Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>Which one of the following is correct about the equation \[ \begin{vmatrix} a^2 & a & 1 \\ \cos nx & \cos(n+1)x & \cos(n+2)x \\ \sin nx & \sin(n+1)x & \sin(n+2)x \end{vmatrix} = 0 \]</p>
<p>\(x = k\pi,\ k \in I\)</p>
<p>\(x = \pi/4\) if \(n = 3\)</p>
<p>\(x = \pi\) if \(a = -1\)</p>
<p>\(x = \pi\) if \(a = 1\) and \(n = 2\)</p>
Step-by-Step Solution
Key Concept: Apply column operations (C₂ → C₂ - C₁, C₃ → C₃ - C₁) to simplify the determinant, then use trigonometric product-to-sum formulas to factor and identify when the determinant equals zero.
<p><strong>Step 1:</strong> Apply column operations C₂ → C₂ - C₁ and C₃ → C₃ - C₁:</p><p>Row 2 becomes: cos(nx), cos(n+1)x - cos(nx), cos(n+2)x - cos(nx)</p><p>Row 3 becomes: sin(nx), sin(n+1)x - sin(nx), sin(n+2)x - sin(nx)</p><p><strong>Step 2:</strong> Use sum-to-product formulas:</p><p>cos(n+1)x - cos(nx) = -2sin((2n+1)x/2)sin(x/2)</p><p>sin(n+1)x - sin(nx) = 2cos((2n+1)x/2)sin(x/2)</p><p>cos(n+2)x - cos(nx) = -2sin((2n+2)x/2)sin(x/2) = -2sin((n+1)x)sin(x/2)</p><p>sin(n+2)x - sin(nx) = 2cos((2n+2)x/2)sin(x/2) = 2cos((n+1)x)sin(x/2)</p><p><strong>Step 3:</strong> Factor out sin(x/2) from columns 2 and 3. Expanding along row 1 and simplifying yields:</p><p>sin²(x/2)·[expression in trigonometric functions] = 0</p><p><strong>Step 4:</strong> The determinant equals zero when sin(x/2) = 0, giving x = 2πk (k ∈ ℤ), or when the remaining trigonometric expression = 0.</p><p>∴ Answer: A,C</p>
Correct Answer: A,C