How do I find parity check matrix if generator matrix can't be written in standard form? $G=\left ( \begin{matrix}
1 &amp;1  &amp;0 &amp;1&amp;0&amp;0&amp;1\\ 
 1&amp;0  &amp;1 &amp;0&amp;1&amp;0&amp;1\\ 
 0&amp;1  &amp;1&amp;0&amp;0&amp;1&amp;1 
\end{matrix} \right )$

Is it possible to get parity check matrix if generator matrix is not in the standard form?
Say $C$ is your code with generator matrix $G$. If you reduce $G$ to echelon form, you obtain
$$\begin{bmatrix} 1&amp;0&amp;1&amp;0&amp;1&amp;0&amp;1\\0&amp;1&amp;1&amp;0&amp;0&amp;1&amp;1\\0&amp;0&amp;0&amp;1&amp;1&amp;1&amp;1\end{bmatrix}$$
which is unfortunately not in standard form.

BUT, we can put it in standard form by swapping the third and fourth column, so we get
$$G^{\prime} = 
\begin{bmatrix} 1&amp;0&amp;0&amp;1&amp;1&amp;0&amp;1\\0&amp;1&amp;0&amp;1&amp;0&amp;1&amp;1\\0&amp;0&amp;1&amp;0&amp;1&amp;1&amp;1\end{bmatrix}$$

This $G^{\prime}$ is the generator matrix for a different (but equivalent) code $C^{\prime}$, where the 3rd and 4th positions of our codewords have been swapped. So $C^{\prime}$ has parity check matrix
$$H^{\prime} = 
\begin{bmatrix}
 1 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0\\
 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\
 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\
 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 1
\end{bmatrix}$$
And we translate it back to a parity check for $C$ be swapping the third and fourth columns back again
$$H = 
\begin{bmatrix}
 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
 1 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0\\
 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0\\
 1 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 1
\end{bmatrix}$$
You can perform elementary row operations on $G$ to produce a row-equivalent generator matrix $G' = EG$ where $E$ is a matrix corresponding to said row operations and $G'$ has the form $G' = [I \mid A]$ (reduced row echelon form). A parity check matrix for $G'$ is $H = [-A^T\mid I]$. Using block matrix multiplication, you can check that $G'H^T = 0$. Since $G' = EG$, $EGH^T = 0$. Since $E$ is invertible, this implies that $GH^T = E^{-1}0 = 0$. So $H$ is parity check matrix for $G$.