| Solve.tridiag {limSolve} | R Documentation |
Solves the linear system of equations
Ax=B
where A has to be square and tridiagonal, i.e with nonzero elements only on, one band above, and one band below the diagonal.
Solve.tridiag ( diam1, dia, diap1, B=rep(0,times=length(dia)))
diam1 |
a vector with (nonzero) elements below the diagonal. |
dia |
a vector with (nonzero) elements on the diagonal. |
diap1 |
a vector with (nonzero) elements above the diagonal. |
B |
Right-hand side of the equations, a vector with length = number of rows of A, or a matrix with number of rows = number of rows of A. |
If the length of the vector dia is equal to N, then the lengths of
diam1 and diap1 should be equal to N-1
matrix with the solution, X, of the tridiagonal system of equations Ax=B.
The number of columns of this matrix equals the number of columns of B.
Karline Soetaert <karline.soetaert@nioz.nl>
Solve.banded, the function to solve a banded system of
linear equations.
Solve.block, the function to solve a block diagonal system of
linear equations.
Solve the generalised inverse solution,
solve the R default
# create tridagonal system: bands on diagonal, above and below nn <- 20 # nr rows and columns of A aa <- runif(nn) bb <- runif(nn) cc <- runif(nn) # full matrix A <- matrix(nrow = nn, ncol = nn, data = 0) diag(A) <- bb A[cbind(1:(nn-1), 2:nn)] <- cc[-nn] A[cbind(2:nn, 1:(nn-1))] <- aa[-1] B <- runif(nn) # solve as full matrix solve(A, B) # same, now using tridiagonal algorithm as.vector(Solve.tridiag(aa[-1], bb, cc[-nn], B)) # same, now with 3 different right hand sides B3 <- cbind(B, B*2, B*3) Solve.tridiag(aa[-1], bb, cc[-nn], B3)