I need to prove that if we concatenate a vector of ones to the bottom and to the right of the identity matrix, the resulting matrix is totally unimodular.
For example, for $2\times2$ identity matrix we get the following matrix:
$$ \left [ \begin{matrix} \color{red}1 & \color{red}0 & 1 \\ \color{red}0 & \color{red}1 & 1 \\ 1 & 1 & 1 \end{matrix} \right ] $$
And for $3\times3$ identity matrix we get:
$$ \left [ \begin{matrix} \color{red}1 & \color{red}0 & \color{red}0 & 1 \\ \color{red}0 & \color{red}1 & \color{red}0 & 1 \\ \color{red}0 & \color{red}0 &\color{red}1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix} \right ] $$
I don't know how to prove it, since each row and column contains more then two non-zero numbers, so I can't use the row/column partition method. I thought about using induction, but I am not sure how.