Matrices & Determinants
System of Linear Equations
Grade None
Question:
<p>The system of linear equations<br>\(x + \mu y - z = 0\)<br>\(\mu x - y - z = 0\)<br>\(x + y - \mu z = 0\)<br>has a non-trivial solution for:</p>
<p>exactly three values of \(\mu\)</p>
<p>infinitely many values of \(\mu\)</p>
<p>exactly one value of \(\mu\)</p>
<p>exactly two values of \(\mu\)</p>
Step-by-Step Solution
Key Concept: A homogeneous system has non-trivial solutions if and only if the determinant of the coefficient matrix equals zero. Set up the determinant and solve for μ using row/column operations to simplify.
<p><strong>Step 1:</strong> For non-trivial solution, we need |A| = 0 where A is the coefficient matrix:</p><p>|A| = |1 μ -1|</p><p> |μ -1 -1|</p><p> |1 1 -μ|</p><p><strong>Step 2:</strong> Apply C₁ → C₁ - C₃:</p><p>|2 μ -1|</p><p>|μ+1 -1 -1|</p><p>|1+μ 1 -μ|</p><p><strong>Step 3:</strong> Apply R₁ → R₁ - 2R₃:</p><p>|-2μ μ-2 2μ-1|</p><p>|μ+1 -1 -1 |</p><p>|1+μ 1 -μ |</p><p><strong>Step 4:</strong> Expanding along first row and simplifying (or using direct expansion):</p><p>Expanding the original determinant:</p><p>1(μ + 1) - μ(−μ² + 1) - 1(μ + 1) = 0</p><p>μ + 1 + μ³ - μ - μ - 1 = 0</p><p>μ³ - μ = 0</p><p>μ(μ² - 1) = 0</p><p>μ(μ - 1)(μ + 1) = 0</p><p><strong>Step 5:</strong> Therefore μ = 0, 1, or -1</p><p>∴ Answer: A (typically μ ∈ {-1, 0, 1})</p>
Correct Answer: A