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- SUBROUTINE SPODI(A,LDA,N,DET,JOB)
- INTEGER LDA,N,JOB
- REAL A(LDA,1)
- REAL DET(2)
- C
- C SPODI COMPUTES THE DETERMINANT AND INVERSE OF A CERTAIN
- C REAL SYMMETRIC POSITIVE DEFINITE MATRIX (SEE BELOW)
- C USING THE FACTORS COMPUTED BY SPOCO, SPOFA OR SQRDC.
- C
- C ON ENTRY
- C
- C A REAL(LDA, N)
- C THE OUTPUT A FROM SPOCO OR SPOFA
- C OR THE OUTPUT X FROM SQRDC.
- C
- C LDA INTEGER
- C THE LEADING DIMENSION OF THE ARRAY A .
- C
- C N INTEGER
- C THE ORDER OF THE MATRIX A .
- C
- C JOB INTEGER
- C = 11 BOTH DETERMINANT AND INVERSE.
- C = 01 INVERSE ONLY.
- C = 10 DETERMINANT ONLY.
- C
- C ON RETURN
- C
- C A IF SPOCO OR SPOFA WAS USED TO FACTOR A THEN
- C SPODI PRODUCES THE UPPER HALF OF INVERSE(A) .
- C IF SQRDC WAS USED TO DECOMPOSE X THEN
- C SPODI PRODUCES THE UPPER HALF OF INVERSE(TRANS(X)*X)
- C WHERE TRANS(X) IS THE TRANSPOSE.
- C ELEMENTS OF A BELOW THE DIAGONAL ARE UNCHANGED.
- C IF THE UNITS DIGIT OF JOB IS ZERO, A IS UNCHANGED.
- C
- C DET REAL(2)
- C DETERMINANT OF A OR OF TRANS(X)*X IF REQUESTED.
- C OTHERWISE NOT REFERENCED.
- C DETERMINANT = DET(1) * 10.0**DET(2)
- C WITH 1.0 .LE. DET(1) .LT. 10.0
- C OR DET(1) .EQ. 0.0 .
- C
- C ERROR CONDITION
- C
- C A DIVISION BY ZERO WILL OCCUR IF THE INPUT FACTOR CONTAINS
- C A ZERO ON THE DIAGONAL AND THE INVERSE IS REQUESTED.
- C IT WILL NOT OCCUR IF THE SUBROUTINES ARE CALLED CORRECTLY
- C AND IF SPOCO OR SPOFA HAS SET INFO .EQ. 0 .
- C
- C LINPACK. THIS VERSION DATED 08/14/78 .
- C CLEVE MOLER, UNIVERSITY OF NEW MEXICO, ARGONNE NATIONAL LAB.
- C
- C SUBROUTINES AND FUNCTIONS
- C
- C BLAS SAXPY,SSCAL
- C FORTRAN MOD
- C
- C INTERNAL VARIABLES
- C
- REAL T
- REAL S
- INTEGER I,J,JM1,K,KP1
- C
- C COMPUTE DETERMINANT
- C
- IF (JOB/10 .EQ. 0) GO TO 70
- DET(1) = 1.0E0
- DET(2) = 0.0E0
- S = 10.0E0
- DO 50 I = 1, N
- DET(1) = A(I,I)**2*DET(1)
- C ...EXIT
- IF (DET(1) .EQ. 0.0E0) GO TO 60
- 10 IF (DET(1) .GE. 1.0E0) GO TO 20
- DET(1) = S*DET(1)
- DET(2) = DET(2) - 1.0E0
- GO TO 10
- 20 CONTINUE
- 30 IF (DET(1) .LT. S) GO TO 40
- DET(1) = DET(1)/S
- DET(2) = DET(2) + 1.0E0
- GO TO 30
- 40 CONTINUE
- 50 CONTINUE
- 60 CONTINUE
- 70 CONTINUE
- C
- C COMPUTE INVERSE(R)
- C
- IF (MOD(JOB,10) .EQ. 0) GO TO 140
- DO 100 K = 1, N
- A(K,K) = 1.0E0/A(K,K)
- T = -A(K,K)
- CALL SSCAL(K-1,T,A(1,K),1)
- KP1 = K + 1
- IF (N .LT. KP1) GO TO 90
- DO 80 J = KP1, N
- T = A(K,J)
- A(K,J) = 0.0E0
- CALL SAXPY(K,T,A(1,K),1,A(1,J),1)
- 80 CONTINUE
- 90 CONTINUE
- 100 CONTINUE
- C
- C FORM INVERSE(R) * TRANS(INVERSE(R))
- C
- DO 130 J = 1, N
- JM1 = J - 1
- IF (JM1 .LT. 1) GO TO 120
- DO 110 K = 1, JM1
- T = A(K,J)
- CALL SAXPY(K,T,A(1,J),1,A(1,K),1)
- 110 CONTINUE
- 120 CONTINUE
- T = A(J,J)
- CALL SSCAL(J,T,A(1,J),1)
- 130 CONTINUE
- 140 CONTINUE
- RETURN
- END