# Solve.block: Solution of an almost block diagonal system of linear... In limSolve: Solving Linear Inverse Models

## Description

Solves the linear system A*X=B where A is an almost block diagonal matrix of the form:

TopBlock

... Array(1) ... ... ...

... ... Array(2) ... ...

...

... ... ... Array(Nblocks)...

... ... ... BotBlock

The method is based on Gauss elimination with alternate row and column elimination with partial pivoting, producing a stable decomposition of the matrix A without introducing fill-in.

uses FORTRAN subroutine colrow

## Usage

 `1` ```Solve.block(Top, AR, Bot, B, overlap) ```

## Arguments

 `Top ` the first block of the almost block diagonal matrix `A`. `AR ` intermediary blocks; `AR`(.,.,K) contains the kth block of matrix `A`. `Bot ` the last block of the almost block diagonal matrix `A`. `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`. `overlap ` the number of columns in which successive blocks overlap, and where `overlap = nrow(Top) + nrow(Bot)`.

## Value

matrix with the solution, X, of the block diagonal system of equations Ax=B, the number of columns of this matrix = number of columns of B.

## Note

A similar function but that requires a totally different input can now also be found in the `Matrix` package

## Author(s)

Karline Soetaert <[email protected]>

## References

J. C. Diaz , G. Fairweather , P. Keast, 1983. FORTRAN Packages for Solving Certain Almost Block Diagonal Linear Systems by Modified Alternate Row and Column Elimination, ACM Transactions on Mathematical Software (TOMS), v.9 n.3, p.358-375

`Solve.tridiag` to solve a tridiagonal system of linear equations.

`Solve.banded` to solve a banded system of linear equations.

`Solve` the generalised inverse solution,

`solve` the R default

## Examples

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48``` ```# Solve the following system: Ax=B, where A is block diagonal, and # 0.0 -0.98 -0.79 -0.15 Top # -1.00 0.25 -0.87 0.35 Top # 0.78 0.31 -0.85 0.89 -0.69 -0.98 -0.76 -0.82 blk1 # 0.12 -0.01 0.75 0.32 -1.00 -0.53 -0.83 -0.98 # -0.58 0.04 0.87 0.38 -1.00 -0.21 -0.93 -0.84 # -0.21 -0.91 -0.09 -0.62 -1.99 -1.12 -1.21 0.07 # 0.78 -0.93 -0.76 0.48 -0.87 -0.14 -1.00 -0.59 blk2 # -0.99 0.21 -0.73 -0.48 -0.93 -0.91 0.10 -0.89 # -0.68 -0.09 -0.58 -0.21 0.85 -0.39 0.79 -0.71 # 0.39 -0.99 -0.12 -0.75 -0.68 -0.99 0.50 -0.88 # 0.71 -0.64 0.0 0.48 Bot # 0.08 100.0 50.00 15.00 Bot B <- c(-1.92, -1.27, -2.12, -2.16, -2.27, -6.08, -3.03, -4.62, -1.02, -3.52, 0.55, 165.08) AA <- matrix (nrow = 12, ncol = 12, 0) AA[1,1:4] <- c( 0.0, -0.98, -0.79, -0.15) AA[2,1:4] <- c(-1.00, 0.25, -0.87, 0.35) AA[3,1:8] <- c( 0.78, 0.31, -0.85, 0.89, -0.69, -0.98, -0.76, -0.82) AA[4,1:8] <- c( 0.12, -0.01, 0.75, 0.32, -1.00, -0.53, -0.83, -0.98) AA[5,1:8] <- c(-0.58, 0.04, 0.87, 0.38, -1.00, -0.21, -0.93, -0.84) AA[6,1:8] <- c(-0.21, -0.91, -0.09, -0.62, -1.99, -1.12, -1.21, 0.07) AA[7,5:12] <- c( 0.78, -0.93, -0.76, 0.48, -0.87, -0.14, -1.00, -0.59) AA[8,5:12] <- c(-0.99, 0.21, -0.73, -0.48, -0.93, -0.91, 0.10, -0.89) AA[9,5:12] <- c(-0.68, -0.09, -0.58, -0.21, 0.85, -0.39, 0.79, -0.71) AA[10,5:12]<- c( 0.39, -0.99, -0.12, -0.75, -0.68, -0.99, 0.50, -0.88) AA[11,9:12]<- c( 0.71, -0.64, 0.0, 0.48) AA[12,9:12]<- c( 0.08, 100.0, 50.00, 15.00) ## Block diagonal input. Top <- matrix(nrow = 2, ncol = 4, data = AA[1:2 , 1:4] ) Bot <- matrix(nrow = 2, ncol = 4, data = AA[11:12, 9:12]) Blk1 <- matrix(nrow = 4, ncol = 8, data = AA[3:6 , 1:8] ) Blk2 <- matrix(nrow = 4, ncol = 8, data = AA[7:10 , 5:12]) AR <- array(dim = c(4, 8, 2), data = c(Blk1, Blk2)) overlap <- 4 # answer = (1, 1,....1) Solve.block(Top, AR, Bot, B, overlap = 4) # Now with 3 different B values B3 <- cbind(B, 2*B, 3*B) Solve.block(Top, AR, Bot, B3, overlap = 4) ```

limSolve documentation built on Aug. 14, 2017, 3:01 p.m.