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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.

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up vote 3 down vote accepted

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

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Sorry, I didn't understand how do you get $\det (B)=\det{(A_4^{-1})}\det{(B_1)}$, could you extend it? –  NGY Aug 26 '11 at 14:23
    
because it is wrong. I'll rewrite in a sec. Sorry for that –  user13838 Aug 26 '11 at 14:35
    
I have edited again for your disappeared second comment :) –  user13838 Aug 26 '11 at 15:19
    
Thanks a lot, then it is clear.(I deleted the second comment since I came up with an idea for it.) –  NGY Aug 26 '11 at 15:35
    
No problem at all. :) –  user13838 Aug 26 '11 at 15:36
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