SS.C40

Suppose that S = {[2, -1, 3, 4], [3, 2, -2, 1]}. Let W = S and let x = [5, 8, -12, -5]. Is x an element of W?

Let’s calculate this out by hand. We will create an augmented matrix and reduce this matrix into its reduced row echelon form.

print(matrix(c(2, -1, 3, 4, 3, 2, -2, 1, 5, 8, -12, -5), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    2    3    5
## [2,]   -1    2    8
## [3,]    3   -2  -12
## [4,]    4    1   -5

Via the Definition EO: Equations Operations

  1. Swap the locations of two equations in the list of equstions.
  2. Multiply each term of an equation by a nonzero quantity.
  3. Multiply each term of one equation by some quantity, and add these terms to a second equation, on both sides of the equality. leave the first equation the same after this operation, but replace the second equation by the new one.

Let’s go ahead and use the 3rd part of the Definition EO to initiate the reduction. We’ll go ahead and perform three steps in one shot.

  1. 2*Row2 + Row1
  2. 3*Row2 + Row3
  3. 4*Row2 + Row4
print(matrix(c(0, -1, 0, 0, 7, 2, 4, 9, 21, 8, 12, 27), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    0    7   21
## [2,]   -1    2    8
## [3,]    0    4   12
## [4,]    0    9   27

Using Part 2 of EO:

  1. (1/3)* Row1
  2. (1/3)* Row3
  3. (1/3)* Row4
print(matrix(c(0, -1, 0, 0, 1, 2, 1, 1, 3, 8, 3, 3), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    0    1    3
## [2,]   -1    2    8
## [3,]    0    1    3
## [4,]    0    1    3

Again, using Part 3 of EO and subsequently rearranging the rows vis Part 1 of EO.

print(matrix(c(1, 0, 0, 0, -2, 1, 0, 0, -8, 3, 0, 0), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    1   -2   -8
## [2,]    0    1    3
## [3,]    0    0    0
## [4,]    0    0    0

Part 3 of EO:

  1. (2)*Row2 + Row1
print(matrix(c(1, 0, 0, 0, 0, 1, 0, 0, -2, 3, 0, 0), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    1    0   -2
## [2,]    0    1    3
## [3,]    0    0    0
## [4,]    0    0    0

Now that we had completely reduced the matrix to its reduced row echelon form, we can easily obtain the scalar values of x1 and x2. In this case, x1 = -2, and x2 = 3. We can confirm this by using R’s lsfit() function.

# Define x and S
# Reference for the functions used.
# https://stat.ethz.ch/R-manual/R-devel/library/stats/html/lsfit.html

x <- c(5, 8, -12, -5)
S <- matrix(c(2, -1, 3, 4, 3, 2, -2, 1), ncol = 2)

# Search for scalars that would determine if x is an element of W
ans <- lsfit(S, x)
print(ans$coefficients)
##     Intercept            X1            X2 
##  1.776357e-15 -2.000000e+00  3.000000e+00

The answer via R is x1 = -2 and x2 = 3.

This confirms that x is indeed an element of W.