# Solution of a tridiagonal system of linear equations Solves the linear system of equations [REMOVE_ME]$$ Ax=B [REMOVE_ME_2]$$ where A has to be square and **tridiagonal**, i.e with nonzero elements only on, one band above, and one band below the diagonal. ## Description 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. ```r Solve.tridiag ( diam1, dia, diap1, B=rep(0,times=length(dia))) ``` ## Arguments - `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. ## Returns 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. ## Author(s) Karline Soetaert ## Examples ```r # 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) ``` ## Details If the length of the vector `dia` is equal to N, then the lengths of `diam1` and `diap1` should be equal to N-1 ## See Also `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

Solve.tridiag function