I have a vector function $F:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^n$, depending upon the variables $(\beta,x_1,\dots,x_n)$ and it is continuously differentiable for any $(\beta,x_1,\dots,x_n)\in\Omega\subset\mathbb{R}^{n+1}$, where $\Omega$ is convex.
I know that, for any $\beta$ there exits uniquely fixed point $(x_1(\beta),\dots,x_n(\beta))$ of $F$, i.e., $$F(\beta,x_1(\beta),\dots,x_n(\beta))=(x_1(\beta),\dots,x_n(\beta))$$
Then, I consider the implicit function $$\Psi(\beta,x_1,\dots,x_n)=(x_1,\dots,x_n)-F(\beta,x_1,\dots,x_n)$$ and I can compute the partial derivative (Jacobian) in the point $(\beta,x_1(\beta),\dots,x_n(\beta))$: $$M(\beta)=\frac{\partial\Psi}{\partial (x_1,\dots,x_n)}(\beta,x_1(\beta),\dots,x_n(\beta))=I-\frac{\partial F}{\partial (x_1,\dots,x_n)}(\beta,x_1(\beta),\dots,x_n(\beta))$$ where $I$ is the identity $n\times n$ matrix.
I know that this Jacobian matrix $M(\beta)$ is invertible for any $\beta$ and I can rewrite it as the block matrix: $$M(\beta)=\begin{bmatrix}a(\beta)&b(\beta)^T\\c(\beta)&D(\beta)\end{bmatrix}$$ where $a(\beta)$ is a number, $b(\beta)$ and $c(\beta)$ are column vectors in $\mathbb{R}^{n-1}$ and $D(\beta)$ is an invertible $(n-1)\times(n-1)$ matrix.
What I want to prove is that $$a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta)>0\text{ for all }\beta$$
This is what I came up with but not sure if it is correct:
Since $M(\beta)$ is invertible for all $\beta$, then by the implicit function theorem, the functions $$(x_1(\beta),\dots,x_n(\beta))$$ are continuously differentiable respect to $\beta$.
Then, by the composition of continuous functions, the Jacobian $$M(\beta)=I-\frac{\partial F}{\partial (x_1,\dots,x_n)}(\beta,x_1(\beta),\dots,x_n(\beta))$$ is continuous respect to $\beta$.
Also, since $M(\beta)$ is invertible for all $\beta$, then, $\det(M(\beta))\ne 0$ for all $\beta$.
We can compute the determinant of the block matrix $M(\beta)$ as follows: \begin{equation}\begin{aligned}\det(M(\beta))&=\det(a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta))\det(D(\beta))\\&=(a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta))\det(D(\beta))\ne 0\end{aligned}\end{equation}
Thus, $a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta)\ne 0$ for all $\beta$.
Finally, I can prove that $a(\beta^*)-b(\beta^*)^TD(\beta^*)^{-1}c(\beta^*)>0$ for a certain value $\beta^*$.
Then, if the function $a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta)$ is continuous, it is strightforward that $a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta)>0$ for all $\beta$.
So I want to know if, in this case, the function $a(\beta)-b(\beta)^TD(\beta)^{-1}c(\beta)$ is continuous given that the Jacobian matrix $$M(\beta)=\begin{bmatrix}a(\beta)&b(\beta)^T\\c(\beta)&D(\beta)\end{bmatrix}$$ is continuous, respect to $\beta$.