Question:
$A$ is an $n$ order square matrix, and $B=A^{(*)}$ is the adjoint matrix of $A$ (i.e. $AB=BA=\det A\cdot I_n$).
Show that
$\det B \begin{pmatrix} 1 & 2 & \cdots & k \\ 1 & 2 & \cdots & k \end{pmatrix} = (\det A)^{k-1}\cdot \det A \begin{pmatrix} k+1 & k+2 & \cdots & n \\ k+1 & k+2 & \cdots & n \end{pmatrix}. $
Here, $B \begin{pmatrix} 1 & 2 & \cdots & k \\ 1 & 2 & \cdots & k \end{pmatrix}$ denotes the $k\times k$ submatrix of $B$ on the top left corner,
$A \begin{pmatrix} k+1 & k+2 & \cdots & n \\ k+1 & k+2 & \cdots & n \end{pmatrix}$ denotes the the $(n-k)\times (n-k)$ submatrix of $A$ on the bottom right corner.
I have tried to use the Laplace expansion theorem but failed.