Dtbcon Dtbtrs Dtpcon Dtptri Dtptrs Dtrcon Dtrtri Dtrtrs

Dtrtrs() Sub Dtrtrs ( Uplo As  String, Trans As  String, Diag As  String, N As  Long, A() As  Double, B() As  Double, Info As  Long, Optional Nrhs As  Long = 1  ) Solution to system of linear equations AX = B or ATX = B for a triangular matrix Purpose This routine solves a triangular system of the form A * X = B or A^T * X = B where A is a triangular matrix of order n and B is an n x nrhs matrix. A check is made to verify that A is nonsingular. Parameters [in] Uplo = "U": A is upper triangular. = "L": A is lower triangular. [in] Trans Specifies the form of the system of equations: = "N": A * X = B. (no transpose) = "T" or "C": A^T * X = B. (transpose) [in] Diag = "N": A is non-unit triangular. = "U": A is unit triangular. (Diagonal elements of A() are not referenced and are assumed to be 1) [in] N Order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in] A() Array A(LA1 - 1, LA2 - 1) (LA1 >= N, LA2 >= N) N x N triangular matrix A. Only the upper or lower triangular part is to be referenced in accordance with Uplo. [in,out] B() Array B(LB1 - 1, LB2 - 1) (LB1 >= max(1, N), LB2 >= Nrhs) (2D array) or B(LB - 1) (LB >= max(1, N), Nrhs = 1) (1D array) [in] N x Nrhs matrix of right hand side matrix B. [out] If Info = 0, the N x Nrhs solution matrix X. [out] Info = 0: Successful exit. = -1: The argument Uplo had an illegal value. (Uplo <> "U" nor "L") = -2: The argument Trans had an illegal value. (Trans <> "N", "T" nor "C") = -3: The argument Diag had an illegal value. (Diag <> "N" nor "U") = -4: The argument N had an illegal value. (N < 0) = -5: The argument A() is invalid. = -6: The argument B() is invalid. = -8: The argument Nrhs had an illegal value. (Nrhs < 0) = i > 0: The i-th diagonal element of A is zero, indicating that the matrix is singular and the solutions have not been computed. [in] Nrhs (Optional) Number of right hand sides, i.e., number of columns of the matrix B. (Nrhs >= 0) (If Nrhs = 0, returns without computation) (default = 1) Reference LAPACK Example Program Solve the system of linear equations Ax = B and estimate the reciprocal of the condition number (RCond) of A, where ( -1.13 0 0 ) ( 0.0452 ) A = ( 0.26 -1.98 0 ), B = ( -0.4856 ) ( -0.96 0.30 -2.32 ) ( 1.2472 ) Sub Ex_Dtrtrs() Const N = 3 Dim A(N - 1, N - 1) As Double, B(N - 1) As Double Dim RCond As Double, Info As Long A(0, 0) = -1.13 A(1, 0) = 0.26: A(1, 1) = -1.98 A(2, 0) = -0.96: A(2, 1) = 0.3: A(2, 2) = -2.32 B(0) = 0.0452: B(1) = -0.4856: B(2) = 1.2472 Call Dtrtrs("L", "N", "N", N, A(), B(), Info) If Info = 0 Then Call Dtrcon("1", "L", "N", N, A(), RCond, Info) Debug.Print "X =", B(0), B(1), B(2) Debug.Print "RCond =", RCond Debug.Print "Info =", Info End Sub Example Results X = -0.04 0.24 -0.49 RCond = 0.314667138698574 Info = 0