|\^/| Maple V Release 3 (SUNY at Albany) ._|\| |/|_. Copyright (c) 1981-1994 by Waterloo Maple Software and the \ MAPLE / University of Waterloo. All rights reserved. Maple and Maple V <____ ____> are registered trademarks of Waterloo Maple Software. | Type ? for help. 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 ] >