Solve and Display Solutions for Systems of Linear Simultaneous Equations

Solve the equation system \(Ax = b\), given the coefficient matrix \(A\) and right-hand side vector \(b\), using gaussianElimination . Display the solutions using showEqn .

Usage

Solve(
 A,
 b = rep (0, nrow (A)),
 vars,
 verbose = FALSE,
 simplify = TRUE,
 fractions = FALSE,
 ...
)

Arguments

A,

the matrix of coefficients of a system of linear equations

b,

the vector of constants on the right hand side of the equations. The default is a vector of zeros, giving the homogeneous equations \(Ax = 0\).

vars

a numeric or character vector of names of the variables. If supplied, the length must be equal to the number of unknowns in the equations. The default is paste0("x", 1:ncol(A).

verbose,

logical; show the steps of the Gaussian elimination algorithm?

simplify

logical; try to simplify the equations?

fractions

logical; express numbers as rational fractions, using the fractions function; if you require greater accuracy, you can set the cycles (default 10) and/or max.denominator (default 2000) arguments to fractions as a global option, e.g., options(fractions=list(cycles=100, max.denominator=10^4)).

...,

arguments to be passed to gaussianElimination and showEqn

Value

the function is used primarily for its side effect of printing the solution in a readable form, but it invisibly returns the solution as a character vector

Details

This function mimics the base function solve when supplied with two arguments, (A, b), but gives a prettier result, as a set of equations for the solution. The call solve(A) with a single argument overloads this, returning the inverse of the matrix A. For that sense, use the function inv instead.

See also

Author

John Fox

Examples

 A1 <- matrix (c (2, 1, -1,
 -3, -1, 2,
 -2, 1, 2), 3, 3, byrow=TRUE)
 b1 <- c (8, -11, -3)
 Solve(A1, b1) # unique solution
#> x1 = 2 
#> x2 = 3 
#> x3 = -1 

 A2 <- matrix (1:9, 3, 3)
 b2 <- 1:3
 Solve(A2, b2, fractions=TRUE) # underdetermined
#> x1 - 1*x3 = 1 
#> x2 + 2*x3 = 0 
#> 0 = 0 

 b3 <- c (1, 2, 4)
 Solve(A2, b3, fractions=TRUE) # overdetermined
#> x1 - 1*x3 = 5/3 
#> x2 + 2*x3 = -1/6 
#> 0 = -1/2