Step 1: Core Idea:
A system of linear equations \( Ax=b \) has a unique solution if and only if the determinant of matrix A is not zero. The values on the right-hand side, such as \( \mu \), only influence the solution's existence, not its uniqueness.
Step 2: Method:
Rewrite the system in matrix form \( Ax=b \) and compute the determinant of the coefficient matrix A.
\[ A = \begin{pmatrix} 1 & 1 & 1
1 & 2 & 5
2 & 3 & \lambda \end{pmatrix}, \quad x = \begin{pmatrix} x
y
z \end{pmatrix}, \quad b = \begin{pmatrix} 6
10
\mu \end{pmatrix} \]For a unique solution, \( \det(A) eq 0 \).
Step 3: Calculation:
Calculate the determinant of A.
\[ \det(A) = 1 \begin{vmatrix} 2 & 5
3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 5
2 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2
2 & 3 \end{vmatrix} \]\[ = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) \]\[ = 2\lambda - 15 - \lambda + 10 - 1 \]\[ = \lambda - 6 \]For a unique solution, \( \det(A) eq 0 \).
\[ \lambda - 6 eq 0 \implies \lambda eq 6 \]If \( \lambda eq 6 \), the system has a unique solution irrespective of the value of \( \mu \). \( \mu \) can be any real number.
Step 4: Conclusion:
The system has a unique solution if \( \lambda eq 6 \) and \( \mu \in \mathbb{R} \).