PARAMETER(N=3)
DIMENSION A(N,N),B(N),X(N)
OPEN(UNIT=13,FILE='8(B).IN',STATUS='OLD')
READ(13,*)((A(I,J),J=1,N),B(I),I=1,N)
DO 7 K=1,N-1
P=A(K,K)
DO 9 I =K+1,N
F=A(I,K)/P
DO 11 J=K+1,N
A(I,J)=A(I,J)-F*A(K,J)
11 CONTINUE
B(I)=B(I)-F*A(K,J)
9 CONTINUE
7 CONTINUE
X(N)= B(N)/A(N,N)
DO 13 K=N-1,1,-1
SUM=0.0
DO 15 J=K+1,N
SUM=SUM+A(K,J)*X(J)
15 CONTINUE
X(K)=(B(K)-SUM)/A(K,K)
13 CONTINUE
WRITE(6,*) 'The Solutions Are'
PRINT*,'X=',X(1)
PRINT*,'Y=',X(2)
PRINT*,'Z=',X(3)
END
005. Transpose and Inverse Matrix
PARAMETER(N=6,M=3)
DIMENSION A(N,M),AI(M,M)
DATA AI/1,2,4,2,1,-2,-3,0,5/
PRINT*,'Matrix A'
PRINT 20,((AI(I,J),J=1,M),I=1,M)
20 FORMAT(3X,3F8.2//)
PRINT*,'Transpose of the Matrix A'
PRINT 20,((AI(I,J),I=1,M),J=1,M)
DO I=1,M
DO J=1,N
IF(J.LE.M)THEN
A(I,J)=AI(I,J)
ELSE
IF(J.EQ.I+M)THEN
A(I,J)=1.0
ELSE
A(I,J)=0.0
ENDIF
ENDIF
ENDDO
ENDDO
WRITE(6,*)
PRINT*,'AUGMENTED MATRIX IS'
WRITE(6,*)
PRINT 10,((A(I,J),J=1,N),I=1,M)
10 FORMAT (3X,3F6.2//)
DO 7 I=1,M
IF(A(I,I).NE.0.0)THEN
P=A(I,I)
ELSE
PRINT*,'PIVOT ELEMENT IS ZERO'
STOP
ENDIF
DO JJ=1,N
A(I,JJ)=A(I,JJ)/P
ENDDO
DO 3 J=1,M
IF(J.EQ.I)GOTO 3
COM=A(J,I)
DO K=1,N
A(J,K)=A(J,K)-A(I,K)*COM
ENDDO
3 CONTINUE
7 CONTINUE
WRITE(6,*)
PRINT*,'AUGMENTED ECHELON MATRIX'
WRITE(6,*)
PRINT 11,((A(I,J),J=1,N),I=1,M)
11 FORMAT(3X,6F8.2//)
DO I=1,M
DO J=M+1,N
K=J-M
AI(I,K)=A(I,J)
ENDDO
ENDDO
WRITE(6,*)
PRINT*,'INVERSE OF THE MATRIX'
PRINT 30,((AI(I,J),J=1,M),I=1,M)
30 FORMAT(3X,3F8.2//)
END
006. Simpsons 3 by 8 rule
PARAMETER (N=12)
REAL I
F(X)=1.0/(1.0+X**2)
A=0.0
B=3.14159/4
H=(B-A)/REAL(N)
SUM=F(A)+F(B)
X=A
DO K =1,N-1
IF (MOD(K,3).EQ.0) THEN
SUM=SUM+2.0*F(X+REAL(K)*H)
ELSE
SUM=SUM+3.0*F(X+REAL(K)*H)
END IF
END DO
I=3.0*SUM*H/8.0
WRITE(6,10)I
10 FORMAT(3X,'Value of the integration=', f15.7/)
print*, 'Actual Value of the integration=',ATAN(3.14158/4)
END
008. Value of f(x) between -10 to 10increments of 0.5
PRINT *,'X F(X)'
WRITE(*,*)
X=-10.0
10 IF(X.LT.0.0)F=1.0+X/SQRT(1.0+X**2)
IF(X.EQ.0.0)F=0.0
IF(X.GT.0.0)F=1.0-X/SQRT(1.0+X**2)
Y=F
PRINT 20,X,Y
20 FORMAT(2X,2F12.2/)
X=X+0.5
IF(X.LE.10.0)GOTO 10
END
010. Prints Inverse Matrix Only
DIMENSION A(3,6), AI(3,3)
DATA AI/1,2,4,2,1,-2,-3,0,5/
PRINT*,'MATRIX A'
WRITE(6,*)
PRINT 20, ((AI(I,J),J=1,3),I=1,3)
20 FORMAT(3X,3F8.2//)
DO I=1,3
DO J=1,6
IF(J.LE.3)THEN
A(I,J)=AI(I,J)
ELSE
IF(J.EQ.I+3)THEN
A(I,J)=1.0
ELSE
A(I,J)=0.0
ENDIF
ENDIF
ENDDO
ENDDO
PRINT*,'AUGMENTED MATRIX'
WRITE(6,*)
PRINT 10, ((A(I,J),J=1,6),I=1,3)
10 FORMAT(3X,6F8.2//)
DO 7 I=1,3
IF(A(I,I).NE.0.0)THEN
P=A(I,I)
ELSE
PRINT*,'PIVOT ELEMENT IS ZERO'
STOP
ENDIF
DO JJ=1,6
A(I,JJ)=A(I,JJ)/P
ENDDO
DO 3 J=1,3
IF(J.EQ.I)GOTO 3
COM=A(J,I)
DO K=1,6
A(J,K)=A(J,K)-A(I,K)*COM
ENDDO
3 CONTINUE
7 CONTINUE
WRITE(6,*)
PRINT*,'AUGMENTED ECHELON MATRIX'
WRITE(6,*)
PRINT 11,((A(I,J),J=1,6),I=1,3)
11 FORMAT(3X,6F8.2//)
DO I=1,3
DO J=4,6
K=J-3
AI(I,K)=A(I,J)
ENDDO
ENDDO
WRITE(6,*)
PRINT*,'INVERSE OF MATRIX A'
WRITE(6,*)
PRINT 30, ((AI(I,J),J=1,3),I=1,3)
30 FORMAT (3X, 3F8.2//)
END