<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>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must equal zero. Find all λ values satisfying det(A) = 0, identify the smallest λ, then find the ratio x:y:z using cofactors or row reduction at that specific λ value.
<p><strong>Step 1:</strong> For non-trivial solutions of the homogeneous system, set the coefficient matrix determinant to zero:</p><p>$$\begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Perform row operations: R₁ → R₁ - R₂ to get:</p><p>$$\begin{vmatrix} 0 & -\lambda+3 & \lambda-3 \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3\lambda-3 \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Expand along the first column:</p><p>$$(\lambda-1)\begin{vmatrix} -\lambda+3 & \lambda-3 \\ 3\lambda+1 & 3\lambda-3 \end{vmatrix} - 2\begin{vmatrix} -\lambda+3 & \lambda-3 \\ 4\lambda-2 & \lambda+3 \end{vmatrix} = 0$$</p><p><strong>Step 4:</strong> Compute the 2×2 determinants and simplify to get: $(\lambda-1)(\lambda-3)(5\lambda-1) = 0$</p><p>So λ = 1, λ = 3, or λ = 1/5. The smallest value is <strong>λ = 1/5</strong>.</p><p><strong>Step 5:</strong> Substitute λ = 1/5 into the original equations and solve using cofactor method or Gaussian elimination:</p><p>The system becomes: $-\frac{4}{5}x + \frac{8}{5}y + \frac{2}{5}z = 0$, etc.</p><p>Solving this system yields: <strong>x : y : z = 1 : 1 : -2</strong></p><p>∴ Answer: A</p>
Correct Answer: A