C14 p56

C14, Page 56

2\(x_1\) + \(x_2\) + 7\(x_3\) - 2\(x_4\) = 4

3\(x_1\)-2\(x_2\) + 11\(x_4\) = 13

\(x_1\) + \(x_2\) + 5\(x_3\) - 3\(x_4\) = 1

A <- matrix(c(2,1,7,-2,3,-2,0,11,1,1,5,-3),byrow =T,nrow=3,ncol=4)
b <- matrix(c(4,13,1),nrow=3,ncol=1)
p <- nrow(A)
U<- cbind(A,b)
U

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    2    1    7   -2    4
## [2,]    3   -2    0   11   13
## [3,]    1    1    5   -3    1

#switch 3rd row to 1st row, since 3rd row has pivot 1
a<-U[3,]
U[3,]<-U[1,]
U[1,]<-a
U

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    5   -3    1
## [2,]    3   -2    0   11   13
## [3,]    2    1    7   -2    4

#make all the number under pivot as 0
U[2,]<-U[2,]-U[1,]*(U[2,1]/U[1,1])
U[3,]<-U[3,]-U[1,]*(U[3,1]/U[1,1])
U

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    5   -3    1
## [2,]    0   -5  -15   20   10
## [3,]    0   -1   -3    4    2

#3rd row * (-1), then switch to 2nd row
U[3,]<-U[3,]*-1
a<-U[3,]
U[3,]<-U[2,]
U[2,]<-a
U

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    5   -3    1
## [2,]    0    1    3   -4   -2
## [3,]    0   -5  -15   20   10

#make all the number under pivot as 0
U[3,]<-U[3,]-U[2,]*(U[3,2]/U[2,2])
U

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    5   -3    1
## [2,]    0    1    3   -4   -2
## [3,]    0    0    0    0    0

Conclusion: There are infinity solutions for this matrix. The result is reduced row-echelon form with last row as zero row, which means there are two free varialbes in the solutions with any value.