General Solutions for three-point parabola

  • Thread starter Thread starter Hyperreality
  • Start date Start date
  • Tags Tags
    General Parabola
Hyperreality
Messages
201
Reaction score
0
I'm having problems solving this particular question
Using matrix calculation to find a general solution for a parabola that passes through points (x1,y1), (x2, y2), (x3, y3).

First I setted my equations

ax1^2+bx2+c=y1

ax2^2+bx2+c=y2

ax3^2+bx2+c=y3

But I've no idea on how I should set up the augmented matrix. I presume the general solution is referring to the solving the constants a, b and c.

Any hints and suggestions is appreciated.
 
Physics news on Phys.org
First of all, your equations are incorrect.
In the first equation, the term "bx2" should read "bx1", while in your third equation the term "bx2" should read "bx3" (I assume these were typos..)

Look at the the left-hand side of your equations.
How can you rewrite this side as a matrix-vector product, where you know the entries in the matrix, while the vector is the unknown?
 


Hi there,

Solving a system of equations involving a parabola can be tricky, but using matrix calculations can make it much simpler. Here are some steps you can follow to find the general solution for a three-point parabola:

1. Set up your equations: As you have already done, start by setting up your equations using the three points given. This will give you three equations with three unknowns (a, b, and c).

2. Write the equations in matrix form: To solve the system of equations using matrices, we need to write them in the form of AX = B, where A is the coefficient matrix, X is the unknowns matrix, and B is the constant matrix. In this case, A will be a 3x3 matrix, X will be a 3x1 matrix, and B will be a 3x1 matrix.

3. Construct the augmented matrix: The augmented matrix is formed by combining the coefficient matrix and the constant matrix, separated by a vertical line. In this case, it will be a 3x4 matrix.

4. Use matrix operations to solve for X: Now that we have our augmented matrix, we can use matrix operations (such as row operations) to solve for X. Once you have the values for a, b, and c, you have found the general solution for the parabola.

I hope this helps! If you need more guidance, you can also try looking up examples of solving systems of equations using matrices. Good luck!
 
##\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