I need to prove that for every binary column vector $a\in\{0, 1\}^n$, the following matrix is totally unimodular:
$$ A =\left [ \begin{matrix} I_n & a \\ a^T & 1 \\ \end{matrix} \right ] $$
Where $I_n$ is the Identity matrix of size $n$.
I noticed that if the i'th element of $a$ is $0$, I can remove the i'th row and the i'th column of $A$, since it contains only a single non-zero element, so I can assume that $a$ contains only ones.
How can I continue from here?
It is not true. Consider $n > 3$ and $a = [1,1,\dots,1]^T$ the all $1$ column vector. Then, $$A = \begin{bmatrix} 1 & 0 &\dots & 0 & 1 \\ 0 & 1 & \dots & 0 & 1 \\ \vdots & & \dots & & \vdots\\ 0 & 0 & \dots & 1 & 1 \\ 1 & 1 & \dots & 1 & 1\end{bmatrix}$$ has determinant $-(n-2) \notin \{0,\pm1\}$ because the row-reduced echelon form of $A$ is $$A' = \begin{bmatrix} 1 & 0 &\dots & 0 & 1 \\ 0 & 1 & \dots & 0 & 1 \\ \vdots & & \dots & & \vdots\\ 0 & 0 & \dots & 1 & 1 \\ 0 & 0 & \dots & 0 & -(n-2)\end{bmatrix} = \underbrace{\begin{bmatrix} 1 & 0 &\dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ \vdots & & \dots & & \vdots\\ 0 & 0 & \dots & 1 & 0 \\ -1 & -1 & \dots & -1 & 1\end{bmatrix}}_{U}A$$ and $det(A') = det(UA) = det(U)det(A) = det(A)$.