Matrices & Determinants
Eigenvalues / non-trivial solution
Grade 12
Question:
<p>The set of all values of \(\lambda\) for which the system of linear equations<br>\(x - 2y - 2z = \lambda x\)<br>\(x + 2y + z = \lambda y\)<br>\(-x - y = \lambda z\)<br>has a non-trivial solution:</p>
<p>is a singleton</p>
<p>contains exactly two elements</p>
<p>is an empty set</p>
<p>contains more than two elements</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. Rewrite the system in the form (A - λI)X = 0 and solve det(A - λI) = 0.
<p><strong>Step 1:</strong> Rewrite the system in standard form by moving $\lambda$ terms to the left:</p>
<p>$(1 - \lambda)x - 2y - 2z = 0$</p>
<p>$x + (2 - \lambda)y + z = 0$</p>
<p>$-x - y - \lambda z = 0$</p>
<p><strong>Step 2:</strong> For non-trivial solutions, $\det(A - \lambda I) = 0$, where:</p>
<p>$$\det \begin{bmatrix} 1-\lambda & -2 & -2 \\ 1 & 2-\lambda & 1 \\ -1 & -1 & -\lambda \end{bmatrix} = 0$$</p>
<p><strong>Step 3:</strong> Expand along the first column:</p>
<p>$(1-\lambda)[(2-\lambda)(-\lambda) + 1] + 1[2\lambda - 2] - 1[-2 - (-2)(2-\lambda)] = 0$</p>
<p>$(1-\lambda)[\lambda^2 - 2\lambda + 1] + [2\lambda - 2] - 1[-2 + 4 - 2\lambda] = 0$</p>
<p>$(1-\lambda)(1-\lambda)^2 + 2\lambda - 2 - 2 + 2\lambda = 0$</p>
<p>$(1-\lambda)^3 + 4\lambda - 4 = 0$</p>
<p><strong>Step 4:</strong> Expand $(1-\lambda)^3 = 1 - 3\lambda + 3\lambda^2 - \lambda^3$:</p>
<p>$-\lambda^3 + 3\lambda^2 - 3\lambda + 1 + 4\lambda - 4 = 0$</p>
<p>$-\lambda^3 + 3\lambda^2 + \lambda - 3 = 0$</p>
<p>$\lambda^3 - 3\lambda^2 - \lambda + 3 = 0$</p>
<p><strong>Step 5:</strong> Factor by grouping:</p>
<p>$\lambda^2(\lambda - 3) - 1(\lambda - 3) = 0$</p>
<p>$(\lambda - 3)(\lambda^2 - 1) = 0$</p>
<p>$(\lambda - 3)(\lambda - 1)(\lambda + 1) = 0$</p>
<p><strong>∴ Answer:</strong> $\lambda \in \{-1, 1, 3\}$</p>
Correct Answer: B