V LI C24

Question asks us to determine if the sets of vectors are linearly independent or linearly dependent. If linearly dependent, exhibit a nontrivial relation of linear dependence

matA <- matrix(c(1,2,-1,0,1,3,2,-1,2,2,4,4,-2,2,3,-1,2,-1,-2,0), ncol = 4)
matA
##      [,1] [,2] [,3] [,4]
## [1,]    1    3    4   -1
## [2,]    2    2    4    2
## [3,]   -1   -1   -2   -1
## [4,]    0    2    2   -2
## [5,]    1    2    3    0

Simplifying the matrix into reduced row echelon form we get

matB <- matrix(c(1,-1,0,0,0,3,-1,1,0,0,4,-2,1,0,0,-1,-1,-1,0,0), ncol = 4)
matB
##      [,1] [,2] [,3] [,4]
## [1,]    1    3    4   -1
## [2,]   -1   -1   -2   -1
## [3,]    0    1    1   -1
## [4,]    0    0    0    0
## [5,]    0    0    0    0

Final RREF

matC <- matrix(c(1,0,0,0,0,0,1,0,0,0,1,1,0,0,0,2,-1,0,0,0), ncol = 4)
matC
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    1    2
## [2,]    0    1    1   -1
## [3,]    0    0    0    0
## [4,]    0    0    0    0
## [5,]    0    0    0    0

We can see that there are 2 pivot columns, so r=2. Since r is less than the number of columns, there are an infinite number or solutions so the vectors are LINEARLY DEPENDENT

Finding a non-trivial solution

X1 and X2 are our dependent variables and X3 and X4 are our free variables.

X1 = (-1)X3 + (-2)X4

X2 = (-1)X3 + (1)X4

By choosing values for our free variables we can find a linear combination of the vectors.

X3 = 2

X4 = 2

And substituting them into the equations, we find that a solution vector is

solVector <- c(6,0,0,0)
solVector
## [1] 6 0 0 0