Problems

Solution: \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 \((x,y,z) = (t,s,1-t+s)\)