Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>If \(\begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{vmatrix} = 0\) and \((1, a, a^2)\), \((1, b, b^2)\), \((1, c, c^2)\) are non-coplanar, then \(abc\) equals:</p>
<p>(1) 0</p>
<p>(2) −1</p>
<p>(3) 1</p>
<p>(4) −2</p>
Step-by-Step Solution
Key Concept: Factor the determinant by splitting the third column as 1 + a³ = 1 + a·a², then use the property that det(A) + det(B) = det(A+B) when columns differ only in one position. Recognize the Vandermonde determinant and the condition that rows are non-coplanar (meaning the Vandermonde determinant is non-zero).
<p><strong>Step 1:</strong> Decompose the given determinant by splitting the third column:</p><p>$$\begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> For the first determinant, expand and rearrange to recognize it as related to the Vandermonde determinant:</p><p>$$\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$$</p><p>For the second determinant, factor out from each row:</p><p>$$\begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = abc\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$$</p><p><strong>Step 3:</strong> Substitute back:</p><p>$$(b-a)(c-a)(c-b) + abc(b-a)(c-a)(c-b) = 0$$</p><p>$$(b-a)(c-a)(c-b)[1 + abc] = 0$$</p><p><strong>Step 4:</strong> Since (1,a,a²), (1,b,b²), (1,c,c²) are non-coplanar, the Vandermonde determinant (b-a)(c-a)(c-b) ≠ 0. Therefore:</p><p>$$1 + abc = 0$$</p><p>∴ Answer: <strong>abc = -1</strong> (Option B)</p>
Correct Answer: B