<p>If \(a - b + c = 0\), then one root of \(\begin{vmatrix} a-x & c & b \\ c & b-x & a \\ b & a & c-x \end{vmatrix} = 0\) is</p>
Step-by-Step Solution
Key Concept: When a - b + c = 0, the matrix has a special structure where the rows satisfy a linear dependency: R₁ + R₂ + R₃ has proportional entries to a vector like (1, 1, 1). Substituting x = a + c reveals this relationship makes the determinant zero.
<p><strong>Step 1:</strong> Use the constraint a - b + c = 0, which gives b = a + c.</p><p><strong>Step 2:</strong> The matrix becomes:</p><p>$$\begin{vmatrix} a-x & c & a+c \\ c & a+c-x & a \\ a+c & a & c-x \end{vmatrix}$$</p><p><strong>Step 3:</strong> Add all three rows: (a - x + c + a + c) + (c + a + c - x + a) + (a + c + a + c - x) = 3(a + c) - 3x</p><p><strong>Step 4:</strong> When x = a + c, the sum of rows becomes 3(a + c) - 3(a + c) = 0, meaning the rows become linearly dependent.</p><p><strong>Step 5:</strong> Linear dependence of rows implies determinant = 0.</p><p>∴ Answer: x = a + c (Option D)</p>
Correct Answer: D