Determinant of a block matrix times the determinant of the inverse of a block Let $A$ be an $n\times n$ invertible matrix. Let $a$ be a number in $\mathbb{F}$, let $\alpha$ be a row $n$-tuple of numbers from $\mathbb{F}$ and let $\beta$ be a column $n$-tuple of numbers from $\mathbb{F}$. Show that $$
|A|^{-1}\left|\begin{array}{cc}
  a & \alpha \\
  \beta & A
\end{array}\right|=a-\alpha A^{-1}\beta
$$
So far, I have the following:
$$
  \begin{pmatrix}
    1 & 0\\
    -\beta a^{-1} & I
  \end{pmatrix}
  \begin{pmatrix}
    a & \alpha\\
    \beta & A
  \end{pmatrix}=\begin{pmatrix}
    a & \alpha\\
    0 & A-a^{-1}\beta\alpha
  \end{pmatrix}\\
  \left|\begin{array}{cc}
  a & \alpha \\
  \beta & A
  \end{array}\right|=\left|\begin{array}{cc}
    a & \alpha\\
    0 & A-a^{-1}\beta\alpha
  \end{array}\right|=|a||A-a^{-1}\beta\alpha|=|aA-\beta\alpha|\\
  |A^{-1}||aA-\beta\alpha|=|a-A^{-1}\beta\alpha|
$$
I don't know where to go from here...

You can use another decomposition.
$$
\begin{pmatrix}
1 & -\alpha A^{-1} \\
0 & A^{-1}
\end{pmatrix}
\begin{pmatrix}
a & \alpha \\
\beta & A
\end{pmatrix}
=
\begin{pmatrix}
a - \alpha A^{-1} \beta & 0 \\
A^{-1} \beta & I
\end{pmatrix}
$$
Taking determinants on both sides gives
$$
|A|^{-1}
\begin{vmatrix}
a & \alpha \\
\beta & A
\end{vmatrix}
=
a - \alpha A^{-1} \beta
,
$$
as desired.
