MHB Can the adj(A) method be simplified for finding inverses of 4x4 matrices?

  • Thread starter Thread starter delgeezee
  • Start date Start date
  • Tags Tags
    Inverse
delgeezee
Messages
12
Reaction score
0
I can find inverses using an adjust for a 3X3 matrix. But My homework book asks us to find the inverse using an adj(A) for a 4x4 matrix. 1 3 1 1
2 5 2 2
1 3 8 9
1 3 2 2

it seems less time efficient to find the inverse using this method. Is it possible to reduce the matrix to a a simpler yet equal form and still come out with the same cofactor matrix?
 
Physics news on Phys.org
delgeezee said:
I can find inverses using an adjust for a 3X3 matrix. But My homework book asks us to find the inverse using an adj(A) for a 4x4 matrix. 1 3 1 1
2 5 2 2
1 3 8 9
1 3 2 2

it seems less time efficient to find the inverse using this method. Is it possible to reduce the matrix to a a simpler yet equal form and still come out with the same cofactor matrix?
Even for a 3x3 matrix, the adjoint method is a very laborious way to find the inverse. For anything larger than that, it rapidly gets far worse. Maybe the purpose of this exercise in the homework book is to get you to see how bad the adjoint method is, so that you will appreciate the value of having other methods.
 
I agree completely that finding the $\text{adj A}$ is very tedious and there isn't any way to simplify the process (row reduction for example will result in a different answer). You will have to calculate 16 determinants for matrices of size 3x3 in this problem. Not fun!
 
##\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...
Back
Top