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5.1.86 lusolve
qcindex Gauss
Syntax:
lusolve ( matrix_expression, matrix_expression,
matrix_expression, matrix_expression )
Type:
- matrix
Purpose:
- Computes all solutions of a linear equation system A*x = b, if solvable
The (m x n matrix A must be given by its LU-decomposition, that is, by
three matrices P, L, and U, in this order, which satisfy
- P * A = L * U,
- P is an (m x m) permutation matrix, i.e., its rows/columns form the
standard basis of K^m,
- L is an (m x m) matrix in lower triangular form with all diagonal
entries equal to 1, and
- U is an (m x n) matrix in upper row echelon form.
The fourth argument, b, is expected to be an (m x 1) matrix.
list Q=lusolve(P,L,U,b); fills the list Q with either one entry = 0
(signaling that A*x=b has no solution), or with the three entries 1, x, H,
where x is any (n x 1) solution of the given linear system, and H is a
matrix the columns of which span the solution space of the homogeneous
linear system. (I.e., ncols(H) is the dimension of the solution space.)
If there is exactly one solution, then H is the 1x1 matrix with entry
zero.
Note:
- The method will give a warning if the matrices violate the above conditions
regarding row and column numbers, or if the number of rows of the vector b
does not equal m.
The method expects matrices with entries coming from the ground field of
the given polynomial ring, only.
Example:
| | ring r=0,(x),dp;
matrix A[4][4]=1,1,1,0,1,2,3,1,1,3,5,2,1,4,7,3;
matrix b[4][1]=2,5,8,11;
list L=ludecomp(A);
list Q=lusolve(L[1],L[2],L[3],b);
if (Q[1] == 1)
{
"one solution:";
print(Q[2]);
"check whether result is correct (iff next is zero vector):";
print(A*Q[2]-b);
if ((nrows(Q[3])==1) and (ncols(Q[3])==1) and (Q[3][1,1]==0))
{ "printed solution is the only solution to given linear system" }
else
{
"homogeneous solution space is spanned by columns of:";
print(Q[3]);
}
}
==> one solution:
==> -1,
==> 3,
==> 0,
==> 0
==> check whether result is correct (iff next is zero vector):
==> 0,
==> 0,
==> 0,
==> 0
==> homogeneous solution space is spanned by columns of:
==> -1,-1,
==> 1, 2,
==> 0, -1,
==> -1,0
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See
ludecomp;
luinverse.
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