Matrices & Determinants
Homogeneous Linear Equations
Grade 12
Question:
<p>The system of homogeneous equations \(\lambda x + (\lambda + 1)y + (\lambda - 1)z = 0\), \((\lambda + 1)x + \lambda y + (\lambda + 2)z = 0\), \((\lambda - 1)x + (\lambda + 2)y + \lambda z = 0\) has non-trivial solution for:</p>
<p>(a) exactly three real values of \(\lambda\)</p>
<p>(b) exactly two real values of \(\lambda\)</p>
<p>(c) exactly one real value of \(\lambda\)</p>
<p>(d) infinitely many real values of \(\lambda\)</p>
Step-by-Step Solution
Key Concept: Non-trivial solutions of homogeneous systems exist when the coefficient matrix determinant is zero, leading to an eigenvalue problem.
<p><strong>Solution:</strong> For non-trivial solutions, the determinant of the coefficient matrix must equal zero:</p><p>$$\begin{vmatrix} \lambda & \lambda+1 & \lambda-1 \\ \lambda+1 & \lambda & \lambda+2 \\ \lambda-1 & \lambda+2 & \lambda \end{vmatrix} = 0$$</p><p>Expanding this determinant yields a polynomial equation in $\lambda$. After simplification, this equation has exactly one real solution.</p>
Correct Answer: C