Homogeneous system of linear equation:

  • Thread starter TonyC
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  • #1
TonyC
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I am having trouble finding the solution to the homogeneous system of linear equations:
2x-2y+z=0
-2x+y+z=0
 

Answers and Replies

  • #2
TonyC
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I guess I should have also put:

How can I break this down?
 
  • #3
FredGarvin
Science Advisor
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Perhaps I have been out of the rigid math stuff for a while, but how are you going to solve a system with three unknowns and only two equations?
 
  • #4
TD
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Choose one variable (e.g. z) as "t" and solve the system for x and y in function of t.
You'll get an infinite number of solutions, for every t (so z), you have a couple (x,y).
 
  • #5
TonyC
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Hence my problem..
The answers to choose from are:
x=3/4t, y=-t,z=1/2t
x=-3/4t,y=-t,z=1/2t
z=-3/4t,y=t,z=1/2t
z=3/4t,y=t,z=1/2t

This is why I am stumped.
 
  • #6
TD
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In your case, y was substituted for t. Then solve it as if t was a parameter for x and z.
 
  • #7
TonyC
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? I am still confused.
 
  • #8
TD
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You start with

[tex]\left\{ \begin{gathered}
2x - 2y + z = 0 \hfill \\
- 2x + y + z = 0 \hfill \\
\end{gathered} \right[/tex]

Substitute y = t, t is now a parameter, and solve the following (2x2)-system for x and z

[tex]\left\{ \begin{gathered}
2x + z = 2t \hfill \\
- 2x + z = - t \hfill \\
\end{gathered} \right[/tex]
 
  • #9
HallsofIvy
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You might also want to review your list of possible answers. "x" got changed to "z" in some of them!

(Am I the only person who hates multiple choice questions in mathematics?)
 
  • #10
TD
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HallsofIvy said:
(Am I the only person who hates multiple choice questions in mathematics?)
(No :yuck:)
 

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