knitr::opts_chunk$set( warning = FALSE, message = FALSE ) options(digits=4) par(mar=c(5,4,1,1)+.1)

library(rgl) library(knitr) knit_hooks$set(webgl = hook_webgl)

This vignette illustrates the ideas behind solving systems of linear equations of the form $\mathbf{A x = b}$ where

- $\mathbf{A}$ is an $m \times n$ matrix of coefficients for $m$ equations in $n$ unknowns
- $\mathbf{x}$ is an $n \times 1$ vector unknowns, $x_1, x_2, \dots x_n$
- $\mathbf{b}$ is an $m \times 1$ vector of constants, the "right-hand sides" of the equations

The general conditions for solutions are:

- the equations are
*consistent*(solutions exist) if $r( \mathbf{A} | \mathbf{b}) = r( \mathbf{A})$- the solution is
*unique*if $r( \mathbf{A} | \mathbf{b}) = r( \mathbf{A}) = n$ - the solution is
*underdetermined*if $r( \mathbf{A} | \mathbf{b}) = r( \mathbf{A}) < n$

- the solution is
- the equations are
*inconsistent*(no solutions) if $r( \mathbf{A} | \mathbf{b}) > r( \mathbf{A})$

We use `c( R(A), R(cbind(A,b)) )`

to show the ranks, and `all.equal( R(A), R(cbind(A,b)) )`

to test
for consistency.

library(matlib) # use the package

Each equation in two unknowns corresponds to a line in 2D space. The equations have a unique solution if all lines intersect in a point.

A <- matrix(c(1, 2, -1, 2), 2, 2) b <- c(2,1) showEqn(A, b) c( R(A), R(cbind(A,b)) ) # show ranks all.equal( R(A), R(cbind(A,b)) ) # consistent?

Plot the equations:

par(mar=c(4,4,0,0)+.1) plotEqn(A,b)

`Solve()`

is a convenience function that shows the solution in a more comprehensible form:

Solve(A, b, fractions = TRUE)

For three (or more) equations in two unknowns, $r(\mathbf{A}) \le 2$, because $r(\mathbf{A}) \le \min(m,n)$.
The equations will be consistent
*if* $r(\mathbf{A}) = r(\mathbf{A | b})$. This means that whatever linear relations
exist among the rows of $\mathbf{A}$ are the *same* as those among the elements of $\mathbf{b}$.

Geometrically, this means that all three lines intersect in a point.

A <- matrix(c(1,2,3, -1, 2, 1), 3, 2) b <- c(2,1,3) showEqn(A, b) c( R(A), R(cbind(A,b)) ) # show ranks all.equal( R(A), R(cbind(A,b)) ) # consistent? Solve(A, b, fractions=TRUE) # show solution

Plot the equations:

par(mar=c(4,4,0,0)+.1) plotEqn(A,b)

Three equations in two unknowns are *inconsistent* when $r(\mathbf{A}) < r(\mathbf{A | b})$.

A <- matrix(c(1,2,3, -1, 2, 1), 3, 2) b <- c(2,1,6) showEqn(A, b) c( R(A), R(cbind(A,b)) ) # show ranks all.equal( R(A), R(cbind(A,b)) ) # consistent?

You can see this in the result of reducing $\mathbf{A} | \mathbf{b}$ to echelon form, where the last row indicates the inconsistency.

echelon(A, b)

`Solve()`

shows this more explicitly:

Solve(A, b, fractions=TRUE)

An approximate solution is sometimes available using a generalized inverse.

x <- MASS::ginv(A) %*% b x

Plot the equations. You can see that each pair of equations has a solution, but all three do not have a common, consistent solution.

par(mar=c(4,4,0,0)+.1) plotEqn(A,b, xlim=c(-2, 4)) points(x[1], x[2], pch=15)

Each equation in three unknowns corresponds to a plane in 3D space. The equations have a unique solution if all planes intersect in a point.

A <- matrix(c(2, 1, -1, -3, -1, 2, -2, 1, 2), 3, 3, byrow=TRUE) colnames(A) <- paste0('x', 1:3) b <- c(8, -11, -3) showEqn(A, b)

Are the equations consistent?

c( R(A), R(cbind(A,b)) ) # show ranks all.equal( R(A), R(cbind(A,b)) ) # consistent?

Solve for $\mathbf{x}$.

solve(A, b) solve(A) %*% b inv(A) %*% b

Another way to see the solution is to reduce $\mathbf{A | b}$ to echelon form. The result is $\mathbf{I | A^{-1}b}$, with the solution in the last column.

echelon(A, b) echelon(A, b, verbose=TRUE, fractions=TRUE)

Plot them. `plotEqn3d`

uses `rgl`

for 3D graphics. If you rotate the figure, you'll see an orientation
where all three planes intersect at the solution point, $\mathbf{x} = (2, 3, -1)$

plotEqn3d(A,b, xlim=c(0,4), ylim=c(0,4))

A <- matrix(c(1, 3, 1, 1, -2, -2, 2, 1, -1), 3, 3, byrow=TRUE) colnames(A) <- paste0('x', 1:3) b <- c(2, 3, 6) showEqn(A, b)

Are the equations consistent? No.

c( R(A), R(cbind(A,b)) ) # show ranks all.equal( R(A), R(cbind(A,b)) ) # consistent?

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