Matrices & Determinants
Determinant Zeros
Grade 12
Question:
<p>If \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)\), where \(a, b\) and \(c\) are all different, then the determinant \(\begin{vmatrix} \frac{1}{(x-a)^2} & \frac{1}{(x-b)^2} & \frac{1}{(x-c)^2} \\ \frac{1}{(x-b)(x-c)} & \frac{1}{(x-c)(x-a)} & \frac{1}{(x-a)(x-b)} \end{vmatrix}\) vanishes when</p>
<p>(a) \(a + b + c = 0\)</p>
<p>(b) \(x = \frac{a+b+c}{3}\)</p>
<p>(c) \(x = 0\)</p>
<p>(d) \(a = b = c\)</p>
Step-by-Step Solution
Key Concept: Linear dependence of rows in rational function determinants occurs at specific symmetric values of the variable.
<p>The given identity establishes a Vandermonde-type relation. For the second determinant, substitute rational functions and use partial fractions. The determinant vanishes when the rows become linearly dependent, which occurs when $x = \frac{a+b+c}{3}$ (the centroid condition).</p>
Correct Answer: B