Dgbcon Dgbtrf Dgbtrs Dgecon Dgetrf Dgetri Dgetrs Dgtcon Dgttrf Dgttrs

Dgttrf() Sub Dgttrf ( N As  Long, Dl() As  Double, D() As  Double, Du() As  Double, Du2() As  Double, IPiv() As  Long, Info As  Long  ) LU factorization of a general tridiagonal matrix Purpose This routine computes an LU factorization of a real tridiagonal matrix A using elimination with partial pivoting and row interchanges. The factorization has the form A = L * U where L is a product of permutation and unit lower bidiagonal matrices and U is upper triangular with nonzeros in only the main diagonal and first two super-diagonals. Parameters [in] N Order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] Dl() Array Dl(LDl - 1) (LDl >= N - 1) [in] N-1 sub-diagonal elements of A. [out] N-1 multipliers that define the matrix L from the LU factorization of A. [in,out] D() Array D(LD - 1) (LD >= N) [in] Diagonal elements of A. [out] N diagonal elements of the upper triangular matrix U from the LU factorization of A. [in,out] Du() Array Du(LDu - 1) (LDu >= N - 1) [in] N-1 super-diagonal elements of A. [out] N-1 elements of the first super-diagonal of U. [out] Du2() Array Du2(LDu2 - 1) (LDu2 >= N - 2) N-2 elements of the second super-diagonal of U. [out] IPiv() Array IPiv(LIPiv - 1) (LIPiv >= N) Pivot indices; for 1 <= i <= N, row i of the matrix was interchanged with row IPiv(i-1). IPiv(i-1) will always be either i or i+1; IPiv(i-1) = i indicates a row interchange was not required. [out] Info = 0: Successful exit. = -1: The argument N had an illegal value. (N < 0) = -2: The argument Dl() is invalid. = -3: The argument D() is invalid. = -4: The argument Du() is invalid. = -5: The argument Du2() is invalid. = -6: The argument IPiv() is invalid. = i > 0: The i-th diagonal element of the factor U is exactly zero. The factorization has been completed, but the factor U is exactly singular, and division by zero will occur if it is used to solve a system of equations. Reference LAPACK Example Program Solve the system of linear equations Ax = B and estimate the reciprocal of the condition number (RCond) of A, where ( -0.58 1.14 0 ) ( -0.5960 ) A = ( -1.56 2.21 0.16 ), B = ( -0.6798 ) ( 0 0.24 0.25 ) ( -0.0406 ) Sub Ex_Dgttrf() Const N = 3 Dim Dl(N - 2) As Double, D(N - 1) As Double, Du(N - 2) As Double Dim Du2(N - 3) As Double, IPiv(N - 1) As Long Dim B(N - 1) As Double, ANorm As Double, RCond As Double, Info As Long Dl(0) = -1.56: Dl(1) = 0.24 D(0) = -0.58: D(1) = 2.21: D(2) = 0.25 Du(0) = 1.14: Du(1) = 0.16 B(0) = -0.596: B(1) = -0.6798: B(2) = -0.0406 ANorm = Dlangt("1", N, Dl(), D(), Du()) Call Dgttrf(N, Dl(), D(), Du(), Du2(), IPiv(), Info) If Info = 0 Then Call Dgttrs("N", N, Dl(), D(), Du(), Du2(), IPiv(), B(), Info) If Info = 0 Then Call Dgtcon("1", N, Dl(), D(), Du(), Du2(), IPiv(), 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 = 5.28453905459942E-02 Info = 0