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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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