# 9. Solving Linear Equations

#### Michael Friendly and John Fox

#### 2025-10-05

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 \(r( \mathbf{A} | \mathbf{b}) = r( \mathbf{A})\)
  - 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

`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

### 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)

## 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"