Let $A$ be an $n\times n$ symmetric matrix with non-negative entries. Let $\mathbf{1}$ be the column vector of dimension $n$ with all entries being $1$. Consider the $(n+1)\times (n+1)$ matrix $$ B= \begin{bmatrix} A & \mathbf{1} \\ \mathbf{1}^T & 0 \end{bmatrix} $$
Question: what is the condition for $A$ so that $B$ is invertible?
Remark: This matrix is related to portfolio optimization problems in finance. I note that when $A$ is a constant matrix, the determinant of $B$ is $0$ and thus $B$ is not invertible.
If $\det(A) \neq 0$ so that $A^{-1}$ exists and the scalar $\alpha = \mathbf{1}^T A^{-1} \mathbf{1} \neq 0$, then we have
$$B^{-1} = \begin{bmatrix} A^{-1} - \alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1} & \alpha^{-1}A^{-1}\mathbf{1} \\ \alpha^{-1}\mathbf{1}^TA^{-1} & -\alpha^{-1} \end{bmatrix}$$
Note that
$$\begin{align}BB^{-1}&= \begin{bmatrix} AA^{-1} - A\alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1}+ \mathbf{1}\alpha^{-1}\mathbf{1}^TA^{-1} & A\alpha^{-1}A^{-1}\mathbf{1}-\alpha^{-1}\mathbf{1} \\ \mathbf{1}^TA^{-1} - \mathbf{1}^T\alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1} + 0\alpha^{-1}A^{-1}\mathbf{1} & \mathbf{1}^T\alpha^{-1}A^{-1}\mathbf{1}-0\alpha^{-1} \end{bmatrix} \\ \\&= \begin{bmatrix} I -\alpha^{-1}\mathbf{1}\mathbf{1}^TA^{-1}+ \alpha^{-1}\mathbf{1}\mathbf{1}^TA^{-1} & \alpha^{-1}\mathbf{1}-\alpha^{-1}\mathbf{1} \\ \mathbf{1}^TA^{-1} - \alpha^{-1}\alpha\mathbf{1}^TA^{-1} & \alpha^{-1}\alpha \end{bmatrix}\\ \\ &= \begin{bmatrix} I & \mathbf{0} \\ \mathbf{0}^T & 1 \end{bmatrix}\end{align} $$
Addendum
In general, for a block matrix
$$B = \begin{bmatrix} A & C \\ E & D \end{bmatrix},$$
if $A^{-1}$ exists , then the Schur complement of $A$ is $D- EA^{-1}C$ and
$$\det(B) = \det(A) \det(D- EA^{-1}C)$$
Thus, $\det(B) \neq 0$ and $B^{-1}$ exists if and only if $\det(D- EA^{-1}C) \neq 0$.
In this case, the Schur complement reduces to a scalar $-\mathbf{1}^TA^{-1} \mathbf{1}$, and the condition $\mathbf{1}^TA^{-1}\mathbf{1} \neq 0$ is necessary and sufficient for $B$ to be invertible.