Simple Linear Algebra problem

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Homework Statement



1. Express the general solution of the following system in vector form.

2x + 4y + 6z + 4w = 4
2x + 5y + 7z + 6w = 3
2x + 3y + 5z + 2w = 5

Clearly identify the particular solution. Also produce a vector with no zero components that satisfies the corresponding homogeneous system.

Homework Equations





The Attempt at a Solution



The solved echelon form is:
2x + 4y + 6z + 4w = 4
0x + 1y + 1z + 2w = -1

x = 4 + 2w -z
y = -1 - z - 2w

solution set: (4-z+2w, -1-z-2w, z, w)

The solution in vector form is displayed as:


x |0 | |4 | | -1 | | 2|
y = |0 |+ |-1|+ | -1 |z + | 2|w
z |0 | |0 | | 1 | | 0|
w |0 | |0 | | 0 | | 1|


The particular solution is the vector:

|4 |
|-1|
|0 |
|0 |

What I'm confused by is this statement: "Also produce a vector with no zero components that satisfies the corresponding homogeneous system.". What is the corresponding homogeneous system? How do I find it? How do I then produce this vector with no zero components?
 
Last edited:

Answers and Replies

  • #2
fzero
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You have a system of equations that is M x = f, where M is a matrix and x, f are vectors. The homogeneous system is M x = 0.
 
  • #3
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You have a system of equations that is M x = f, where M is a matrix and x, f are vectors. The homogeneous system is M x = 0.

So how do I take the solutions that I have to show a homogeneous system? My equations do not equal to zero. Is there a conversion process to do this?
 
  • #4
fzero
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So how do I take the solutions that I have to show a homogeneous system? My equations do not equal to zero. Is there a conversion process to do this?

The corresponding homogeneous system is just


2x + 4y + 6z + 4w = 0
2x + 5y + 7z + 6w = 0
2x + 3y + 5z + 2w = 0

You can solve this by using the same techniques as for the inhomogeneous system.
 
  • #5
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The corresponding homogeneous system is just


2x + 4y + 6z + 4w = 0
2x + 5y + 7z + 6w = 0
2x + 3y + 5z + 2w = 0

You can solve this by using the same techniques as for the inhomogeneous system.

What process did you use to arrive at this equation set? I'd like to do so on my own. Thanks for the help!

*edit* I see that you simply took all of the solutions and replaced them with zero's.
 
  • #6
fzero
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What process did you use to arrive at this equation set? I'd like to do so on my own. Thanks for the help!

*edit* I see that you simply took all of the solutions and replaced them with zero's.

Yes, as I said before, given a vector x of unknowns and a matrix A of coefficients, you can write the linear system

A x = 0.

Because of the 0 vector on the RHS, this is called a homogenous equation. Technically the reason for the terminology is that the equation is invariant under a rescaling of x by a scalar multiple t,

A (t x) = t ( A x ) = t 0= 0.

If we are given an additional vector y, we can construct the inhomogenous system

A x = y.
 
  • #7
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So solving for that homogeneous system produces the following result:

2 4 6 4 = 0
0 1 1 2 = 0
0 0 0 0 = 0

with x=-z+2w and y=-z-2w, solution set is (-z+2w, -z-2w, z, w)

and a solution vector of:

0
0
0
0

Which is not what the problem is asking for. How do I find a solution vector that has no zero components? Is this even possible if the solutions for a homogeneous system of equations are zero's by definition?
 
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  • #8
fzero
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z and w don't have to be zero.
 
  • #9
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z and w don't have to be zero.


Ah! That's right! I can just make them any number, and then add the resulting vectors together correct? Thank you so much!

Would I right that as such? :

when z=w=1;

solution vector:

1
-3
1
1
 

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