<p>The number of values of \(k\) for which the linear equations:<br>\(4x + ky + 2z = 0\)<br>\(kx + 4y + z = 0\)<br>\(2x + 2y + z = 0\)<br>possess a non-zero solution is</p>
Step-by-Step Solution
Key Concept: A homogeneous system of linear equations has non-zero solutions if and only if the determinant of the coefficient matrix equals zero. Set up the determinant and solve for values of k.
<p><strong>Step 1:</strong> For non-zero solutions to exist in the homogeneous system, the coefficient matrix determinant must equal 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 minors:</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>∴ The number of values of k is <strong>2</strong></p>
Correct Answer: A