Question:

Step 1: Identify the condition for a non-trivial solution. A system of homogeneous linear equations has a non-trivial solution if and only if the determinant of the coefficient matrix equals zero. We need to find all values of $\mu$ that satisfy this condition. Step 2: Write the coefficient matrix. The coefficient matrix for the given system is: $$A = \begin{pmatrix} 1 & \mu & -1 \\ \mu & -1 & -1 \\ 1 & 1 & -\mu \end{pmatrix}$$ Step 3: Calculate the determinant using cofactor expansion along the first row. We expand the determinant along the first row: $$\det(A) = 1 \begin{vmatrix} -1 & -1 \\ 1 & -\mu \end{vmatrix} - \mu \begin{vmatrix} \mu & -1 \\ 1 & -\mu \end{vmatrix} + (-1) \begin{vmatrix} \mu & -1 \\ 1 & 1 \end{vmatrix}$$ Step 4: Evaluate each $2 \times 2$ minor. Computing the three minors: - First minor: $\begin{vmatrix} -1 & -1 \\ 1 & -\mu \end{vmatrix} = (-1)(-\mu) - (-1)(1) = \mu + 1$ - Second minor: $\begin{vmatrix} \mu & -1 \\ 1 & -\mu \end{vmatrix} = \mu(-\mu) - (-1)(1) = -\mu^2 + 1$ - Third minor: $\begin{vmatrix} \mu & -1 \\ 1 & 1 \end{vmatrix} = \mu(1) - (-1)(1) = \mu + 1$ Step 5: Substitute the minors back into the determinant expression. $$\det(A) = 1(\mu + 1) - \mu(-\mu^2 + 1) - 1(\mu + 1)$$ Step 6: Expand and simplify. $$\det(A) = \mu + 1 + \mu^3 - \mu - \mu - 1$$ Combining like terms: $$\det(A) = \mu^3 - 3\mu$$ Step 7: Set the determinant equal to zero and solve for $\mu$. For a non-trivial solution to exist: $$\mu^3 - 3\mu = 0$$ Factoring out $\mu$: $$\mu(\mu^2 - 3) = 0$$ Step 8: Find all solutions. This equation yields three solutions: $$\mu = 0, \quad \mu = \sqrt{3}, \quad \mu = -\sqrt{3}$$ Step 9: State the final answer. The system of linear equations has a non-trivial solution for exactly three values of $\mu$: $\mu = 0$, $\mu = \sqrt{3}$, and $\mu = -\sqrt{3}$. **Answer: Option 1** (exactly three values of $\mu$)