Matrices & Determinants
Properties of Determinants
Grade None
Question:
<p>If \(D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}\), for \(x \neq 0,\, y \neq 0\) then \(D\) is</p>
<p>divisible by neither \(x\) nor \(y\)</p>
<p>divisible by both \(x\) and \(y\)</p>
<p>divisible by \(x\) but not \(y\)</p>
<p>divisible by \(y\) but not \(x\)</p>
Step-by-Step Solution
Key Concept: Apply row/column operations to simplify the determinant: subtract the first row from rows 2 and 3 to isolate x and y terms, then expand along the first column to reveal the factored form xy.
<p><strong>Step 1:</strong> Apply row operations. Subtract Row 1 from Row 2 and Row 3:</p><p>R₂ → R₂ - R₁ and R₃ → R₃ - R₁</p><p>$$D = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix}$$</p><p><strong>Step 2:</strong> This is now an upper triangular matrix. The determinant equals the product of diagonal elements:</p><p>$$D = 1 \cdot x \cdot y = xy$$</p><p><strong>Step 3:</strong> Verify: Since x ≠ 0 and y ≠ 0, we have D = xy ≠ 0, confirming the matrix is invertible.</p><p>∴ Answer: <strong>D = xy</strong> (Option B)</p>
Correct Answer: B