Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>The system of linear equations</p><p>\(x + \lambda y - z = 0\)</p><p>\(\lambda x - y - z = 0\)</p><p>\(x + y - \lambda z = 0\)</p><p>has a non-trivial solution for</p>
<p>(1) Exactly one value of \(\lambda\)</p>
<p>(2) Exactly two values of \(\lambda\)</p>
<p>(3) Exactly three values of \(\lambda\)</p>
<p>(4) Infinitely many values of \(\lambda\)</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 λ values that make det(A) = 0.
<p><strong>Step 1:</strong> For non-trivial solution of homogeneous system Ax = 0, we need det(A) = 0</p><p><strong>Step 2:</strong> Set up determinant:</p><p>det(A) = |1 λ -1|</p><p> |λ -1 -1|</p><p> |1 1 -λ|</p><p><strong>Step 3:</strong> Expand along row 1:</p><p>= 1(-λ + 1) - λ(-λ² + 1) - 1(λ + 1)</p><p>= 1 - λ + λ³ - λ - 1 - λ</p><p>= λ³ - 3λ</p><p>= λ(λ² - 3)</p><p><strong>Step 4:</strong> For det(A) = 0:</p><p>λ(λ² - 3) = 0</p><p>λ = 0, λ = √3, λ = -√3</p><p><strong>Step 5:</strong> Non-trivial solution exists when λ ∈ {-√3, 0, √3}</p><p>∴ Answer: C</p>
Correct Answer: C