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

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

2 Answers 2

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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.

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

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

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