On calculating the minor determinant of adjoint matrix 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.

It is actually more related to this. Maybe it is better to express it by partitioning. 
Given an "invertible" square matrix $A$,
$$
A = \left(\begin{array}{c|c}A_1 &A_2\\\hline A_3 &A_4\end{array}\right), A^{-1} =  \left(\begin{array}{c|c}\hat A_1 &\hat A_2\\\hline \hat A_3 &\hat A_4\end{array}\right)
$$
and by stealing the latex code from wikipedia
$$
\begin{multline}
\det{(A^{-1})}=\det\left[ \begin{matrix} A & B \\ C & D \end{matrix}\right]^{-1} = \underbrace{\det\left[ \begin{matrix} I & 0 \\ -D^{-1}C & I \end{matrix}\right]}_1\det\left[ \begin{matrix} (A-BD^{-1}C)^{-1} & 0 \\ 0 & D^{-1} \end{matrix}\right]\\ \underbrace{\det\left[ \begin{matrix} I & -BD^{-1} \\ 0 & I \end{matrix}\right]}_{1}
\end{multline}
$$
another detail is that $(A-BD^{-1}C)^{-1} = \hat{A}_1$. Therefore $\det{(A^{-1})}=\det{(\hat{A}_1)}\det{(A_4^{-1})} $also 
$B$ is defined as
$$
B = \det{(A)} A^{-1} = \left(\begin{array}{c|c}B_1 &B_2\\\hline B_3 &B_4\end{array}\right)
$$
partitioned accordingly. From the Schur Complement formula and from $\det{(\alpha A) = \alpha^n\det{(A)}}$, we know that 
$$
\begin{align}
\det{(A^{-1})} &= \det{(A_4^{-1})}\det{(\hat A_1)}\\
&= \det{(A_4^{-1})}\left[ (\det(A))^{-k}\det{(B_1)}\right] \\
&=   (\det(A))^{-k}\frac{1}{\det{(A_4)}}\det{(B_1)}\\ 
&=\frac{1}{\det{(A)}}
\end{align}
$$
Then, we have the desired result,
$$
\det{(A_4)} = (\det{(A)})^{k-1}\det{(B_1)}
$$
