 Solution

Subtracting the second equation from the first gives $2x-\lambda z=-4$. With $\lambda =2$ this is $x-z=-2$, whereas the third equation requires $x-z=-2\mu$. Consistency therefore requiresAt that value one parameter remains free:Thus the solution is not unique. For any other $\mu$ the two required values of $x-z$ conflict and the linear system has no solution.