<p>The number of real values of \(\lambda\), for which the system of linear equations:<br>\(2x + 4y - \lambda z = 0\)<br>\(4x + \lambda y + 2z = 0\)<br>\(\lambda x + 2y + 2z = 0\)<br>has infinitely many solutions, is</p>
Step-by-Step Solution
Key Concept: A homogeneous system has infinitely many solutions (non-trivial solutions) if and only if the determinant of the coefficient matrix equals zero. Set det(A) = 0 and solve for λ.
<p><strong>Step 1:</strong> Write the coefficient matrix and note this is a homogeneous system (all equations = 0).</p><p>A = [2, 4, -λ; 4, λ, 2; λ, 2, 2]</p><p><strong>Step 2:</strong> For infinitely many solutions, det(A) = 0. Expand the determinant:</p><p>det(A) = 2(λ·2 - 2·2) - 4(4·2 - 2·λ) + (-λ)(4·2 - λ·λ)</p><p>= 2(2λ - 4) - 4(8 - 2λ) - λ(8 - λ²)</p><p>= 4λ - 8 - 32 + 8λ - 8λ + λ³</p><p>= λ³ + 4λ - 40</p><p><strong>Step 3:</strong> Solve λ³ + 4λ - 40 = 0</p><p>Testing λ = 2: 8 + 8 - 40 = -24 ≠ 0</p><p>Testing λ = -2: -8 - 8 - 40 = -56 ≠ 0</p><p>By careful analysis or numerical methods, this cubic has <strong>exactly 1 real root</strong> (the cubic has negative discriminant indicating one real and two complex roots).</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: B