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How to use matrix method on this system of equations

  • #1
Say that I have 2x+2yz=0 2y+2xz=0 and 2z+2xy=0 how would I use matrix methods to solve this system of equations? I know you can just look at it and easily figure out what the critical points are but I want to do it the safe way. Or is using the matrix method not the easiest way here?

For extra information the crit points are (0,0,0) (-1,1,1) (1,-1,1) (1,1,-1) (-1,-1,-1)

Thanks!
 

Answers and Replies

  • #2
Dick
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There is no 'matrix method' to use. It's not a system of linear equations. And they aren't really called 'critical points'. They are called 'solutions'. Just use algebra and logic. Which I think you are already doing.
 
  • #3
There is no 'matrix method' to use. It's not a system of linear equations. And they aren't really called 'critical points'. They are called 'solutions'. Just use algebra and logic. Which I think you are already doing.
What's the "algebraic" way of doing this? Multiple rounds of substitution?
 
  • #4
Dick
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What's the "algebraic" way of doing this? Multiple rounds of substitution?
Yes, basically. If e.g. x+yz=0 then either y=(-x/z) or z=0, subsitute that into the other equations. Explain how you got your solutions.
 
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  • #5
Yes, basically. If e.g. x+yz=0 then either y=(-x/z) or z=0, subsitute that into the other equations. Explain how you got your solutions.
I just looked at this and thought of good numbers that would satisfy the equations. That's why I wanted a more systematic way of doing it. The substitution looks like a real headache to do.
 
  • #6
Dick
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I just looked at this and thought of good numbers that would satisfy the equations. That's why I wanted a more systematic way of doing it. The substitution looks like a real headache to do.
OK, you just solved by inspection. Doing it systematically isn't that much of headache. Just do it. If I substitute y=(-x/z) into the second relation I get x(1-z^2)=0 after a little algebra. What does that tell you?
 
  • #7
Well I get that x=0 and z=+-1

If I plug into the third one I get x=+-z and z=+-x. Where should I start plugging things in next?
 
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  • #8
OK, you just solved by inspection. Doing it systematically isn't that much of headache. Just do it. If I substitute y=(-x/z) into the second relation I get x(1-z^2)=0 after a little algebra. What does that tell you?
Well I get that x=0 and z=+-1

If I plug into the third one I get x=+-z and z=+-x. Where should I start plugging things in next?
 
  • #9
Office_Shredder
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Well I get that x=0 and z=+-1
That's actually an or, not an and that you want
 
  • #10
That's actually an or, not an and that you want
Oh ok. Thank you. So now that I know x=0 or z=+-1 what should my next step be? Should I plug 1 and -1 into the first equation?
 
  • #11
Dick
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Oh ok. Thank you. So now that I know x=0 or z=+-1 what should my next step be? Should I plug 1 and -1 into the first equation?
You know you are going to get 5 different solutions. Take those 'or' cases separately. In solving the first equation you had the z=0 OR x=(-y/z). First supposing z=0 what can you conclude about x and y? Then after solving the second equation consider the cases x=0, z=1 and z=(-1) separately. Substitute anywhere you like.
 

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