For what values of $a$ does the following systems of equations $\;$ $x + ay = 1, \;\; ax + y = a^2$ $\;$ have solutions? Find those solutions.
Given system of equations:
$x + ay = 1$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $ax + y = a^2$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$, $\;\;$ $x = 1 - ay$ $\;\;\; \cdots \; (3)$
Then, in view of equation $(3)$, equation $(2)$ becomes
$a \left(1 - ay\right) + y = a^2$
i.e. $\;$ $a - a^2 y + y = a^2$
i.e. $\;$ $y \left(1 - a^2\right) = a^2 - a$ $\;\;\; \cdots \; (4)$
Case 1:
If $\; \;$ $1 - a^2 \neq 0$, $\;\;$ i.e $\;$ if $\;$ $a^2 \neq 1$ $\;\;$ i.e. $\;$ if $\;$ $a \neq \pm 1$, $\;$ then we have from equation $(4)$
$y = \dfrac{a \left(a - 1\right)}{1 - a^2}$
i.e. $\;$ $y = \dfrac{a \left(a - 1\right)}{\left(1 + a\right) \left(1 - a\right)}$
i.e. $\;$ $y = \dfrac{-a}{1 + a}$ $\;\;\;$ when $\;$ $a \neq \pm 1$
Substituting the value of $y$ in equation $(3)$ gives
$x = 1 - a \left(\dfrac{-a}{1 + a}\right)$
i.e. $\;$ $x = \dfrac{1 + a + a^2}{1 + a}$ $\;\;\;$ when $\;$ $a \neq \pm 1$
Case 2:
When $\;$ $a = + 1$, $\;$ we have from equations $(1)$ and $(2)$
$x + y = 1$ $\;\;\; \cdots \; (1a)$ $\;\;\;$ and $\;\;\;$ $x + y = 1$ $\;\;\; \cdots \; (2a)$
$\therefore \;$ If $x$ is any number $c, \; c \in R$, then from equation $(1a)$,
$y = 1 -x = 1- c$
Case 3:
When $\;$ $a = -1$, $\;$ we have from equations $(1)$ and $(2)$
$x - y = 1$ $\;\;\; \cdots \; (1b)$ $\;\;\;$ and $\;\;\;$ $-x + y = 1$ $\;\;\; \cdots \; (2b)$
Adding equations $(1b)$ and $(2b)$ gives $\;\;$ $0 = 2$, $\;\;$ which is not possible.
$\therefore \;$ The solution to the given system of equations is
$\left(x, y\right) = \phi$ $\;$ when $\;$ $a = -1$
$\left(x, y\right) = \left\{\left(c, 1 - c\right) \mid c \in R \right\}$ $\;$ when $\;$ $a = +1$
and $\;$ $\left(x, y\right) = \left\{\left(\dfrac{1 + a + a^2}{1 + a}, \dfrac{-a}{1 + a}\right) \right\}$ $\;$ when $\;$ $a \neq \pm 1$