REAL*8 MAT(10,10)   
	CHARACTER*40 INFILE,FILENAME
	LOGICAL EFLG

5	WRITE(6,*) 'ENTER NAME OF FILE WITH MATRIX TO BE INVERTED:'
	READ(5,1200) INFILE
1200	FORMAT(A)
	IF (INFILE(:5).EQ.'     ') STOP
	  INQUIRE(FILE=INFILE, NAME=FILENAME,EXIST=EFLG)
	  IF (.NOT.EFLG) GOTO 5
        OPEN (UNIT=2,FILE=INFILE,STATUS='OLD')
        READ (2,*)DIM
	READ (2,*) M
        READ (2,*) N
	DO 8 I=1,M
8	READ (2,*) (MAT(I,J),J=1,N)
	CALL MATPRN (MAT,M,N)
	IF (N.NE.N) STOP 'MATRIX IS NOT SQUARE!'
	CALL MATINV (MAT,N,DET)
	WRITE(6,*)' THE DETERMINANT IS:',DET
	CALL MATPRN (MAT,N,N)
	WRITE(6,*) 'INVERTING AGAIN TO TEST:'
	CALL MATINV (MAT,N,DET)
	WRITE(6,*)' THE DETERMINANT IS:',DET
	CALL MATPRN (MAT,N,N)

	END

	SUBROUTINE MATPRN(ARRAY,M,N)
C	DOUBLE PRECISION ARRAY
	REAL*8 ARRAY(10,10)
	DO 10 I=1,M
10	WRITE (6,*) (ARRAY(I,J),J=1,N)
  	RETURN
        END


	SUBROUTINE MATINV (ARRAY, NORDER, DET)
	DOUBLE PRECISION ARRAY, AMAX, SAVE
2	DIMENSION ARRAY(10,10), IK (10), JK (10)
10	DET = 1
11 	DO 100  K=1, NORDER
C
C	FIND LARGEST ELEMENT ARRAY(I,J) IN REST OF MATRIX
C
	AMAX=0
21 	DO 30 I=K, NORDER
     	DO 30 J=K, NORDER
23 	IF (DABS (AMAX) - DABS (ARRAY(I,J) ) ) 24, 24, 30
24 	AMAX=ARRAY(I,J)
     	IK(K) =I
     	JK(K) =J
30 	CONTINUE
C
C INTERCHANGE ROWS AND COLUMNS TO PUT AMAX IN ARRAY(K,K)
C
31 	IF (AMAX) 41, 32, 41
32 	DET=0.
     	GO TO 140
41 	I=IK (K)
     	IF (I-K) 21, 51, 43
43 	DO 50 J=1, NORDER
     	SAVE= ARRAY(K,J)
      	ARRAY(K,J)= ARRAY(I,J)
50 	ARRAY(I,J)= -SAVE
51 	J=JK (K)
     	IF (J-K) 21, 61, 53
53 	DO 60 I=1, NORDER
     	SAVE= ARRAY(I,K)
     	ARRAY(I,K)= ARRAY(I,J)
60 	ARRAY(I,J)= -SAVE
C
C	ACCUMULATE ELEMENTS OF INVERSE MATRIX
C
61 	DO 70 I=1, NORDER
     	IF(I-K) 63, 70, 63
63 	ARRAY(I,K) = -ARRAY(I,K)  / AMAX
70 	CONTINUE
71 	DO 80 I=1, NORDER
     	DO 80 J=1, NORDER
     	IF (I-K) 74, 80, 74
74 	IF (J-K) 75, 80, 75
75 	ARRAY(I,J) = ARRAY(I,J) + ARRAY(I,K) *ARRAY(K,J)
80 	CONTINUE
81 	DO 90 J=1, NORDER
     	IF (J-K) 83, 90, 83
83 	ARRAY(K,J) = ARRAY(K,J) / AMAX
90 	CONTINUE
     	ARRAY(K,K) = 1. / AMAX
100 	DET=DET *AMAX
C
C 	RESTORE ORDERING OF MATRIX
C 
101 	DO 130 L=1, NORDER
       	K= NORDER  - L +1
       	J= IK (K)
       	IF (J-K) 111, 111, 105
105 	DO 110  I=1, NORDER
       	SAVE= ARRAY(I,K)
        ARRAY(I,K)= -ARRAY(I,J)
110 	ARRAY(I,J)= SAVE
111 	I= JK(K)
       	IF (I-K) 130, 130, 113
113 	DO 120  J=1, NORDER
       	SAVE= ARRAY(K,J)
       	ARRAY(K,J) = -ARRAY(I,J)
120 	ARRAY(I,J)= SAVE
130 	CONTINUE
140  	RETURN
	END