<s>[INST] Describe how the number of solutions to an equation system depend on the variable $a$. Describe how the number of solutions to the system:
     \begin{eqnarray*}
                x+y+z &=&1 \\
                2x+y+a^2z &=&a \\
                x+3y+3z &=&3
        \end{eqnarray*}
        is dependant on the value of $a$.
My attempt at a solution:
First we write the the equation system in matrix form:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\2 & 1 & a^2 & a\\1 & 3 & 3 & 3\end{array}\right]$$
Now, after subtracting the two times first row from the second we get:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\0 & -1 & a^2-2 & a-2\\1 & 3 & 3 & 3\end{array}\right]$$
And subtract one time the first row from the third:
$$\left[\begin{array}{ccc|c}1 & 1 & 1 & 1\\0 & -1 & a^2-2 & a-2\\0 & 2 & 2 & 2\end{array}\right]$$
Then we subtract one half the third row from the first:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 2 & 2 & 2\end{array}\right]$$
We continue with adding two times the second row to the third:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 0 & 2+2(a^2-2) & 2+2(a-2)\end{array}\right]$$
We divide the third row by $2$ and simplify the expressions.
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & a^2-2 & a-2\\0 & 0 & a^2-1 & a-1\end{array}\right]$$
We subtract one time the third row from the second row:
$$\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & -1 & -1 & -1\\0 & 0 & a^2-1 & a-1\end{array}\right]$$
This result gives me the following conclusion:
If $a = 1$ the system will have infinitely many solutions.
If $a = -1$ the system will not have any solutions, 
For all other values of $a$ the system will have a unique solution.
Am I thinking correctly?
 [/INST] If $A$ denotes the $3\times 3$-matrix of your equation $Ax=b$, then $\det(A)=-2(a-1)(a+1)$. Hence if the field has not characteristic two, and $(a-1)(a+1)\neq 0$, then we have a unique solution. You have already solved this case correctly. Note that you have divided by $2$ here.
For $char(K)=2$ we always have $\det(A)=0$. The system then is given by
\begin{align*}
x+y+z& =1, \\
y+a^2z & = a. \\
\end{align*}
We can solve this by setting $x=1-y-z$ and $y=a(1-az)$. If $K$ has infinitely many elements, then we have infinitely many solutions, independent of the value of $a$.
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