I am currently looking into linear codes and thought I had understood building a parity-check matrix for a given generator matrix - but the presented solution is anything but what I have.
Given a generator matrix $G= \begin{pmatrix}I_3 \\ A\end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}$
for a linear code in $\mathbb{F_3}$,the check matrix is $$H = (-A^{T(?)} | I_{n-m}) = (-A | I_{6-3}) =\begin{pmatrix}2&2&0&1&0&0 \\2&0&2&0&1&0\\0&2&2&0&0&1\end{pmatrix}.$$
I calculated the check matrix and only got this:
$$H_{wrong?} = \begin{pmatrix}1&1&0&1&0&0 \\1&0&1&0&1&0\\0&1&1&0&0&1\end{pmatrix}. $$
The right-hand $3\times 3$ square is obviously the identity matrix, the left-hand $3\times 3$ square is my transposed matrix $A$ given from $G$.
Do I have to do further calculations on $A^T$ because of the linear code being in $\mathbb{F_3}$? Is the actual transposed matrix where I transposed 1's have to be -1's for the check matrix, to which I then add the 3? I am completely lost and just spitballing.
$$A=\begin{pmatrix} 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}\Rightarrow A^T=\begin{pmatrix} 1 & 1 & 0 \\1 & 0 & 1\\0&1&1\end{pmatrix}\Rightarrow-A^T=\begin{pmatrix} -1 & -1 & 0 \\-1 & 0 & -1\\0&-1&-1\end{pmatrix}.$$ Then, since $-1\equiv 2\bmod 3$; we conclude $$-A^T=\begin{pmatrix} 2 & 2 & 0 \\2 & 0 & 2\\0&2&2\end{pmatrix}.$$