Question

I just need a way to express this in a more solvable format...\[\\textup{find}A^{-1}\textup{if}A=begin{pmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&\vdots&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{pmatrix}\]

Answer 1

\[begin{pmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&\vdots&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{pmatrix}\]

Answer 2

:| Give me a while to solve these TeX issues,

Answer 3

I'll wait. :)

Answer 4

I don't know what's wrong... does this site not have the environment to support the argument pmatrix? stackexchange doesn't, so I suppose it doesn't surprise me that much. :(

Answer 5

hey bad maybe you should report that into the feedback section

Answer 6

Yay! I got it!\[A=\begin{bmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&\vdots&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{bmatrix}\]

Answer 7

The question is "find the inverse" but I'd like to solve it myself. Right now, though, I don't know where to start.

Answer 8

you sure that pattern is right?
those numbers in the matrix don't seem to have a rule about them

Answer 9

oh nevermind, I see it

Answer 10

I'll work on this tomorrow, and get back to this if I figure out how to do it.

Answer 11

I'm glad you feel pressured to solve this. :P

Answer 12

Blah, I'm going to sleep, then coming back later.