$\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 & 1\\ \end{array} } \right) $ in $\mathbb{Z}_{2}$

Question

\begin{align} &\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 & 1\\ \end{array} } \right) \text{in}\ \mathbb{Z}_{2} \\ &=\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{array} } \right) \end{align}

I set $z=t=0 (x,y,z,t)$ and I got a partial solution $(0,1,0,0).$

I solved two homogeneous matrices once for $z = 1$ and $t = 0$, then for $z = 0$ and $t = 1$ and I got two solutions $(1,1,1,0)$ and $(1,1,0,1).$

Then, I got $(0,1,0,0) + a*(1,1,1,0)+b*(1,1,0,1$)

Therefore, all possible results are $(0,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,1)$

Would this be correct?

Answer 1

Not right. The reduced row echelon form turns into two equations: \begin{align*} x_1 + x_3 &= 0 \\ x_2 + x_3 &= 1 \end{align*} (not pictured: $x_4$). We indeed have $x_3$ and $x_4$ as free variables, but something seems to have gone wrong when you figured out $x_1$ and $x_2$ based on these free variables.

The first equation rearranges to $x_1 = x_3$, i.e. $x_1$ must be the same as $x_3$. On the other hand, $x_2 + x_3 = 1$ if and only if $x_2 \neq x_3$ That is, $x_2$ must be the opposite value to both $x_1$ and $x_3$. On the other hand, $x_4$ is totally free, not affecting any other variable. So, our solutions are: $$(x_1, x_2, x_3, x_4) = (1, 0, 1, 0), (1, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1).$$

EDIT: If you specifically wish to write it as a particular solution plus a homogeneous solution, you can follow the usual method. The variable $x_4$ is completely free, so we shall call it $t$. The variable $x_3$ is also free, but $x_1$ and $x_2$ depend on it. Let's call it $s$, and note $x_1 = -s = s$ and $x_2 = 1 - s = 1 + s$. Thus, $$(x_1, x_2, x_3, x_4) = (s, s + 1, s, t) = (0, 1, 0, 0) + s(1, 1, 1, 0) + t(0, 0, 0, 1).$$ The first term is a particular solution, while the other two terms form the full homogeneous solution.

Answer 2

By equalizing, see Calculating basis of null space, one gets immediately the solutions without any further calculations. From $$\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 1\\ \end{array} } \right) $$ equalizing gives $$\left( {\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 1\\ 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 \end{array} } \right), $$ hence the solutions are given by $$\begin{pmatrix} 0\\1\\0\\0\end{pmatrix} +r\begin{pmatrix}1\\1\\-1\\0\end{pmatrix} +s\begin{pmatrix}0\\0\\0\\-1\end{pmatrix}. $$ Remember that $-1=1$.