What is the branch number of this matrix? We have the following matrix:
$$\begin{pmatrix}0&amp;1&amp;1&amp;1\\
1&amp;0&amp;1&amp;1\\
1&amp;1&amp;0&amp;1\\
1&amp;1&amp;1&amp;0\end{pmatrix}$$

What is the branch number?

Is this a MDS marix?
The matrix is not MDS over $GF(2)$; No binary MDS codes exist and non nonbinary (over $GF(2^n)$ MDS codes would have this generator whose scalar entries are in the field $GF(2)$). Over $GF(2^n)$ The branch number, which is the minimum weight of the corresponding linear code is 4, in $GF(2^n)$ for all $n$. This covers all possible fields of interest for crypto.

Note that you just need to take the given matrix, call it $A,$ form $G=[I | A]$ over the relevant field and determine the minimum weight of the resulting code. Magma (and GAP, Pari, Python) can do this for you easily. This code turns out to be a quasi-cyclic code over the fields $GF(2^n)$ with minimum distance 4.

Edit: It seems that 4 is the limit for the minimum distance of this type of code. In particular a $16\times16$ version, still has minimum distance only 4, for the fields $GF(2^n)$ where $n=1,2,\ldots,8.$ I assume $n=8$ is of interest to the OP, since it corresponds to having bytes as code symbols, as in AES.

This was checked on Magma as well.