I'm trying to solve $y = xA$ where $y = (0, 1, 0, 1)$ and
$$ A = \left(\begin{array} & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array}\right) $$
also everything is in $\mathbb{Z}_2$.
I have tried to invert it using Gauss-Jordan elimination by expanding $A$ to $7 \times 7$ matrix with this result:
$$ A^{-1} = \left(\begin{array} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ \end{array}\right) $$
Am I in correct path? Is there any other easier way to solve it?
Thank you.
If you want "normal" Gauss-Jordan elimination to happen, you should rewrite $y = xA$ into $A^{\mathsf T} x^{\mathsf T} = y^{\mathsf T}$, giving you $$\begin{bmatrix}1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7\end{bmatrix} = \begin{bmatrix}0 \\ 1 \\ 0 \\ 1\end{bmatrix}.$$ Now you can row-reduce and solve for example for $x_1,x_2,x_3,x_4$ in terms of $x_5,x_6,x_7$.
Technically you can do the same thing to the original matrix by just tilting your head and doing column reduction instead, but that's confusing to think about.