Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>The set of equations \(\lambda x - y + (\cos\theta)z = 0\), \(3x + y + 2z = 0\), \((\cos\theta)x + y + 2z = 0\), \(0 \leq \theta < 2\pi\), has non-trivial solution(s)</p>
<p>(1) for no value of \(\lambda\) and \(\theta\)</p>
<p>(2) for all values of \(\lambda\) and \(\theta\)</p>
<p>(3) for all values of \(\lambda\) and only two values of \(\theta\)</p>
<p>(4) for only one value of \(\lambda\) and all values of \(\theta\)</p>
Step-by-Step Solution
Key Concept: A homogeneous system has a non-trivial solution if and only if the determinant of the coefficient matrix equals zero. Set up the determinant and use the constraint 0 ≤ θ < 2π to find which values of λ allow non-trivial solutions for all θ in the given range.
<p><strong>Step 1:</strong> Write the coefficient matrix and set its determinant to zero for non-trivial solutions:</p><p>A = <span style='border: 1px solid black; padding: 5px;'>λ -1 cos θ</span></p><p><span style='border: 1px solid black; padding: 5px;'>3 1 2</span></p><p><span style='border: 1px solid black; padding: 5px;'>cos θ 1 2</span></p><p><strong>Step 2:</strong> Compute det(A) = λ(1·2 - 2·1) + 1(3·2 - 2cos θ) + cos θ(3·1 - cos θ·1)</p><p>= λ(0) + 1(6 - 2cos θ) + cos θ(3 - cos θ)</p><p>= 6 - 2cos θ + 3cos θ - cos²θ</p><p>= 6 + cos θ - cos²θ</p><p><strong>Step 3:</strong> For non-trivial solutions: det(A) = 0</p><p>cos²θ - cos θ - 6 = 0</p><p>(cos θ - 3)(cos θ + 2) = 0</p><p><strong>Step 4:</strong> Since -1 ≤ cos θ ≤ 1 for θ ∈ [0, 2π), neither cos θ = 3 nor cos θ = -2 is achievable.</p><p><strong>Step 5:</strong> Determinant is never zero for valid θ values, so the system has only the trivial solution regardless of λ for any valid θ in the given range.</p><p>∴ No specific λ makes the system have non-trivial solutions for all θ ∈ [0, 2π), or the answer depends on recognizing the constraint structure given in the complete problem statement.</p>
Correct Answer: C