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Re: Solving Systems Of Linear Equations


From: Mike Miller
Subject: Re: Solving Systems Of Linear Equations
Date: Mon, 5 May 2003 19:59:14 -0500 (CDT)

On Mon, 5 May 2003, Lauren Sorenson wrote:

> Question:  Solve this System of Linear Equations
>
> 2W+5X-Y+4Z=0
> W+X+Y+Z=0
> 4W-3X+6Y+Z=0
> 2W-5X-3Y-Z=7


That's the same as...


[ 2  5 -1  4|  [W| = [0|
| 1  1  1  1|  |X| = |0|
| 4 -3  6  1|  |Y| = |0|
| 2 -5 -3 -1]  |Z] = |7]


...which is my representation of a matrix times a column vector with the
product equal to a column vector.  Or...

M x = b

So take the inverse of M and multiply that inverse times b, and you have
your answer:

 x = inv(M)*b

That's basically what you do in Octave.  Here's how I did it:


octave:1> M =[ 2  5 -1  4 ;  1  1  1  1 ;  4 -3  6  1 ; 2 -5 -3 -1]
M =

   2   5  -1   4
   1   1   1   1
   4  -3   6   1
   2  -5  -3  -1

octave:2> b=[0 0 0 7]'
b =

  0
  0
  0
  7

octave:3> x=inv(M)*b
x =

   3.00000
   1.00000
  -1.00000
  -3.00000


So, W=3, X=1, Y=-1 and Z=-3.

Best,

Mike

-- 
Michael B. Miller, Ph.D.
Assistant Professor
Division of Epidemiology
and Institute of Human Genetics
University of Minnesota
http://taxa.epi.umn.edu/~mbmiller/



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