To get a matrix into Reduced-Row-Echelon-Form do:
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Warning: new definition for norm
Warning: new definition for trace
> A := matrix(4,5,[100,10,1,-1,0, 15,15,15,-1,0, 1,10,100,-1,396, -1,-1,1,0,1]);
[ 100 10 1 -1 0 ]
[ ]
[ 15 15 15 -1 0 ]
A := [ ]
[ 1 10 100 -1 396 ]
[ ]
[ -1 -1 1 0 1 ]
> A1 := swaprow(A,1,4);
[ -1 -1 1 0 1 ]
[ ]
[ 15 15 15 -1 0 ]
A1 := [ ]
[ 1 10 100 -1 396 ]
[ ]
[ 100 10 1 -1 0 ]
> A2 := mulrow(A1,1,-1);
[ 1 1 -1 0 -1 ]
[ ]
[ 15 15 15 -1 0 ]
A2 := [ ]
[ 1 10 100 -1 396 ]
[ ]
[ 100 10 1 -1 0 ]
> A3 := addrow(A2,1,2,-15);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 0 30 -1 15 ]
A3 := [ ]
[ 1 10 100 -1 396 ]
[ ]
[ 100 10 1 -1 0 ]
> A4 := addrow(A3,1,3,-1);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 0 30 -1 15 ]
A4 := [ ]
[ 0 9 101 -1 397 ]
[ ]
[ 100 10 1 -1 0 ]
> A5 := addrow(A4,1,4,-100);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 0 30 -1 15 ]
A5 := [ ]
[ 0 9 101 -1 397 ]
[ ]
[ 0 -90 101 -1 100 ]
> A6 := swaprow(A5,2,3);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 9 101 -1 397 ]
A6 := [ ]
[ 0 0 30 -1 15 ]
[ ]
[ 0 -90 101 -1 100 ]
> A7 := mulrow(A6,2,1/9);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
A7 := [ ]
[ 0 0 30 -1 15 ]
[ ]
[ 0 -90 101 -1 100 ]
> A8 := addrow(A7,2,4,90);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
A8 := [ ]
[ 0 0 30 -1 15 ]
[ ]
[ 0 0 1111 -11 4070 ]
> A9 := mulrow(A8,3,1/30);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
A9 := [ ]
[ 0 0 1 -1/30 1/2 ]
[ ]
[ 0 0 1111 -11 4070 ]
> A10 := addrow(A9,3,4,-1111);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
[ ]
A10 := [ 0 0 1 -1/30 1/2 ]
[ ]
[ 781 ]
[ 0 0 0 --- 7029/2 ]
[ 30 ]
> A11 := mulrow(A10,4,30/781);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
A11 := [ ]
[ 0 0 1 -1/30 1/2 ]
[ ]
[ 0 0 0 1 135 ]
> # The solution can now be obtained by "backsub" (backsubstitution),
> # or you may continue with more elementary operations to put zeros
> # on top of the leading ones.
Continue to Reduced-Row-Echelon-Form