<s>[INST] Matrix equation where some entries are solution to a polynome 
Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
  \begin{bmatrix}1&1&1&3\\1&1&1&-1\\1&z&z^2&0\\1&z^2&z&0\end
{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} =\begin{bmatrix}9\\1\\0\\0\end{bmatrix}

This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it; 

1) The equation $z^2+z+1=0$ ensures that every vector of the form
  \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} = \begin{bmatrix}a\\a\\a\\b\end{bmatrix}
  solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and 
  \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end
{bmatrix} = \begin{bmatrix}1\\1\\1\\2\end{bmatrix} is indeed a solution

I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?
 [/INST] $$\begin{bmatrix}1&1&1&3\\1&1&1&-1\\1&z&z^2&0\\1&z^2&z&0\end
{bmatrix}\begin{bmatrix}a\\a\\a\\b\end
{bmatrix} =\begin{bmatrix}3a+3b\\3a-b\\a(1+z+z^2)\\a(1+z+z^2)\end{bmatrix}=\begin{bmatrix}3a+3b\\3a-b\\0\\0\end{bmatrix}$$
We just have to solve for $3a+3b=9$ and $3a-b=1$.
We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.
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