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

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 $$x = (1,1,0)$$.

• No, $x=(1,1,0)$ is not a solution, since $Ax=(1,1,1)$ then, which is different from $(1,0,0)$. Apr 26, 2019 at 16:47

## 2 Answers

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.

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

• I do it with Gauss, where I get all zeros from the left diagonal. But then my solution is (0,0,1). Apr 26, 2019 at 13:12
• That appears to be correct.
– user403337
Apr 26, 2019 at 13:14
• The textbook provides us with the solution: x = (1, 1, 0). What am I missing? Apr 26, 2019 at 13:15
• That appears to be an error. The matrix is invertible, so there is a unique solution. $(0,0,1)^t$ is clearly it.
– user403337
Apr 26, 2019 at 13:18