Problem source

For all values of $a$ and $b,$ either solve the simultaneous equations
\begin{alignat*}{1}
x+y+az & =2\\
x+ay+z & =2\\
2x+y+z & =2b
\end{alignat*}
or prove that they have no solution.

Solution source

Consider the matrix system:

\begin{align*}
\left(\begin{array}{ccc|c}
  1 & 1 & a & 2 \\
  1 & a & 1 & 2 \\
  2 & 1 & 1 & 2b \\
\end{array}\right) \\
\left(\begin{array}{ccc|c}
  1 & 1 & a & 2 \\
  0 & a-1 & 1-a & 0 \\
  0 & -1 & 1-2a & 2b-4 \\
\end{array}\right)\\
\left(\begin{array}{ccc|c}
  1 & 1 & a & 2 \\
  0 & a-1 & 1-a & 0 \\
  0 & 0 & -2a & 2b-4 \\
\end{array}\right)  \\
\end{align*}

Assuming that $a \neq 1, 0$ all steps are fine and:

$z = \frac{2-b}{a}, y = \frac{2-b}{a}, x +(1+a)y = 2, x = 2 - \frac{(2-b)(1+a)}{a} = \frac{ab+b-2}{a}$

If $a = 0$, $y = z$ and $\begin{cases} x + y &= 2 \\ 2x + 2y &= 2b \end{cases} \Rightarrow b= 2, x = t, y = 2-t, z = 2-t$

If $a = 1$, $x = 2b-2, y = t, z = 4-t-2b$, where $t \in \mathbb{R}$