Let $A$ be a generic, possibly non symmetric, invertible matrix of order $n$ and ${A}^{-1}$ its known inverse. Let $b$ and $d$ be vectors of length $n$ and $c$ a scalar. Compute the inverse of the block matrix:

with $X$ matrix of order $n$, $y$ and $w$ vectors of length $n$ and $z$ a scalar, all to be determined so that:

$$\left[\begin{array}{cc}A& b\\ {d}^{T}& c\end{array}\right]\left[\begin{array}{cc}X& y\\ {w}^{T}& z\end{array}\right]=\left[\begin{array}{cc}I& 0\\ {0}^{T}& 1\end{array}\right]$$

where $I$ is the identity matrix of order $n$ and $0$ a vector of length $n$ with all components $0$.

Remark that by taking the transpose, since $A$ is not assumed to be symmetric, this is written as:$$w=-z{\left({A}^{-1}\right)}^{T}d$$

It is also possible to compute an inverse matrix of order $n$ from a known matrix of order n+1. The problem is still stated as:

$$\left[\begin{array}{cc}A& b\\ {d}^{T}& c\end{array}\right]\left[\begin{array}{cc}X& y\\ {w}^{T}& z\end{array}\right]=\left[\begin{array}{cc}I& 0\\ {0}^{T}& 1\end{array}\right]$$

However in this case $X$, $y$ , $w$ and $z$ are known, and ${A}^{-1}$has to be computed. Use:$${A}^{-1}b=-\frac{1}{z}y$$

Finally, it is important to remark that here the last row and column are attached or eliminated, but this is only for convenience of presentation. The inverse matrix can be updated in this way when any row (and column) is eliminated. When a row (and column) is attached in the last position, row and columns can be sorted afterwards as desired, in both matrices accordingly.