Determinant of a block matrix times the determinant of the inverse of a block

Question

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...

Answer 1

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.