Is there an assignment of the values 0 and 1 to the variables
\(x_1,x_2,x_3\) so that the following is satisfied over \(GF(2)\)?
\[
x_1\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} +
x_2\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} +
x_3\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
=
\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\]
(Recall that in \(GF(2)\), we have \(1+1=0\).)
The answer is “No”. Note that in
order to obtain a tuple that contains a
\(1\) in the third component on the left-hand side, either \(x_2\) or \(x_3\)
must be 1, but not both. We now do a case analysis based on this fact.
Suppose that \(x_2 = 1\), and \(x_3 = 0\).
If \(x_1 = 1\), the left-hand side gives
\(\begin{bmatrix}0 \\0\\1\end{bmatrix}\). If \(x_1 = 0\), the left-hand side
is simply \(\begin{bmatrix}1 \\1\\1\end{bmatrix}\). In neither case is
the left-hand side equal to the right-hand side.
Suppose that \(x_2 = 0\), and \(x_3 = 1\).
If \(x_1 = 1\), the left-hand side gives
\(\begin{bmatrix}1 \\1\\1\end{bmatrix}\). If \(x_1 = 0\), the left-hand side
is simply \(\begin{bmatrix}0 \\0\\1\end{bmatrix}\). In neither case is
the left-hand side equal to the right-hand side.