Q1a, Fortran - rextester

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! Creating a program to solve a numerical by gauss elimination method (forward substitution)
  Real a(10,10),b(10),x(10),pivot,sum
  Integer i,j,k,n
  WRITE(*,*)'enter here the dimensions of the matrix'
   Read(*,*)n
  Write(*,*)'enter here the coefficients of the given equations in matrix form ,inputting the values row wise'
  Do i=1,n
   Read(*,*)(a(i,j),j=1,n),b(i)     !read the matrix
    End do
     Write(*,*)'So the matrix form by the given coefficient of the equations is'                            ! Entered matrix and vector
   Do i=1,n
    WRITE(*,*) (a(i,j),j=1,n)      
    End do
     Write(*,*)' hence the B vector is'
      Do i=1,n
       Write(*,*)b(i)
     End do
    Do i=n,1,-1       ! lower  triangle matrix
      Do j=i-1,1,-1
   pivot=a(j,i)/a(i,i)
     Do k=1,n
    a(j,k)=a(j,k)-pivot*a(i,k)
   End do
    b(j)=b(j)-pivot*b(i)
   End do
  End do
   Write(*,*)' and the lower triangular matrix right after gauss elimination'
    Do i=1,n
      Write(*,*)(a(i,j),j=1,n)
     End do
       Write(*,*)'Corresponding B vector:'
   Do I=1,n
      WRITE(*,*) b(i)
    End do
   Do i=1,n              ! Forward substitution
     sum=0
      Do j=1,n-1
     sum=sum+a(i,j)*x(j)
    End do
   x(i)=(b(i)-sum)/a(i,i)
    End do 
    Write(*,*)' So The solution will be as follows :'
   Do i=1,n
    WRITE(*,10)i,x(i)
  End do
10  format('X(',I3,')=',f10.5)
Stop
End

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Compilation time: 0.23 sec, absolute running time: 0.17 sec, cpu time: 0.02 sec, memory peak: 5 Mb, absolute service time: 0,49 sec