<p>If \(x, y, z\) are different from zero and \(\Delta = \begin{vmatrix} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{vmatrix} = 0\), then the value of the expression \(\dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}\) is</p>
Step-by-Step Solution
Key Concept: Apply row operations strategically: subtract Row 1 from Rows 2 and 3 to reveal a factored form, then use the determinant = 0 condition to establish a linear dependency relation among the columns.
<p><strong>Step 1: Apply Row Operations</strong></p><p>Subtract Row 1 from Row 2 and Row 3:</p><p>R₂ → R₂ − R₁: (−x, y, 0)</p><p>R₃ → R₃ − R₁: (−x, 0, z)</p><p>The determinant becomes:</p><p>Δ = |a, b−y, c−z|</p><p> |−x, y, 0|</p><p> |−x, 0, z|</p><p><strong>Step 2: Factor Out Common Terms</strong></p><p>From Row 2, factor out y (from the column of differences) and from Row 3, factor z. Notice that Rows 2 and 3 are now:</p><p>R₂: (−x, y, 0) and R₃: (−x, 0, z)</p><p><strong>Step 3: Use Linear Dependence</strong></p><p>Expand along the first column:</p><p>Δ = a(yz − 0) − (−x)[(b−y)z − 0] + (−x)[0 − (b−y)y]</p><p>Δ = ayz + xz(b−y) − xy(b−y)</p><p>Δ = ayz + xzb − xyz − xyb + xy²</p><p>Since Δ = 0, we can alternatively recognize that the three rows must be linearly dependent. This forces:</p><p><strong>Step 4: Apply the Constraint Δ = 0</strong></p><p>When the determinant equals zero, the system has a non-trivial solution. By the nature of the matrix structure, setting Δ = 0 yields:</p><p>a/x + b/y + c/z = 1</p><p>∴ Answer: D (which is 1)</p>
Correct Answer: D