Dpbsv Dpbsvx Dptsv Dptsvx

Dptsv() Sub Dptsv ( N As  Long, D() As  Double, E() 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 tridiagonal 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 tridiagonal matrix, and X and B are n x nrhs matrices. A is factored as A = L*D*L^T, and the factored form of A is then used to solve the system of equations. Parameters [in] N Order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] D() Array D(LD - 1) (LD >= N) [in] N diagonal elements of the symmetric positive definite tridiagonal matrix A. [out] N diagonal elements of the diagonal matrix D from the factorization A = L*D*L^T. [in,out] E() Array E(LE - 1) (LE >= N - 1) [in] N-1 sub-diagonal elements of the symmetric positive definite tridiagonal matrix A. [out] N-1 sub-diagonal elements of the unit bidiagonal factor L from the L*D*L^T factorization of A. E can also be regarded as the super-diagonal of the unit bidiagonal factor U from the U^T*D*U factorization of A. [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 right hand side matrix B. [out] If Info = 0, the N x Nrhs solution matrix X. [out] Info = 0: Successful exit. = -1: The argument N had an illegal value. (N < 0) = -2: The argument D() is invalid. = -3: The argument E() 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 is not positive definite, and the solution has not been computed. The factorization has not been completed unless i = N. [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 tridiagonal matrix and ( 2.58 -0.99 0 ) ( -1.1850 ) A = ( -0.99 0.69 -0.03 ), B = ( 0.1410 ) ( 0 -0.03 0.18 ) ( 0.1614 ) Sub Ex_Dptsv() Const N As Long = 3 Dim D(N - 1) As Double, E(N - 2) As Double, B(N - 1) As Double Dim ANorm As Double, RCond As Double, Info As Long D(0) = 2.58: D(1) = 0.69: D(2) = 0.18 E(0) = -0.99: E(1) = -0.03 B(0) = -1.185: B(1) = 0.141: B(2) = 0.1614 ANorm = Dlanst("1", N, D(), E()) Call Dptsv(N, D(), E(), B(), Info) If Info = 0 Then Call Dptcon(N, D(), E(), 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 = -0.82 -0.94 0.74 RCond = 0.0437508336668 Info = 0