Dspcon Dsptrf Dsptri Dsptrs Dsycon Dsytrf Dsytri Dsytrs

Dsptri() Sub Dsptri ( Uplo As  String, N As  Long, Ap() As  Double, IPiv() As  Long, Info As  Long  ) Inverse of a symmetric matrix in packed form Purpose This routine computes the inverse of a real symmetric indefinite matrix A in packed form using the factorization A = U*D*U^T or A = L*D*L^T computed by Dsptrf. This method inverts U and then computes inv(A) by solving the system inv(A)*L = inv(U) for inv(A). Parameters [in] Uplo Specifies whether the details of the factorization are stored as an upper or lower triangular matrix. = "U": Upper triangular, form is A = U*D*U^T. = "L": Lower triangular, form is A = L*D*L^T. [in] N Order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] Ap() Array Ap(LAp - 1) (LAp >= N(N + 1)/2) [in] The block diagonal matrix D and the multipliers used to obtain the factor U or L as computed by Dsptrf, stored as a packed triangular matrix. [out] If Info = 0, the (symmetric) inverse of the original matrix, stored as a packed triangular matrix. The upper or lower triangular part of the inverse is stored in accordance with Uplo. [in] IPiv() Array IPiv(LIPiv - 1) (LIPiv >= N) Details of the interchanges and the block structure of D as determined by Dsytrf. [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 Ap() is invalid. = -4: The argument IPiv() is invalid. = i > 0: The i-th element of D is exactly zero; the matrix is singular and its inverse could not be computed. Reference LAPACK Example Program Compute the inverse matrix of A, where ( 2.2 -0.11 -0.32 ) A = ( -0.11 2.93 0.81 ) ( -0.32 0.81 -2.37 ) Sub Ex_Dsptri() Const N = 3 Dim Ap(N * (N + 1) / 2) As Double, IPiv(N - 1) As Long, Info As Long Ap(0) = 2.2 Ap(1) = -0.11: Ap(3) = 2.93 Ap(2) = -0.32: Ap(4) = 0.81: Ap(5) = 2.37: Call Dsptrf("L", N, Ap(), IPiv(), Info) If Info = 0 Then Call Dsptri("L", N, Ap(), IPiv(), Info) Debug.Print "Inv(A) =" Debug.Print Ap(0) Debug.Print Ap(1), Ap(3) Debug.Print Ap(2), Ap(4), Ap(5) Debug.Print "Info =", Info End Sub Example Results Inv(A) = 0.463651335375572 1.10603849087679E-04 0.376908423331071 6.25649106339333E-02 -0.128801869057578 0.474409403096834 Info = 0