<p>If \(p = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A| = 4\), then \(\alpha\) is equal to</p>
Step-by-Step Solution
Key Concept: If p is the adjoint of matrix A, then p = adj(A). We use the fundamental property: A·adj(A) = |A|·I, which means adj(A) has a special relationship with A's determinant. Additionally, |adj(A)| = |A|^(n-1) for an n×n matrix.
<p><strong>Step 1:</strong> Recall that if p = adj(A), then for a 3×3 matrix: |adj(A)| = |A|^(3-1) = |A|^2</p><p><strong>Step 2:</strong> Given |A| = 4, we have: |adj(A)| = 4^2 = 16</p><p><strong>Step 3:</strong> Calculate the determinant of p using cofactor expansion along the first row:</p><p>p = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}</p><p>= 1·\begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} - \alpha·\begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + 3·\begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix}</p><p><strong>Step 4:</strong> Evaluate the 2×2 determinants:</p><p>\begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = 3(4) - 3(4) = 12 - 12 = 0</p><p>\begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} = 1(4) - 3(2) = 4 - 6 = -2</p><p><strong>Step 5:</strong> Substitute back into the determinant expression:</p><p>|p| = 1(0) - \alpha(-2) + 3(-2) = 0 + 2\alpha - 6 = 2\alpha - 6</p><p><strong>Step 6:</strong> Set |p| = 16 (from Step 2):</p><p>2\alpha - 6 = 16</p><p>2\alpha = 22</p><p>\alpha = 11</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A