Matrices & Determinants
Eigenvalues / Non-trivial Solutions
Grade 12
Question:
<p>The set of all values of \(\lambda\) for which the system of linear equations</p><p>\(2x_1 - 2x_2 + x_3 = \lambda x_1\)</p><p>\(2x_1 - 3x_2 + 2x_3 = \lambda x_2\)</p><p>\(-x_1 + 2x_2 = \lambda x_3\)</p><p>has a non-trivial solution,</p>
<p>(1) is an empty set</p>
<p>(2) is a singleton set</p>
<p>(3) contains two elements</p>
<p>(4) contains more than two elements</p>
Step-by-Step Solution
Key Concept: This is a homogeneous system of linear equations in the form (A - λI)x = 0. A non-trivial solution exists if and only if det(A - λI) = 0, making λ an eigenvalue of matrix A.
<p><strong>Step 1:</strong> Rewrite the system in matrix form (A - λI)x = 0:</p><p>$$\begin{pmatrix} 2-\lambda & -2 & 1 \\ 2 & -3-\lambda & 2 \\ -1 & 2 & -\lambda \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$</p><p><strong>Step 2:</strong> For non-trivial solution, det(A - λI) = 0:</p><p>$$\det\begin{pmatrix} 2-\lambda & -2 & 1 \\ 2 & -3-\lambda & 2 \\ -1 & 2 & -\lambda \end{pmatrix} = 0$$</p><p><strong>Step 3:</strong> Expand along Row 3:</p><p>$$-(-1)\begin{vmatrix} -2 & 1 \\ -3-\lambda & 2 \end{vmatrix} + 2\begin{vmatrix} 2-\lambda & 1 \\ 2 & 2 \end{vmatrix} - \lambda\begin{vmatrix} 2-\lambda & -2 \\ 2 & -3-\lambda \end{vmatrix} = 0$$</p><p>$$(-4 + 3 + \lambda) + 2(4 - 2\lambda - 2) - \lambda(-6 - 3\lambda - 4 + 2\lambda) = 0$$</p><p>$$(\lambda - 1) + 2(2 - 2\lambda) - \lambda(-\lambda - 10) = 0$$</p><p>$$\lambda - 1 + 4 - 4\lambda + \lambda^2 + 10\lambda = 0$$</p><p>$$\lambda^2 + 7\lambda + 3 = 0$$</p><p><strong>Step 4:</strong> The values of λ are the roots of this characteristic equation.</p><p>∴ Answer: C</p>
Correct Answer: C