Full-rank matrix I have a sparse square matrix and want to see if it is full rank (so that I can apply the implicit function theorem).
$$\left[\begin{array}{cccccccccc}
 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
 x_{1}^{2} & Nx_{1} & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\
 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{1} & 0 & 0\\
 x_{2}^{2} & 0 & Nx_{2} & 0 & 0 & -1 & 0 & 0 & 0 & 0\\
 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{2} & 0\\
  0 & 0 & 0 & 0 & z_1 & 0 & 0 & -1 & 1 & 0\\
  x_{3}^{2} & 0 & 0 & Nx_{3} & 0 & 0 & -1 & 0 & 0 & 0\\
  0 & 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{3}\\
  0 & 0 & 0 & 0 & z_2 & z_2 & 0 & 0 & -1 & 1
\end{array}\right]$$
where all variables are strictly positive, and $\sum x_i=1$.
Given that it is sparse, one approach that we considered is to do row-reductions and rearranging to reduce it to a block matrix. This is possible and yields:
$$\left[\begin{array}{ccc}
A & B & 0 \\
0 & C & D \\
E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc}
 N & 0 & 0 & x_{1}-1 & x_{1} & x_{1} & 0 & 0 & 0\\
 0 & N & 0 & x_{2} &x_2-1 & x_{2} & 0 & 0 & 0\\
 0 & 0 & N & x_{3} & x_{3} & x_3-1 & 0 & 0 & 0\\ \hline
 0 & 0 & 0 & x_1z_1 & 0 & 0 & -1 & 1 & 0\\
 0 & 0 & 0 & x_1z_2 & x_2z_2 & 0 & 0 & -1 & 1\\
 0 & 0 & 0 & x_1 & x_2 & x_3 & 0 & 0 & 0\\ \hline
 c & 0 & 0 & 0 & 0 & 0 & -x_{1}^2 & 0 & 0\\
 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{2}^2 & 0\\
 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{3}^2
\end{array}\right]$$

OK, let's call the block matrix above $M$. First eliminate $N$ by a substitution $c\mapsto d N$. Then substitute $z_i \mapsto d y_i$ to eliminate $d$. Then you can construct the Schur complement w.r.t. the first and last rows/columns of $M$ to get $\det(M) = -c^2 N x_1^2 x_2^2 x_3^2 \det P$, with
$$
P = 
\begin{pmatrix}
x_1 y_1 + x_1^{-1}-x_2^{-1}-{x_1^{-2}} & x_1^{-1}-x_2^{-1}+x_2^{-2} &
   x_1^{-1}-x_2^{-1} \\
x_1 y_2+x_2^{-1}-x_3^{-1} & x_2 y_2+x_2^{-1}-x_3^{-1}-x_2^{-2} &
   x_2^{-1}-x_3^{-1}+x_3^{-2} \\
 x_1 & x_2 & x_3 \\
\end{pmatrix}
$$
It might be easier to discuss this determinant.
