Problem5:Use GaussJordan elimination to find the solution to: 
> ;
5 x  2 y + 6 z = 0 2 x + y + 3 z = 1

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 rowechelon form. All the points on the line, (x,y,z)=(212t,527t,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 oneline solutions? 