Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>The system of linear equations:<br>\(x + \lambda y - z = 0\)<br>\(\lambda x - y - z = 0\)<br>\(x + y - \lambda z = 0\)<br>has a non-trivial solution for</p>
<p>exactly three values of \(\lambda\).</p>
<p>infinitely many values of \(\lambda\).</p>
<p>exactly one value of \(\lambda\).</p>
<p>exactly two 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 it zero.
<p><strong>Step 1:</strong> Write the coefficient matrix for the system:</p>
<p>$$A = \begin{bmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{bmatrix}$$</p>
<p><strong>Step 2:</strong> For non-trivial solutions to exist, $\det(A) = 0$</p>
<p><strong>Step 3:</strong> Calculate $\det(A)$ by expanding along Row 1:</p>
<p>$$\det(A) = 1(-\lambda - (-1)) - \lambda(\lambda(-\lambda) - (-1)(1)) + (-1)(\lambda(1) - (-1)(1))$$</p>
<p>$$= 1(1 - \lambda) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1)$$</p>
<p>$$= 1 - \lambda + \lambda^3 - \lambda - \lambda - 1$$</p>
<p>$$= \lambda^3 - 3\lambda$$</p>
<p>$$= \lambda(\lambda^2 - 3)$$</p>
<p><strong>Step 4:</strong> Set $\det(A) = 0$:</p>
<p>$$\lambda(\lambda^2 - 3) = 0$$</p>
<p>$$\therefore \lambda = 0, \quad \lambda = \sqrt{3}, \quad \text{or} \quad \lambda = -\sqrt{3}$$</p>
<p><strong>Answer:</strong> Non-trivial solutions exist for exactly three values of $\lambda$: $\lambda \in \{0, \sqrt{3}, -\sqrt{3}\}$</p>
Correct Answer: A