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$A=(a_{ij})$ is an $n\times n$ matrix and $\det(A)=d\neq 0$. Removing the last line and the last column of $\text{adj}(A),$ we get the $(n-1) \times (n-1)$ matrix $B$.

Find $\det(B)$.

Thanks.

Yuxuan Dong
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    Looks like the answer can be obtained by a theorem of Jacobi which can be found on page 98 here: http://www.maths.ed.ac.uk/~aar/papers/aitken.pdf – Casteels Oct 29 '13 at 11:22
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    If $n=2$, doesn't $\det(B)$ just turn out to be the lower right entry of $A$? How can you know what that is, if all you know is $\det A$? – Gerry Myerson Oct 29 '13 at 11:32
  • I was thinking the same thing. It's not clear whether or not $A$ is given. – Casteels Oct 29 '13 at 11:35
  • Observing Casteels's comment, it seems to be we'd get that $$\det B= A_{nn}\cdot\left(\det A\right)^{n-1}$$ – DonAntonio Oct 29 '13 at 12:23
  • Hi,@DonAntonio,this seems not to work for $A=\left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \ \end{array} \right)$ – Yuxuan Dong Oct 29 '13 at 12:36
  • I know, @Aikilis: I got the same with some examples I did. I think that book is pretty weird, as it talks of he "complementary signed minor of $;A';$ "...and I haven't yet found what the heck $;A';$ is! There's something that we're misunderstanding... – DonAntonio Oct 29 '13 at 13:13
  • I found the following: http://mathworld.wolfram.com/JacobisTheorem.html , but since I can't parse it at once and I've not much time to dive into it now I'll leave it to you to make some sense, if possible, out of it. This seems to be clearer than in Aitken's book, yet still pretty hard to grasp. I wonder how hard could it be to give one or two GD examples...! – DonAntonio Oct 29 '13 at 13:16
  • Ok, from here: http://books.google.co.il/books?id=W9qP4ahS0YEC&pg=PR7&lpg=PR7&dq=Jacobi+theorem+minors+determinant&source=bl&ots=vmzMcoLdBs&sig=nFPqE9Z3BNkKr-blX_cyXXZzW5k&hl=en&sa=X&ei=rLRvUszoL-HQ7AbIl4D4CA&ved=0CFIQ6AEwBQ#v=onepage&q=Jacobi%20theorem%20minors%20determinant&f=false I think I've succeeded to make some sense out of all this mess, @Aikilis . In your example: the adjugate matrix is $$\begin{pmatrix}6&0&0\0&3&0\0&0&2\end{pmatrix}$$ so for example the $;2\times 2;$ principal minor is, according to Jacobi $$|A|^1\cdot 3=18$$...and it certainly is! – DonAntonio Oct 29 '13 at 13:29
  • @DonAntonio Oh,I see!Thanks for taking time!I'll summarize our discucssion as an answer later. – Yuxuan Dong Oct 29 '13 at 14:18
  • And I'm appreciative of @Casteels 's hint for I learned a new theorem(Jacobi) from it.What a happy day. – Yuxuan Dong Oct 29 '13 at 14:19
  • Guys..maybe we can make life easier...$det(B)$ is just the right-bottom entry of $adj(adj(A))$. $adj(adj(A))=d^{n-2}A$.It's done...I just learned it from somebody.@DonAntonio – Yuxuan Dong Oct 29 '13 at 15:05

2 Answers2

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$\det(B)$ is just the right-bottom entry of $adj(adj(A))$. And $adj(adj(A))=d^{n-2}A$.So $\det(B)=d^{n-2}A_{n,n}$.

I learned it from somebody just now.

And in another hand,we can use Jacobi's theorem.(see comments)

Yuxuan Dong
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Here is the algebraic derivation. First write $$\bf{A} = \left(\matrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right) \,\,\,\text{and}\,\,\,\, \bf{A}^{-1}=\pmatrix{\bf{E} & \bf{f} \\ \bf{g}^\top & h}$$ and define determinant $\Delta = \det\bf{A}$. The dimension of $\bf{A}$ is $n$. The question then is the determinant of the matrix $\left(\Delta\bf{I}\cdot\bf{E}\right)$ which has dimension $n-1$. Let $k$ be non-zero (otherwise the result is trivial; multiplying the terms of $A^{-1}A$ shows $E$ to be singular).

Rewriting the determinant: $$\Delta = \det\left[\overbrace{\pmatrix{\bf{I} & -\mathbf{b}\frac{1}{k} \\ \mathbf{0}^\top & 1\\}}^{\det = 1}\pmatrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right]$$ or $$\Delta = \det\left(\mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top\right)\cdot k$$ Since the $\bf{b}$ column was zeroed out, the determinant is the multiplication of the blocks' determinants. Also the unimodular row operation matrix does not change the determinant (standard algebra).

Now with that form of the determinant handy, we focus on the $\bf{E}$ matrix. Using a previously derived formula (also worked out algebraically) we write: $$\mathbf{E}^{-1} = \mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top$$

Hmm, that looks familiar, it is the matrix that determined $\Delta$. So now we have

$$\Delta = \det\left(\mathbf{E}^{-1}\right)\cdot k$$ or $$\frac{1}{\det\left(\mathbf{E}^{-1}\right)}= \det(\mathbf{E}) = \frac{k}{\Delta}$$

And we finally arrive at the formula of interest (so long as we remember that the dimension of $\mathbf{E}$ is $n-1$)

$$\det\left(\Delta\bf{I}\cdot\bf{E}\right) = \Delta^{n-1}\frac{k}{\Delta} = \Delta^{n-2}k$$

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