How Can You Find Infinite Solutions for a System of Linear Equations?

  • Thread starter Thread starter unscientific
  • Start date Start date
  • Tags Tags
    Infinite
unscientific
Messages
1,728
Reaction score
13

Homework Statement



I worked out until the last part of the question and 3 equations with 3 unknowns got reduced to this:

x - 2y + 3z = 1
x + 3z = 3



The Attempt at a Solution




y = 1,
x = 3 -3z

Letting x = λ where λ is any real number,

(x,y,z) = (3,1,0) + λ(-3,0,1)


It wouldn't make a difference if i let z be λ instead right?
 
Physics news on Phys.org
The solution you give is the same as x= 3- 3\lambda, y= 1, and z= \lambda. Since z= \lambda, it doesn't matter which you use. You can write the solution as (x, y, z)= (3-\lambda, 1, \lambda) or as (x, y, z)= (3-z, 1, z).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top