MHB Determinant Property: Seen it Before? True?

Dustinsfl
Messages
2,217
Reaction score
5
Has anyone seen this before? Is this true?
$$
\begin{vmatrix}
a & b+c & 1\\
b & a+c & 1\\
c & a+b & 1
\end{vmatrix} =
\begin{vmatrix}
a & b & 1\\
b & a & 1\\
c & a & 1
\end{vmatrix} +
\begin{vmatrix}
a & c & 1\\
b & c & 1\\
c & b & 1
\end{vmatrix}
$$
In this example this works but I don't know if this just a coincidence.
 
Physics news on Phys.org
Well, determinants are linear w.r.t. addition in any single row or column. (Wink)
 
Determinants are multilinear, alternating functions of row or column vectors. If one adds the stipulation that:

$\det(I_n) = 1$

then these properties completely determine the determinant function.

Clearly one such function (the determinant function) with these properties exists. For a proof that the determinant function is the ONLY function with these properties, see:

http://www.millersville.edu/~bikenaga/linear-algebra/det-unique/det-unique.html
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\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...

Similar threads

Back
Top