<p>If \(P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}\) is the adjoint of a 3 × 3 matrix <em>A</em> and |<em>A</em>| = 4, then \(\alpha\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the fundamental property that A·adj(A) = |A|·I, which means adj(A) must satisfy this matrix equation. Since P = adj(A) and |A| = 4, we can verify which value of α makes P·A = 4I solvable, or equivalently, use the fact that |adj(A)| = |A|^(n-1) for an n×n matrix.
<p><strong>Step 1:</strong> Use the property that for a 3×3 matrix: |adj(A)| = |A|^(3-1) = |A|^2</p><p><strong>Step 2:</strong> Since P is the adjoint of A and |A| = 4, we have |P| = 4^2 = 16</p><p><strong>Step 3:</strong> Calculate the determinant of P:</p><p>|P| = 1(3·4 - 3·4) - α(1·4 - 3·2) + 3(1·4 - 3·2)</p><p>= 1(12 - 12) - α(4 - 6) + 3(4 - 6)</p><p>= 0 - α(-2) + 3(-2)</p><p>= 2α - 6</p><p><strong>Step 4:</strong> Set |P| = 16:</p><p>2α - 6 = 16</p><p>2α = 22</p><p>α = 11</p><p>∴ Answer: B</p>
Correct Answer: B