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? |