# Problem5:

Use Gauss-Jordan elimination to find the solution to:

> ;

```                              5 x - 2 y + 6 z = 0

-2 x + y + 3 z = 1```

### SOLUTION:

 Step by step. First get the augmented matrix for the system,

> A := matrix(2,4,[5,-2,6,0, -2,1,3,1]);

```                                [ 5    -2    6    0]
A := [                  ]
[-2     1    3    1]```
 then make that first 5 into a 1 by multiplying the first row by 1/5,

> A1 := mulrow(A,1,1/5);

```                              [ 1    -2/5    6/5    0]
A1 := [                      ]
[-2     1       3     1]```
 then put a 0 where there is a -2. We do that by adding a multiple (2) of row1 to row2,

> A2 := addrow(A1,1,2,2);

```                              [1    -2/5    6/5     0]
A2 := [                      ]
[0    1/5     27/5    1]```
 now we get that 1/5 into a 1 by multiplying the second row by 5,

> A3 := mulrow(A2,2,5);

```                               [1    -2/5    6/5    0]
A3 := [                     ]
[0     1      27     5]```
 now we can turn the -2/5 into a 0 by adding to row1, (2/5)*row2,

> A4 := addrow(A3,2,1,2/5);

```                                 [1    0    12    2]
A4 := [                 ]
[0    1    27    5]```
 so bingo! we now have A in reduced row-echelon form. All the points on the line, (x,y,z)=(2-12t,5-27t,t) for t free scalar parameter, are solutions to the original system of equations. As usual there are many easier ways to get these solutions with maple. Can you find two one-line solutions?

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Tue Feb 8 17:40:39 EST 2000