Click here if you are seeking to calculate Norms and Condition Numbers
> with(linalg):
> ?linalg
> ?det
> ?mulrow
> ?addrow
> ?matrix
> ?vector
> ?multiply
> ?augment
> A := matrix(6,4);
A = | 6 | 3 | -2 | 1 |
8 | 0 | 7 | 2 | |
9 | 5 | 4 | -3 |
> A := matrix( [ [6,3,-2,1], [8,0,7,2], [9,5,4,-3] ] ) ;
Spaces are allowed but not required. Also, the command may span several lines.
> A[3,4];
shows that -3 is stored in row 3 column 4 of A
> A[3,4] := 9 ;
stores 9 in row 3 column 4 of A
|
|
|
> A := matrix( [ [6,3,-2,1], [8,0,7,2], [9,5,4,-3] ] ) ;
> B := matrix( [ [7,2,0], [-5,3,4], [2,1,9], [6,-7,2] ] ) ;
> C := matrix( [ [7,2,0], [-5,3,4], [2,1,9] ] ) ;
> det(C);
gives the determinant of square matrix C
> inverse(A);
gives the inverse of matrix A
> rank(A);
gives the rank of matrix A
> gaussjord(A);
gives the Gauss-Jordan canonical form of matrix A
> augment(A,C);
augments two matrices A and C
> multiply(A,B);
matrix product AB
> multiply(B,A);
matrix product BA
> evalm(A);
> A := mulrow( A, r, k ) ;
> A := mulcol( A, c, k ) ;
> A := addrow( A, r1, r2, k ) ;
> p := charpoly(C, x) ;
calls the resulting polynomial p
> eigenvals(C,'radical');
> eigenvects(C,'radical');
8x + 7z = 2
9x + 5y + 4z = -3
> M := matrix( [ [6,3,-2,1], [8,0,7,2], [9,5,4,-3] ] ) ;
gives
M = | 6 | 3 | -2 | 1 |
8 | 0 | 7 | 2 | |
9 | 5 | 4 | -3 |
> gaussjord(M);
1 | 0 | 0 | 142 / 197 |
0 | 1 | 0 | -289 / 197 |
0 | 0 | 1 | -106 / 197 |
As long as the identity matrix appears in the coefficient matrix portion, then the rightmost column gives the solution x, y, z.
If you use a decimal point anywhere, the solution will be given in decimal form.
Alternatively, you can solve the linear system Ax = b where A is an n x n coefficient matrix and b is the n-term constant vector by:
> A := matrix( [ [6,3,-2], [8,0,7], [9,5,4] ] ) ;
defines coefficient matrix A
A = | 6 | 3 | -2 |
8 | 0 | 7 | |
9 | 5 | 4 |
> b := vector( [1,2,-3] ) ;
defines constant vector b
> linsolve(A,b);
gives the solution of system
Ax = b in vector form
> det(A);
gives the determinant of coefficient matrix
A
> AI := inverse(A);
gives the inverse A-1 of
coefficient matrix A and calls it AI
> multiply(AI,b);
multiplies A-1b, which is the
solution x of Ax = b
> ?linalg;
Written and Maintained by
Last modified: 06/07/2012
Copyright © 19972023 Kevin G. TeBeest. All rights reserved.
Maple® is a registered trademark of Waterloo Maple Software.
Prof. Kevin G. TeBeest
Applied Mathematics
Kettering University