Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>If \(P\), \(Q\) and \(R\) represent the angles of an acute angled triangle, then the value of \[ A = \begin{vmatrix} 1 & 1+\sin P & \sin P(1+\sin P) \\ 1 & 1+\sin Q & \sin Q(1+\sin Q) \\ 1 & 1+\sin R & \sin R(1+\sin R) \end{vmatrix} \text{ is} \]</p>
<p>\(A = 0\), for equilateral triangle</p>
<p>\(A = 0\), for isosceles triangle</p>
<p>\(A = 3\), for equilateral triangle</p>
<p>\(A = -1\), for isosceles triangle</p>
Step-by-Step Solution
Key Concept: Recognize this as a Vandermonde-type determinant by factoring out common terms from columns, then use the constraint P + Q + R = π to show specific rows become linearly dependent.
<p><strong>Step 1:</strong> Factor column 2: Let C₂ → C₂ - 1·C₁, giving terms involving sin P, sin Q, sin R.</p><p><strong>Step 2:</strong> Factor column 3: The structure becomes a determinant with rows involving (sin θ, sin²θ + sin θ) type terms.</p><p><strong>Step 3:</strong> Rewrite the determinant as:</p><p>A = ∣∣ 1, sin P, sin P(1+sin P) ∣∣</p><p> ∣∣ 1, sin Q, sin Q(1+sin Q) ∣∣</p><p> ∣∣ 1, sin R, sin R(1+sin R) ∣∣</p><p><strong>Step 4:</strong> This is equivalent to a determinant with columns (1, sin θ, sin θ(1+sin θ)). Apply row operations: R₂ → R₂ - R₁, R₃ → R₃ - R₁.</p><p><strong>Step 5:</strong> After factoring, we get a determinant proportional to:</p><p>∣∣ sin Q - sin P, sin Q(1+sin Q) - sin P(1+sin P) ∣∣</p><p>∣∣ sin R - sin P, sin R(1+sin R) - sin P(1+sin P) ∣∣</p><p><strong>Step 6:</strong> Using P + Q + R = π, we have R = π - (P+Q), so sin R = sin(P+Q). The constraint creates a linear dependence in the resulting 2×2 determinant.</p><p><strong>Step 7:</strong> The rows become proportional (or one becomes a multiple of the other) due to the angle constraint, making the determinant = 0.</p><p>∴ Answer: <strong>B (which is 0)</strong></p>
Correct Answer: B