Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 5 / Solution

The required matrix form of the linear system isSubtract the first row from the third and expand the determinant along the resulting row:Thus the linear system has a unique solution, for every real $\mu$, precisely when $\boxed{\lambda \ne 1,2}$.

For $\lambda =0$, the second equation gives $x=\mu$. Subtracting the first equation from the third gives $-y=1$, and the first equation then determines $z$:For $\lambda =1$, the first and third left-hand sides are identical but their right-hand sides are $1$ and $2$. HenceFor $\lambda =2$, subtracting the first equation from the third gives $y=1$. The first equation now gives $x+2z=0$, while the second requires $x+2z=\mu$. ConsequentlyThe free variable in a linear system in the consistent singular case describes a line of solutions. These direct reductions distinguish an inconsistent linear system from one with infinitely many solutions, which a zero determinant alone cannot do.