Find all vectors that suit $Ax = (1, 0, 0)$ where $A$ is linear mapping

Question

We have linear mapping give with a matrix from standard basis to basis $X$. Basis $X = ((1,0,1),(0,1,0),(0,1,1))$.

The matrix looks like the following

\begin{bmatrix}1&0&1\\0&1&0\\1&0&0\\\end{bmatrix}

I need to find all of the vectors $x$ (over the field of $\mathbb{Z}_2$, meaning it has only $0$ and $1$ as numbers).

I could use this: $(Ax)y = xAy \cdot (x)x$. Where $xAy$ I mean matrix $A$ from basis $X$ to basis $Y$. But I keep getting incorrect result. The solution is $x = (1,1,0)$.

Answer 1

Since $A\in M_3(\Bbb F_2)$ is invertible, $$ x=A^{-1}(1,0,0)^T=\begin{pmatrix}0&0&1\\0&1&0\\1&0&1\\ \end{pmatrix} \begin{pmatrix} 1 \cr 0 \cr 0 \end{pmatrix}= \begin{pmatrix} 0 \cr 0 \cr 1 \end{pmatrix} $$ is the unique solution.

Answer 2

Hint: Row-reduce the augmented matrix $\left (\begin{array}{rrr|r}1&0&1&1\\0&1&0&0\\1&0&0&0\end{array}\right )$.