Linear Algebra - System of 2 Equations with 3 Variables-possible?

chrisdapos
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Linear Algebra - System of 2 Equations with 3 Variables--possible?

Homework Statement


Solve: x1-3x2+4x3=-4
3x1-7x2+7x3=-8
-4x1+6x2-x3=7


The Attempt at a Solution


I was able to make it to:
1 -3 4 -4
0 -10 25 -11
0 0 0 0
So the third row goes away, and I am left with:
1 -3 4 -4
0 -10 25 -11
I am pretty sure that cannot be solved, or am I overlooking something? Thank you in advance!
 
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It looks like you have a free variable. So you can solve for the other two in terms of it.
 
chrisdapos said:
I am pretty sure that cannot be solved, or am I overlooking something? Thank you in advance!

The solution isn't unique, you can try plugging in a value for x_1 and solving the rest.
x_1=1,2,3,4,5,6...\pi, e,...
 
whenever you have more unknowns than equations, you get infinitely many solutions and one or more variables become free varaibles
 
proton said:
whenever you have more unknowns than equations, you get infinitely many solutions and one or more variables become free varaibles

Sometimes there can still be zero solutions:
x+y+z=0
x+y+z=1
 
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...
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