9. Solving Linear Equations

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

or, spelled out,

$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{pmatrix} \quad=\quad \begin{pmatrix} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \\ \end{pmatrix}$ For three equations in three unknowns, the equations look like this:

 A <- matrix(paste0("a_{", outer(1:3, 1:3, FUN = paste0), "}"),   nrow=3)  b <- paste0("b_", 1:3) x <- paste0("x", 1:3) showEqn(A, b, vars = x, latex=TRUE)

\begin{array}{lllllll} a_{11} \cdot x_1 &+& a_{12} \cdot x_2 &+& a_{13} \cdot x_3 &=& b_1 \\ a_{21} \cdot x_1 &+& a_{22} \cdot x_2 &+& a_{23} \cdot x_3 &=& b_2 \\ a_{31} \cdot x_1 &+& a_{32} \cdot x_2 &+& a_{33} \cdot x_3 &=& b_3 \\ \end{array}

Conditions for a solution

The general conditions for solutions are:

the equations are consistent (solutions exist) if r(𝐀|𝐛)=r(𝐀)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 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

Equations in two unknowns

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.

Two consistent equations

 A <- matrix(c(1, 2, -1, 2), 2, 2) b <- c(2,1) showEqn(A, b)

## 1*x1 - 1*x2 = 2  ## 2*x1 + 2*x2 = 1

Check whether they are consistent:

 c( R(A), R(cbind(A,b)) ) # show ranks

## [1] 2 2

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

## [1] TRUE

Plot the equations:

 plotEqn(A,b)

## x[1] - 1*x[2] = 2  ## 2*x[1] + 2*x[2] = 1

Plot of two consistent equations which plot as lines intersecting in a point

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

 Solve(A, b, fractions = TRUE)

## x1 = 5/4  ## x2 = -3/4

Three consistent equations

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)

## 1*x1 - 1*x2 = 2  ## 2*x1 + 2*x2 = 1  ## 3*x1 + 1*x2 = 3

 c( R(A), R(cbind(A,b)) ) # show ranks

## [1] 2 2

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

## [1] TRUE

 Solve(A, b, fractions=TRUE) # show solution

## x1 = 5/4  ## x2 = -3/4  ## 0 = 0

Plot the equations:

 plotEqn(A,b)

## x[1] - 1*x[2] = 2  ## 2*x[1] + 2*x[2] = 1  ## 3*x[1] + x[2] = 3

Plot of three consistent equations which plot as three lines intersecting in a point

Three inconsistent equations

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)

## 1*x1 - 1*x2 = 2  ## 2*x1 + 2*x2 = 1  ## 3*x1 + 1*x2 = 6

 c( R(A), R(cbind(A,b)) ) # show ranks

## [1] 2 3

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

## [1] "Mean relative difference: 0.5"

You can see this in the result of reducing $\mathbf{A} | \mathbf{b}$ to echelon form, where the last row indicates the inconsistency because it represents the equation $0 x_1 + 0 x_2 = -3$ .

 echelon(A, b)

## [,1] [,2] [,3] ## [1,] 1 0 2.75 ## [2,] 0 1 -2.25 ## [3,] 0 0 -3.00

Solve() shows this more explicitly, using fractions where possible:

 Solve(A, b, fractions=TRUE)

## x1 = 11/4  ## x2 = -9/4  ## 0 = -3

An approximate solution is sometimes available using a generalized inverse. This gives $\mathbf{x} = (2, -1)$ as a best close solution.

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

## [,1] ## [1,] 2 ## [2,] -1

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))

## x[1] - 1*x[2] = 2  ## 2*x[1] + 2*x[2] = 1  ## 3*x[1] + x[2] = 6

 # add the ginv() solution points(x[1], x[2], pch=15)

Plot of the lines corresponding to three inconsistent equations. They do not all intersect in a point, indicating that there is no common solution.

Equations in three unknowns

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.

