Basic Solutions and Linear Combinations

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SUMMARY

This discussion focuses on the concepts of basic solutions and linear combinations in the context of a homogeneous system of linear equations with four variables: z1, z2, z3, and z4. The example provided illustrates that the basic solutions are the direction vectors represented by the linear combination of parameters t and s. The specific basic solutions identified are the vectors [1, 2, 1, 0] and [0.5, -1, 0, 1]. Additionally, a request for assistance with a Fortran 77 subroutine for reading data from a configuration file is included.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically homogeneous systems of linear equations.
  • Familiarity with reduced row echelon form and its implications for linear equations.
  • Knowledge of vector representation and linear combinations in mathematics.
  • Basic proficiency in Fortran 77 programming for data handling.
NEXT STEPS
  • Study the properties of homogeneous systems of linear equations in detail.
  • Learn about the geometric interpretation of basic solutions and direction vectors.
  • Explore advanced topics in linear algebra, such as vector spaces and span.
  • Review Fortran 77 programming techniques, focusing on file I/O operations and error handling.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as programmers working with Fortran 77 who need to manage data input and output effectively.

lapo3399
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Just to clarify these concepts: if a homogeneous system of linear equations with four variables z1, z2, z3, and z4 yields a matrix in reduced row echelon form that defines (as an arbitrary example) the linear equations

z1 = z3 + 0.5z4 = t + 0.5s
z2 = 2z3 - z4 = 2t - s
z3 = t
z4 = s

then the linear combination should be:

[tex] \left[ \begin{array}{ c } z1 & z2 & z3 & z4 \end{array} \right] = t \left[ \begin{array}{ c } 1 & 2 & 1 & 0 \end{array} \right] + s \left[ \begin{array}{ c } 0.5 & -1 & 0 & 1 \end{array} \right][/tex]

and the basic solutions are:

[tex] \left[ \begin{array}{ c } 1 & 2 & 1 & 0 \end{array} \right] , \left[ \begin{array}{ c } 0.5 & -1 & 0 & 1 \end{array} \right][/tex]

My main problem is understanding exactly what the basic solutions are. I'm not sure whether they're just the direction vectors multiplied by each of the parameters or whether they include the parameters as well. Please clarify this for me.

Thanks.
 
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pls i need some one to help me correct my read subroutine in fortran 77.i must have made some errors and i need to be corrected.Thanks
walterman

c**********************************************************
subroutine read_data(a,b,n,nsys)
c**********************************************************
implicit double precision (a-h,o-z)
dimension a(n,n),b(n)
charactername*20, form*50
c
write(*,*)
write(*,*) 'Enter configuration file name: '
read (*,'(a)')
open(unit=10,file='fname',status='New',iostat=non)
100 continue
read(unit=*,fmt=*) nsys
if(nsys.eq.5) then
form='(5f10.5)'
else
form='(6f10.5)'
end if
read(unit=*,fmt=*,err=999) ((a(i,j), j=1,nsys), i=1,nsys)
read(unit=*,fmt=*,err=999) (b(i), i=1,nsys)
999 write(*,*)
write(*,'(a,i3,a,f16.8)') ' System size : ',nsys,' x',nsys
write(*,*)
write(*,*) 'xmulticient matrix : '
write(20,fmt=form) ((a(i,j), j=1,nsys), i=1,nsys)
write(*,fmt=form) ((a(i,j), j=1,nsys), i=1,nsys)
write(*,*)
write(*,*) 'constant vector: '
write(21,fmt=form) (b(i), i=1,nsys)
write(*,fmt=form) (b(i), i=1,nsys)
write(*,*)

return
end
 

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