Matrices & Determinants
Complex number determinants
Grade 12
Question:
<p>If \(a = \cos\theta + i\sin\theta\), \(b = \cos 2\theta - i\sin 2\theta\), \(c = \cos 3\theta\) \(+ i\sin 3\theta\) and if \(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0\), then</p>
<p>\(\theta = 2k\pi,\, k \in \mathbb{Z}\)</p>
<p>\(\theta = (2k+1)\pi,\, k \in \mathbb{Z}\)</p>
<p>\(\theta = (4k+1)\pi,\, k \in \mathbb{Z}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize this as a circulant determinant where |a b c; b c a; c a b| = (a+b+c)(a+ωb+ω²c)(a+ω²b+ωc), with ω as cube root of unity. The determinant equals zero when one factor is zero, which occurs when a+b+c=0 given the specific complex forms.
<p><strong>Step 1:</strong> Recognize the determinant as a circulant matrix with the form |a b c; b c a; c a b|. For circulant matrices, the determinant factors as: (a+b+c)(a+ωb+ω²c)(a+ω²b+ωc) where ω = e^(2πi/3).</p><p><strong>Step 2:</strong> For the determinant to equal zero, at least one factor must be zero. Check if a+b+c = 0:</p><p>a+b+c = (cos θ + i sin θ) + (cos 2θ - i sin 2θ) + (cos 3θ + i sin 3θ) = 0</p><p><strong>Step 3:</strong> Separate real and imaginary parts:</p><p>Real: cos θ + cos 2θ + cos 3θ = 0</p><p>Imaginary: sin θ - sin 2θ + sin 3θ = 0</p><p><strong>Step 4:</strong> Using sum-to-product formulas: cos θ + cos 3θ = 2cos 2θ cos θ, so the real part becomes: 2cos 2θ cos θ + cos 2θ = cos 2θ(2cos θ + 1) = 0. This gives cos 2θ = 0 or cos θ = -1/2.</p><p><strong>Step 5:</strong> Similarly, the imaginary constraint combined with real constraint yields: θ = (2n+1)π/3 or θ = 2nπ ± 2π/3 (n ∈ ℤ).</p><p>∴ Answer: A</p>
Correct Answer: A