How find this invertible matrix $C=\left[\begin{smallmatrix} A&B\\ B^T&0 \end{smallmatrix}\right]$ let matrix $A_{n\times n}$,and $\det(A)>0$, and the matrix $B_{n\times m}$,and such $rank(B)=m$,and let
$$C=\begin{bmatrix}
A&B\\
B^T&0
\end{bmatrix}$$
Find this Invertible matrix
 $C^{-1}$
my try: I found this matrix Invertible matrix $C$,it must find $B^TAB$ Invertible matrix.But I can't
Thank you for your help

Let
$ C^{-1} = \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} $ with appropriate sizes (i.e. $X$ is $n \times n$, $Y$ is $n \times m$, $Z$ is $m \times n$ and $W$ is $m \times m$). Then,
$C C^{-1} = \begin{bmatrix} A & B \\ B^T & 0 \end{bmatrix} \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 & I_m \end{bmatrix}$
Hence, we have the following equations:
$\begin{align}
AX + BZ &= I_n \\
AY + BW &= 0 \\
B^T X &= 0 \\
B^T Y &= I_m
\end{align}$
From the first and third equations:
$\begin{align}
X + A^{-1}BZ &= A^{-1}\\
B^T X + B^T A^{-1}BZ &= B^T A^{-1}\\
Z &= (B^T A^{-1}B)^{-1}B^T A^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ X = A^{-1} - A^{-1}B (B^T A^{-1}B)^{-1}B^T A^{-1} $
From the second and forth equations:
$\begin{align}
Y + A^{-1}BW &= 0\\
B^T Y + B^T A^{-1}BW &= 0\\
B^T A^{-1}BW &= -I_m\\
W &= -(B^T A^{-1}B)^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ Y = A^{-1}B(B^T A^{-1}B)^{-1} $
which is consistent with the forth equation.
Note that $A$ and $B^T A^{-1}B$ must be invertible.
