library(pracma)

Page 444

C31 † For the linear transformation S compute the pre-images.

The first pre iamge is the span of vectors x such that \[\mathbf{T(x)} = \left[\begin{array} {rrr} -2 \\ 5 \\ 3 \end{array}\right] \] which is $$ =

$$

#Matrix for first preimage
pre_image1 <- matrix(c(
    1,-2,-1,-2,
    3,-1,2,5,
    1,1,2,3
    
    ),nrow=3, byrow = TRUE)
pre_image1
##      [,1] [,2] [,3] [,4]
## [1,]    1   -2   -1   -2
## [2,]    3   -1    2    5
## [3,]    1    1    2    3
pre_image1r <- rref(pre_image1)
pre_image1r
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    1    0
## [2,]    0    1    1    0
## [3,]    0    0    0    1

We can see from row reduced from the system of equation is inconsistent hence there are no preimages.

The second pre iamge is the span of vectors x such that \[\mathbf{T(x)} = \left[\begin{array} {rrr} -5 \\ 5 \\ 7 \end{array}\right] \]

pre_image2 <- matrix(c(
    1,-2,-1,-5,
    3,-1,2,5,
    1,1,2,7
    
    ),nrow=3, byrow = TRUE)
pre_image2
##      [,1] [,2] [,3] [,4]
## [1,]    1   -2   -1   -5
## [2,]    3   -1    2    5
## [3,]    1    1    2    7
#Matrix pi2 in row reduced-echelon form
pre_image2r = rref(pre_image2)
pre_image2r
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    1    3
## [2,]    0    1    1    4
## [3,]    0    0    0    0

From here we can see that the system of equations is consistent. Due to the existence of the bottom row there are infinite solutions. We can define the span of pre image vector as: \[ T^{-1}(\left[\begin{array} {rrr} -5 \\ 5 \\ 7 \end{array}\right]) \]

$$

{

} ,c \[ \]

{ - + c }, c $$