Matrices & Determinants
Evaluation of Determinants
Grade 12
Question:
<p>If \(D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}\), then \(D\) is divisible by:</p>
<p>Both \(x\) and \(y\)</p>
<p>\(x\) only</p>
<p>\(y\) only</p>
<p>Neither \(x\) nor \(y\)</p>
Step-by-Step Solution
Key Concept: Factor out common terms by performing row/column operations: subtract Row 1 from Rows 2 and 3 to create a structure where x and y appear as isolated factors in the resulting determinant.
<p><strong>Step 1:</strong> Apply row operations. Subtract Row 1 from Row 2 and Row 3:</p><p>$$D = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix}$$</p><p><strong>Step 2:</strong> Expand along the first column (only non-zero entry is 1):</p><p>$$D = 1 \cdot \begin{vmatrix} x & 0 \\ 0 & y \end{vmatrix} = 1 \cdot (xy - 0) = xy$$</p><p><strong>Step 3:</strong> Therefore D = xy, which is divisible by both x and y individually, and by xy.</p><p>∴ Answer: A (D is divisible by xy)</p>
Correct Answer: A