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.
 A: 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,
$$
\det{(A_4)} = (\det{(A)})^{k-1}\det{(B_1)}
$$