Three consistent equations

An example:

 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)

## 2*x1 + 1*x2 - 1*x3 = 8  ## -3*x1 - 1*x2 + 2*x3 = -11  ## -2*x1 + 1*x2 + 2*x3 = -3

Are the equations consistent?

 c( R(A), R(cbind(A,b)) ) # show ranks

## [1] 3 3

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

## [1] TRUE

Solve for $\mathbf{x}$ .

 solve(A, b)

## x1 x2 x3  ## 2 3 -1

Other ways of solving:

 solve(A) %*% b

## [,1] ## x1 2 ## x2 3 ## x3 -1

 inv(A) %*% b

## [,1] ## [1,] 2 ## [2,] 3 ## [3,] -1

Yet another way to see the solution is to reduce $\mathbf{A | b}$ to echelon form. The result of this is the matrix $[\mathbf{I \quad | \quad A^{-1}b}]$ , with the solution in the last column.

 echelon(A, b)

## x1 x2 x3  ## [1,] 1 0 0 2 ## [2,] 0 1 0 3 ## [3,] 0 0 1 -1

`echelon() can be asked to show the steps, as the row operations necessary to reduce $\mathbf{X}$ to the identity matrix $\mathbf{I}$ .

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

##  ## Initial matrix:

## x1 x2 x3  ## [1,] 2 1 -1 8 ## [2,] -3 -1 2 -11 ## [3,] -2 1 2 -3 ##  ## row: 1  ##  ## exchange rows 1 and 2

## x1 x2 x3  ## [1,] -3 -1 2 -11 ## [2,] 2 1 -1 8 ## [3,] -2 1 2 -3 ##  ## multiply row 1 by -1/3

## x1 x2 x3  ## [1,] 1 1/3 -2/3 11/3 ## [2,] 2 1 -1 8 ## [3,] -2 1 2 -3 ##  ## multiply row 1 by 2 and subtract from row 2

## x1 x2 x3  ## [1,] 1 1/3 -2/3 11/3 ## [2,] 0 1/3 1/3 2/3 ## [3,] -2 1 2 -3 ##  ## multiply row 1 by 2 and add to row 3

## x1 x2 x3  ## [1,] 1 1/3 -2/3 11/3 ## [2,] 0 1/3 1/3 2/3 ## [3,] 0 5/3 2/3 13/3 ##  ## row: 2  ##  ## exchange rows 2 and 3

## x1 x2 x3  ## [1,] 1 1/3 -2/3 11/3 ## [2,] 0 5/3 2/3 13/3 ## [3,] 0 1/3 1/3 2/3 ##  ## multiply row 2 by 3/5

## x1 x2 x3  ## [1,] 1 1/3 -2/3 11/3 ## [2,] 0 1 2/5 13/5 ## [3,] 0 1/3 1/3 2/3 ##  ## multiply row 2 by 1/3 and subtract from row 1

## x1 x2 x3  ## [1,] 1 0 -4/5 14/5 ## [2,] 0 1 2/5 13/5 ## [3,] 0 1/3 1/3 2/3 ##  ## multiply row 2 by 1/3 and subtract from row 3

## x1 x2 x3  ## [1,] 1 0 -4/5 14/5 ## [2,] 0 1 2/5 13/5 ## [3,] 0 0 1/5 -1/5 ##  ## row: 3  ##  ## multiply row 3 by 5

## x1 x2 x3  ## [1,] 1 0 -4/5 14/5 ## [2,] 0 1 2/5 13/5 ## [3,] 0 0 1 -1 ##  ## multiply row 3 by 4/5 and add to row 1

## x1 x2 x3  ## [1,] 1 0 0 2 ## [2,] 0 1 2/5 13/5 ## [3,] 0 0 1 -1 ##  ## multiply row 3 by 2/5 and subtract from row 2

## x1 x2 x3  ## [1,] 1 0 0 2 ## [2,] 0 1 0 3 ## [3,] 0 0 1 -1

Now, let’s 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))

Three inconsistent equations

 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)

## 1*x1 + 3*x2 + 1*x3 = 2  ## 1*x1 - 2*x2 - 2*x3 = 3  ## 2*x1 + 1*x2 - 1*x3 = 6

Are the equations consistent? No.

 c( R(A), R(cbind(A,b)) ) # show ranks

## [1] 2 3

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

## [1] "Mean relative difference: 0.5"

Michael Friendly and John Fox

2024-11-24

Conditions for a solution

Equations in two unknowns

Two consistent equations

Three consistent equations

Three inconsistent equations

Equations in three unknowns

Three consistent equations

Three inconsistent equations