Past exam of the mathematical methods I course of the University of Oxford / 2017 / 4

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.

For $\lambda =1$ and $\mu =1/2$, the third equation gives $x=z-1$. Combining it with $2x-z=-4$ gives $z=-2$ and $x=-3$, after which the second equation gives $y=12$. HenceThe coefficient matrix has determinant $\lambda -2=-1\ne 0$, proving uniqueness. Direct substitution gives the required right-hand sides $1$, $5$ and $-1$.