Problem source

Find the simultaneous solutions of the three linear equations
\begin{alignat*}{1}
a^{2}x+ay+z & =a^{2}\\
ax+y+bz & =1\\
a^{2}bx+y+bz & =b
\end{alignat*}
for all possible real values of $a$ and $b$.

Solution source

\begin{align*}
&& a^{2}x+ay+z & =a^{2} \tag{1}\\
&& ax+y+bz & =1 \tag{2}\\
&& a^{2}bx+y+bz & =b \tag{3} \\
\\
(1) - a(2): && (1-ba)z &= a^2-a \\
\Rightarrow && z &= \frac{a^2-a}{1-ab} \tag{if $ab \neq 1$} \\
\\
(2) - (3): && (a-a^2b)x &= b - 1 \\
\Rightarrow && x &= \frac{b-1}{a(1-ab)} \tag{if $a \neq 0, ab \neq 1$} \\
\\
b(1) - (3): && (ab-1)y &= a^2 - b^2 \\
\Rightarrow && y &= \frac{a^2-b^2}{ab-1}
\end{align*}

Let's consider the cases where $a = 0$, then

\begin{align*}
&& z &= 0 \\
&& y + bz &= 1 \\
&& y+bz &= b \\
\Rightarrow && y &= 1 = b 
\end{align*}

So if $a = 0$ then $b = 1$ and $x \in \mathbb{R}, y = 1, z = 0$.

If $a \neq 0, ab = 1$, then

\begin{align*}
&& a^2 x + ay + z &= a^2 \\
&& ax + y + \frac1{a}z &= a \\
&& ax + y + \frac{1}{a}z &= b \\
\end{align*}

The last two equations imply $a = b = \pm 1$.

$a = 1 \Rightarrow x+y+z = 1$, so we have a lot of solutions.

$a = -1 \Rightarrow x -y +z = 1$ so again, lots of solutions.

Conclusion:

If $ab \neq 1, a \neq 0$, we have:

\[ (x,y,z) = \left (\frac{b-1}{a(1-ab)},  \frac{a^2-b^2}{ab-1},   \frac{a^2-a}{1-ab} \right)\]

If $a = 0$ then $b = 1$ and we have: $(x,y,z) = (t, 1, 0)$.

If $ab = 1$ then $a = 1$ or $a = -1$.

If $a = 1$ then $(x,y,z) = (t, s, 1-t-s)$
If $a = -1$ then $(x,y,z) = (t,s,1-t+s)$