$ \mathbf A = \begin{bmatrix} \begin{array}{c | c | c} \mathbf D &\pmb 1 &\mathbf x \\ \hline -\pmb 1^\top &0 &0 \\ \hline -\mathbf x^\top &0 &0 \end{array} \end{bmatrix}, $ where $\mathbf D \in \Bbb R^{l\times l}$ is a symmetric positive definite matrix, $\pmb 1\in\Bbb R^{l}=[1,1,\dots,1]^\top$ and $\mathbf x\in\Bbb R^{l}$.
I have tried with some random $x$, $A$ is always full rank. I have tried it in Matlab $100000$ times.
If $\mathbf A = \begin{bmatrix} \mathbf D&\mathbf x \\ -\mathbf x^\top &0 \end{bmatrix}$, then we can easily prove it.
Since $\mathbf D$ has inverse, for any $\mathbf x$, we have $ \begin{bmatrix} \mathbf D&\mathbf x \\ -\mathbf x^\top &0 \end{bmatrix} \begin{bmatrix} \mathbf I &-\mathbf D^{-1}\mathbf x \\ \varnothing &1 \end{bmatrix} =\begin{bmatrix} \mathbf D&\varnothing \\ -\mathbf x^\top &\mathbf x^\top\mathbf D^{-1}\mathbf x. \end{bmatrix} $
Taking the determinant of both sides, we obtain $ \begin{vmatrix} \mathbf D&\mathbf x \\ -\mathbf x^\top &0 \end{vmatrix} =|\mathbf D|\cdot |\mathbf x^\top\mathbf D^{-1}\mathbf x|. $
Since $\mathbf D^{-1}$ is positive definite, $|\mathbf x^\top\mathbf D^{-1}\mathbf x|$ is nonzero. This means the determinant of $\mathbf A$ is nonzero, and then $rank(\mathbf A)=l+1$ without any other condition.
But after adding the term $\pmb 1$, I don't know how to do it.
If $ \mathbf A = \begin{bmatrix} \begin{array}{c | c | c} \mathbf D &\pmb 1 &\mathbf x \\ \hline -\pmb 1^\top &0 &0 \\ \hline -\mathbf x^\top &0 &0 \end{array} \end{bmatrix}, $ Similarly, we can deduce that $ |\mathbf D|\cdot \begin{vmatrix} \pmb 1^\top \mathbf D^{-1}\pmb 1 &\pmb 1^\top \mathbf D^{-1}\mathbf x \\ \pmb 1^\top \mathbf D^{-1}\mathbf x &\mathbf x^\top \mathbf D^{-1}\mathbf x \end{vmatrix} =|\mathbf D|\cdot|\pmb 1^\top \mathbf D^{-1}(\pmb 1 \mathbf x^\top-\mathbf x\pmb 1^\top)\mathbf D^{-1}\mathbf x|. $
Hence we need $ \pmb 1^\top \mathbf D^{-1}(\pmb 1 \mathbf x^\top-\mathbf x\pmb 1^\top)\mathbf D^{-1}\mathbf x \neq 0 $.
Can we say that as long as $\mathbf x\neq k \pmb 1$ $(k\in \Bbb R)$, then A is full rank?