Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 17 / b

The matrix form is

Subtract the first row from the third. Expansion along the resulting row $(a-1,0,0)$ gives the determinantHence

For $a\ne 1,5/2$, the matrix is invertible, so there is exactly one solution. To exhibit it, subtract the first linear equation from the third to obtain $(a-1)x=2$, then subtract twice the second from the first to obtain $-x+(5-2a)z=-3$. Back-substitution givesAt $a=1$, the first and third linear equations have identical left-hand sides but right-hand sides one and three, so there is no solution.

At $a=5/2$, subtracting the first from the third forces $x=4/3$, while the first minus twice the second forces $x=3$. These contradict each other, so there is again no solution. Thus there is one solution for $a\ne 1,5/2$, no solution for either exceptional value, and no parameter value with infinitely many solutions.