Singular Manual: lusolve

  | ### 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
See `system("rref",A);`. See ludecomp; luinverse; system.

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