Determinant of complex block matrix

Question

Let $A$ be an $n\times n$ invertible matrix. Let $a \in \Bbb C$, let $\alpha$ be a row $n$-tuple of complex numbers and let $\beta$ be a column $n$-tuple of complex numbers. Show that

$$(\det(A))^{-1}\, \det\left(\begin{bmatrix}a & \alpha\\ \beta & A \end{bmatrix}\right)=\det\left(a-\alpha A^{-1}\beta\right).$$

Can anyone show me how to prove this?

Answer 1

Hint: $$ \begin{align} (\det A)^{-1} \det \begin{bmatrix} a & \alpha\\ \beta & A \end{bmatrix} &= \det\left( \begin{bmatrix} 1 & \left(\vec 0\right)^T\\ \vec 0 & A^{-1} \end{bmatrix} \right)\det\left( \begin{bmatrix} a & \alpha\\ \beta & A \end{bmatrix} \right) \\ &= \det\left( \begin{bmatrix} 1 & \left(\vec 0\right)^T\\ \vec 0 & A^{-1} \end{bmatrix} \begin{bmatrix} a & \alpha\\ \beta & A \end{bmatrix} \right) \\ &= \det \begin{bmatrix} a & \alpha \\ A^{-1}\beta & I \end{bmatrix} \end{align} $$ Where $\vec 0$ is the column $n$-tuple of zeros.