- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
I try to solve a simple case AX=B with subroutine GBTRS.
The matrix B is 6x1: BT=(3, 1, 11, -3, -5, -1)
The matrix A 6x6 containing 2 subdiagonals and 1 superdiagonal is:
A=2 -1
-3 1 -2
0 -1 2
2 1 1
1 -1 -2 0
-3 -3 1
Doesthe matrix A in band matrixbe arranged right?
Is the following codecorrect?I can't get the correct answer...:(
REAL::A(6,6),B(6)
INTEGER::IPIV(6), INFO
DATA A/0.,0.,0.,2.,-3.,0., 0.,0.,-1.,1.,0.,0., 0.,0.,-2.,-1.,2.,1., &
& 0.,0.,2.,1.,-1.,-3., 0.,0.,1.,-2.,-3.,0., 0.,0.,0.,1.,0.,0. /
DATA B/3., 1., 11., -3., -5., -1./CALL GBTRF(A,2,6,IPIV)
CALL GBTRS(A, B, IPIV)
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Sorry, I was wrong. It is really KL=2. In such a case you should put KL into the second interface:
CALL GBTRS(A, B, IPIV, 2)
Otherwise it is assumed KL=KU=5/3
-Vladimir
Link Copied
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
I try to solve a simple case AX=B with subroutine GBTRS.
The matrix B is 6x1: BT=(3, 1, 11, -3, -5, -1)
The matrix A 6x6 containing 2 subdiagonals and 1 superdiagonal is:
A=2 -1
-3 1 -2
0 -1 2
2 1 1
1 -1 -2 0
-3 -3 1
Doesthe matrix A in band matrixbe arranged right?
Is the following codecorrect?I can't get the correct answer...:(
REAL::A(6,6),B(6)
INTEGER::IPIV(6), INFO
DATA A/0.,0.,0.,2.,-3.,0., 0.,0.,-1.,1.,0.,0., 0.,0.,-2.,-1.,2.,1., &
& 0.,0.,2.,1.,-1.,-3., 0.,0.,1.,-2.,-3.,0., 0.,0.,0.,1.,0.,0. /
DATA B/3., 1., 11., -3., -5., -1./CALL GBTRF(A,2,6,IPIV)
CALL GBTRS(A, B, IPIV)
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Sorry, I was wrong. It is really KL=2. In such a case you should put KL into the second interface:
CALL GBTRS(A, B, IPIV, 2)
Otherwise it is assumed KL=KU=5/3
-Vladimir
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Sorry, I was wrong. It is really KL=2. In such a case you should put KL into the second interface:
CALL GBTRS(A, B, IPIV, 2)
Otherwise it is assumed KL=KU=5/3
-Vladimir
Hi, Vladimir.
I paid too much attention to the GBTRF to make such a stupid question.
I should read the document more carefully.
Thank you very much :)

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page