<p>The number of values of <em>k</em> for which the linear equations</p><p>\(4x + ky + 2z = 0\)</p><p>\(kx + 4y + z = 0\)</p><p>\(2x + 2y + z = 0\)</p><p>possess a non-zero solution is</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must equal zero. Set up and solve det(A) = 0 to find all values of k.
<p><strong>Step 1:</strong> For non-zero solutions to exist in a homogeneous system, the coefficient matrix determinant must be zero:</p><p>$$\begin{vmatrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand along the first row:</p><p>$$4\begin{vmatrix} 4 & 1 \\ 2 & 1 \end{vmatrix} - k\begin{vmatrix} k & 1 \\ 2 & 1 \end{vmatrix} + 2\begin{vmatrix} k & 4 \\ 2 & 2 \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Calculate 2×2 determinants:</p><p>$$4(4-2) - k(k-2) + 2(2k-8) = 0$$</p><p>$$8 - k^2 + 2k + 4k - 16 = 0$$</p><p>$$-k^2 + 6k - 8 = 0$$</p><p>$$k^2 - 6k + 8 = 0$$</p><p><strong>Step 4:</strong> Factor the quadratic:</p><p>$$(k-2)(k-4) = 0$$</p><p>$$k = 2 \text{ or } k = 4$$</p><p><strong>Step 5:</strong> Verify both values yield linearly dependent equations (non-zero solutions exist).</p><p>∴ The number of values of k is <strong>2</strong></p>
Correct Answer: C