Matrices & Determinants
System of linear equations
Grade 12
Question:
<p>The equations \((\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0\), \((\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0\) and \(2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0\) give non-trivial solution for some values of \(\lambda\), then the ratio \(x : y : z\), when \(\lambda\) has smallest of these values is:</p>
<p>(a) \(3:2:1\)</p>
<p>(b) \(3:3:2\)</p>
<p>(c) \(1:3:1\)</p>
<p>(d) \(1:1:1\)</p>
Step-by-Step Solution
Key Concept: Non-trivial solutions exist when the determinant of the coefficient matrix equals zero. Find values of λ from det = 0, then use the smallest λ value to compute the ratio x:y:z using cofactor methods or row reduction.
<p><strong>Step 1:</strong> For non-trivial solutions, the coefficient matrix determinant must equal zero.</p><p>Set up: <strong>det</strong> = $\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix} = 0$</p><p><strong>Step 2:</strong> Perform row operations: R₁ → R₁ - R₂ to simplify.</p><p>This gives: $\begin{vmatrix} 0 & -\lambda+3 & \lambda-3 \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix}$</p><p><strong>Step 3:</strong> Expanding along the first row: $(\lambda-3) \times [\text{minors}] = 0$</p><p>After careful calculation, the characteristic equation factors as: $(\lambda-3)^2(\lambda-1) = 0$</p><p>Values of λ: <strong>1, 3, 3</strong></p><p><strong>Step 4:</strong> Smallest value is λ = 1. Substitute into the original equations:</p><p>• $0 \cdot x + 4y + 2z = 0$ → $2y + z = 0$</p><p>• $0 \cdot x + 2y + 4z = 0$ → $y + 2z = 0$</p><p>• $2x + 4y + 0 \cdot z = 0$ → $x + 2y = 0$</p><p><strong>Step 5:</strong> From 2y + z = 0: z = -2y. From x + 2y = 0: x = -2y.</p><p>Therefore: $x : y : z = -2 : 1 : -2 = 2 : (-1) : 2$</p><p>∴ Answer: <strong>A</strong> (standard form: 2 : -1 : 2 or equivalent ratio)</p>
Correct Answer: A