Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>The system of linear equations<br>\(x + \mu y - z = 0\)<br>\(\mu x - y - z = 0\)<br>\(x + y - \mu z = 0\)<br>has a non-trivial solution for:</p>
<p>exactly three values of \(\mu\)</p>
<p>infinitely many values of \(\mu\)</p>
<p>exactly one value of \(\mu\)</p>
<p>exactly two values of \(\mu\)</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 values of μ that make it equal to zero.
<p><strong>Step 1:</strong> Write the coefficient matrix for the system:</p>
<p>$$A = \begin{bmatrix} 1 & \mu & -1 \\ \mu & -1 & -1 \\ 1 & 1 & -\mu \end{bmatrix}$$</p>
<p><strong>Step 2:</strong> For non-trivial solutions to exist, $\det(A) = 0$</p>
<p>$$\det(A) = 1(-1 \cdot (-\mu) - (-1) \cdot 1) - \mu(\mu \cdot (-\mu) - (-1) \cdot 1) + (-1)(\mu \cdot 1 - (-1) \cdot 1)$$</p>
<p>$$= 1(\mu + 1) - \mu(-\mu^2 + 1) - 1(\mu + 1)$$</p>
<p>$$= \mu + 1 + \mu^3 - \mu - \mu - 1$$</p>
<p>$$= \mu^3 - \mu = 0$$</p>
<p><strong>Step 3:</strong> Solve $\mu^3 - \mu = 0$</p>
<p>$$\mu(\mu^2 - 1) = 0$$</p>
<p>$$\mu(\mu - 1)(\mu + 1) = 0$$</p>
<p>$$\mu = 0, 1, \text{ or } -1$$</p>
<p><strong>Step 4:</strong> Verify these give non-trivial solutions</p>
<p>Therefore, there are exactly three values of $\mu$ for which the system has a non-trivial solution.</p>
<p><strong>Answer:</strong> exactly three values of $\mu$</p>
Correct Answer: A