Question on reducing 3 equations with 3 unknowns

  • Thread starter Thread starter majon
  • Start date Start date
  • Tags Tags
    Unknowns
majon
Messages
8
Reaction score
0
Say I have a system of equations of the following form:

a_1 A^2 + b_1 B^2 + c_1 C^2 + d_1 = f_1
a_2 A^2 + b_2 B^2 + c_2 C^2 + d_2 = f_2
a_3 A^2 + b_3 B^2 + c_3 C^2 + d_3 = f_3

Where a,b,c,d,f are coefficients, and A,B and C are unknown variables.

1. Can we write them in terms of two variables only?
2. Can we write them in terms of one variable only?
 
Physics news on Phys.org
If you replace the A^2 with AA and similarly with B and C then you have 3 linear eqns in AA, BB, and CC that can be easily solved using matrix methods.
 
I was asking because I was reading a document where the author deals with 3 equations and 3 unknown parameters, each equation has quadratic dependence on each of the 3 parameters. Then the author says: with these three equations and three parameters we can reduce the equations to two second order equations with respect to two unknowns (call this case 1), or to one quartic equation with respect to one unknown (call this case 2). So I wanted to know how to do this. I'll do the exercise by eliminating one of the variable and see what I get.
 
Ad jedishrfu said, in the given equation, since only the squares of the unknown values occur, we can treat them as, say X=A^2, Y= B^2, and Z= C^2. Then the equations become a_1X+ b_1Y+ c_1Z= f_1, a_2X+ b_2Y+ c_2Z= f_2, and a_3X+ b_3Y+ c_3Z= f_3. (There is no need for the separate 'd_1' and 'f_1'- we can always subtract d_1 from both sides of the equation.)

There are many ways to solve a "system of equations" but, yes, they all basically involve reducing from three equations in three unknowns to two equations in two unknowns and then to one equation in one unknown. For example, if a_2 is non-zero, I can solve the first equation for X: a_1X= f_1- b_1Y- c_1Z so X= (f_1-b_1X- c_1Z)/a_1. Now replace "X" in the other two equations by that to get two equations in Y and Z only.

Or:
1) multiply the first equation by a_2 to get
a_1a_2X+ b_1a_2Y+ c_1a_2Z= f_1a_2 and
2) multiply the second equation by a_1 to get
a)a_1a_2X+ a_1b_2Y+ a_2c_3Z= a_2f_3. Now
3) subtract those two equation. Since the X term in each equation has the same coefficient, a_1a_2, they cancel leaving a single equation in Y and Z.

Do the same with, say, the first and third equation, to get a second equation in Y and Z.
 
##\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