Solving Linear Equations (original) (raw)

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 \(\mathbf{x}\).

## x1 x2 x3 
##  2  3 -1

Other ways of solving:

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

##      [,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.

##      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"