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"