Solving a 4x4 Determinant for Math Projects

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I was doing a math project, and I was wondering if anyone knew how to solve a 4x4 determinant.
 
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Convert it to it's upper triangular form using Gaussian Elimination and the determinant should be the product of it's diagonal elements. I think.
 
Aplying the permutation definition, you will have 24 terms to sum. Its much easier to do by minors.
 
The DEFINITION of a genereal n by n determinant is this: form all possible products taking one number from each row and column. There will be n! ways to do this. If you write the terms so that the numbers are in order of the columns, the row numbers will be a permutation of 1,2,3...n. Multiply each product by -1 if this is an odd permutation, 1 if even permutation, and add.

The simplest way to calculate it is row reduce as Jhageb suggested.

Second simplest way is to "expand by minors".
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

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