where \(a\) and \(b\) are constants and \(a \neq 1\).
The second equation tells us that \(z = -3y\).
Now substituting in to the first equation, we have that \(2x -y-3y=0\), and so \(x = 2y\).
Now substituting for \(x\) and \(z\) in the third equation, we have \[2y+y-3ay=b,\] which is \[3(1-a)y=b.\]
So if \(a \neq 1\), we can divide by \(3(1-a)\) to give \[y=\frac{b}{3(1-a)},\] and so \[x = \frac{2b}{3(1-a)}\] and \[z=-\frac{b}{1-a}.\]
Thus given any \(b, a\) \((a \neq 1)\), we have a unique solution.
What happens if \(a=1\)?
If \(a=1\), then our equation \(3(1-a)y=b\) tells us that \[0=b.\] So if \(b \neq 0\), there are no solutions.
If, however, \(a=1\) and \(b=0\), the original equations become \[2x-y+z=0,\] \[3y+z=0,\] \[x+y+z=0.\]
Summing the first two equations gives \[2x+2y+2z=0,\] and hence \[x+y+z=0,\] which is the third equation.
So the third equation gives us no more information in this case, and we effectively have only two independent equations.
Therefore there are infinitely many solutions to the equations in the case \(a=1\), \(b=0\).
They all satisfy \(x+y+z=0\) and \(3y+z=0\). So if we take \(y=t\), then we get \(z=-3t\) and \(x=2t\), giving the general solution \[x=2t,\quad y=t,\quad z=-3t.\]
This is a line in three dimensions, through the origin; we have \[\begin{pmatrix}x \\ y \\ z\end{pmatrix}=t\begin{pmatrix}2 \\ 1 \\ -3\end{pmatrix}\].
Here is a geometrical view…
Each of the equations represents a plane in 3-dimensional space. The first two planes meet in a line.
In the case \(a\ne 1\), the third plane crosses this line in just one point (we have a unique solution).
In the case \(a=1\), the third plane either contains this line itself, giving the whole line as a solution (the case \(b=0\)), or it lies parallel to this line, so there are no solutions (the case \(b\ne0\)).
Alternatively, we can think about the solution to the equations via matrices. The equations become
\[\begin{pmatrix}2&-1&1 \\ 0&3&1 \\ 1&1&a\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ b\end{pmatrix}.\]
The matrix \(A\) on the left can be pre-multipied by its inverse to give a unique solution for \(\begin{pmatrix}x \\ y \\ z\end{pmatrix}\) UNLESS \(a = 1\), in which case \(A\) is singular and has no inverse.
