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Warning: new definition for norm
Warning: new definition for trace
> read `gauss.mpl`;
> evalm(A11);
[ 1 1 -1 0 -1 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
[ ]
[ 0 0 1 -1/30 1/2 ]
[ ]
[ 0 0 0 1 135 ]
> # At this point the matrix is in row-echelon form
> # The solution cam be read directly from here.
> #The last equation is: N = 135 and therefore x = 1, y = 3 and z = 5.
> # It is an accident that we can read x, y, and z, directly from N.
> # In general we would have to solve the (triangular) system of
> # linear equations:
> pivot(A11,2,2);
[ 1 0 -110/9 1/9 -406/9 ]
[ ]
[ 0 1 101/9 -1/9 397/9 ]
[ ]
[ 0 0 1 -1/30 1/2 ]
[ ]
[ 0 0 0 1 135 ]
> pivot(",3,3);
[ 1 0 0 -8/27 -39 ]
[ ]
[ 71 ]
[ 0 1 0 --- 77/2 ]
[ 270 ]
[ ]
[ 0 0 1 -1/30 1/2 ]
[ ]
[ 0 0 0 1 135 ]
> pivot(",4,4);
[ 1 0 0 0 1 ]
[ ]
[ 0 1 0 0 3 ]
[ ]
[ 0 0 1 0 5 ]
[ ]
[ 0 0 0 1 135 ]
> # Ofcourse Maple can get to the row-echelon form in a single step!
> gausselim(A);
[ 1 10 100 -1 396 ]
[ ]
[ 0 9 101 -1 397 ]
[ ]
[ 0 0 30 -1 15 ]
[ ]
[ 781 ]
[ 0 0 0 --- 7029/2 ]
[ 30 ]
> # Well almost in row-echelon form, except for the leading ones...
> # Back substitution on this (upper-triangular) matrix gives the answer:
> backsub(");
[ 1, 3, 5, 135 ]
> # i.e.: x =1, y=3, z=5, N=135.
> # The Maple command that gives the Reduced-Row-Echelon-Form is:
> gaussjord(A);
[ 1 0 0 0 1 ]
[ ]
[ 0 1 0 0 3 ]
[ ]
[ 0 0 1 0 5 ]
[ ]
[ 0 0 0 1 135 ]
>