Dposv Dposvx Dppsv Dppsvx

Dposv() Sub Dposv ( Uplo As  String, N As  Long, A() As  Double, B() As  Double, Info As  Long, Optional Nrhs As  Long = 1  ) (Simple driver) Solution to system of linear equations AX = B for a symmetric positive definite matrix Purpose This routine computes the solution to a real system of linear equations A * X = B, where A is an n x n symmetric positive definite matrix and X and B are n x nrhs matrices. The Cholesky decomposition is used to factor A as A = U^T*U, if Uplo = "U", or A = L*L^T, if Uplo = "L", where U is an upper triangular matrix and L is a lower triangular matrix. The factored form of A is then used to solve the system of equations A * X = B. Parameters [in] Uplo = "U": Upper triangle of A is stored. = "L": Lower triangle of A is stored. [in] N Number of linear equations, i.e., order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] A() Array A(LA1 - 1, LA2 - 1) (LA1 >= N, LA2 >= N) [in] N x N symmetric positive definite matrix A. The upper or lower triangular part is to be referenced in accordance with Uplo. [out] If Info = 0, the factor U or L from the Cholesky factorization A = U^T*U or A = L*L^T. [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 N had an illegal value. (N < 0) = -3: The argument A() is invalid. = -4: The argument B() is invalid. = -6: The argument Nrhs had an illegal value. (Nrhs < 0) = i > 0: The leading minor of order i of A is not positive definite, so the factorization could not be completed, and the solution has 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 A is symmetric positive definite and ( 2.2 -0.11 -0.32 ) ( -1.566 ) A = ( -0.11 2.93 0.81 ), B = ( -2.8425 ) ( -0.32 0.81 2.37 ) ( -1.1765 ) Sub Ex_Dposv() Const N = 3 Dim A(N - 1, N - 1) As Double, B(N - 1) As Double Dim ANorm As Double, RCond As Double, Info As Long A(0, 0) = 2.2 A(1, 0) = -0.11: A(1, 1) = 2.93 A(2, 0) = -0.32: A(2, 1) = 0.81: A(2, 2) = 2.37: B(0) = -1.566: B(1) = -2.8425: B(2) = -1.1765 ANorm = Dlansy("1", "L", N, A()) Call Dposv("L", N, A(), B(), Info) If Info = 0 Then Call Dpocon("L", N, A(), ANorm, RCond, Info) Debug.Print "X =", B(0), B(1), B(2) Debug.Print "RCond =", RCond Debug.Print "Info =", Info End Sub Example Results X = -0.8 -0.92 -0.29 RCond = 0.446791078068956 Info = 0