 2α - 6</p><p><strong>Step 4:</strong> Set 2α - 6 = 4 or 2α - 6 = -4</p><p>From first: α = 5</p><p>From second: α = 1</p><p><strong>Step 5:</strong> Verify which satisfies the context (typically α = 5 for standard form, but both are mathematically valid unless additional constraints exist).</p><p>∴ Answer: <strong>α = 5</strong> (or α = 1)</p>
Correct Answer: A