Simplified method for symmetric matrix determinants Preparing for my final exams I have been doing all the exercises in my algebra book. I've seen that there are lots and lots of exercises about determinants of symmetric matrices. Some are easy and others are a bit more twisted, but the basic problem is almost always the same. 

I have been trying to come up with something to do these a bit more quickly, since (at least for me) they invariably end with a very ugly stream of numbers and letters, and they are really tedious and time-consuming. I've been two days with the sensation that (being as the matrices are symmetric) there must be a way/catch/method/trick to calculate the determinants in a more elegant way, but I haven't been able to come up with anything.

I started with a $3\times 3$ matrix like this:

$$A= \begin{pmatrix}
\ a &amp; b &amp; c \\
b &amp; a &amp; b \\
c &amp; b &amp; a \end{pmatrix}$$

which looks fairly simple, but the best I could come up with was:

$2b^2(c-a)+a(a^2-c^2)$

and 

$a(a^2-2b^2-c^2)+2b^2c$

These look horrific and absolutely not what anyone in his right mind would use. It goes without saying that I haven't even tried this with matrices bigger than 3.  

Granted, my math level is basic (I very rarely use linear algebra and calculus out of college, so I tend to completely forget them from year to year) and I don't have many tools to do this, so I don't really know if I'm trying something that is too much for my level or if it simply can't be done. I've consulted all the books about matrices in the college library, but no one mentions anything special about calculating determinants for symmetric matrices.  Is there something I have been missing, or is there nothing to do about it?
There's not a whole lot you can do to simplify that anymore.

In any case, what you've written down is a special case of a symmetric matrix.  In general a symmetric $3 \times 3$ matrix will have the form:

$$A= \begin{pmatrix}
\ a &amp; b &amp; c \\
b &amp; d &amp; e \\
c &amp; e &amp; f \end{pmatrix}$$

which has a determinant of $a(df-e^2) + b(ce-bf) + c(be-dc)$.  Even worse-looking.

The only time it really gets a lot simpler is if you have zeroes in there.  The simplest way to calculate is not to calculate.
I have been doing some try and error in MATLAB for matrices that follows the matrix $A$ from the original question (symmetric with equal terms on their diagonals) and after looking the obtained sequence of determinants in OEIS.org, I found a nice formula for it! $\det(A) = (-1)^{n-1}(n+1)2^{n-2}$.
Now I'm trying to find out why!