C31 For the linear transformations \(S\) compute the pre-images:
\(S:\mathbb{C}^3 \rightarrow \mathbb{C}^3\), \(S \Bigg ( \begin{bmatrix}a \\ b \\ c\\ \end{bmatrix} \Bigg ) = \begin{bmatrix}a-2b-c \\ 3a-b+2c\\ a+b+2c \end{bmatrix}\)
\(S^{-1}\Bigg ( \begin{bmatrix}-2 \\ 5 \\3 \end{bmatrix} \Bigg)\)
Given the pre-image \(\begin{bmatrix}-2\\ 5\\ 3 \end{bmatrix}\), we set it equal to \(S \Bigg( \begin{bmatrix}a \\ b \\c \end{bmatrix}\Bigg)\), which was given as \(\begin{bmatrix}a-2b-c \\ 3a-b+2c\\ a+b+2c \end{bmatrix}\), which gives us the matrix \(\begin{bmatrix}1 & -2 & -1 & -2\\ 3 & -1 & 2 & 5\\ 1 & 1 & 2 & 3\end{bmatrix}\)
In order to find the pre-image, we need to row-reduce. We can do \(R2 - 3(R1) = R2\) and \(R3 - R1 = R3\) to give us \(\begin{bmatrix}1 & -2 & -1 & -2\\0 & 5 & 5 & 11\\ 0 & 3 & 3 & 5 \end{bmatrix}\). We can then do \(\frac{R2}{5} = R2\), which lets us then do \(R1 + 2(R2) = R2\) and \(R3 - 3(R2) = R3\) to give us this matrix \(\begin{bmatrix}1 & 0 & 1 & -\frac{12}{11}\\ 0 & 1 & 1 & \frac{5}{11}\\ 0 & 0 & 0 & \frac{40}{11} \end{bmatrix}\). Finally, we can divide \(\frac{R3}{\frac{40}{11}} = R3\), and then eliminate the values on the last column to give us the row-reduced form of \(\begin{bmatrix}1 & 0 & 1 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\)
Because of the last row we see that there are no values of a, b, and c that will create the pre-image, therefore the pre-image is an empty set.
\(S^{-1}\Bigg ( \begin{bmatrix}-5 \\ 5\\ 7 \end{bmatrix}\Bigg )\)
Since we did it the hard way for the previous question, let’s use R for this row-reduction.
library(pracma)
m <- matrix(c(1, 3, 1, -2, -1, 1, -1, 2, 2, -5, 5, 7), nrow = 3, ncol = 4)
rref.m <- rref(m)
rref.m## [,1] [,2] [,3] [,4]
## [1,] 1 0 1 3
## [2,] 0 1 1 4
## [3,] 0 0 0 0This matrix does have a solution, so we solve for a & b & c, which gives us \(\begin{bmatrix}a \\ b \\ c \end{bmatrix} \Bigg| a = 3 - c, b = 4 - c\)
This then becomes \(\begin{bmatrix}3 - c\\ 4 - c\\ c\end{bmatrix} \Bigg|c \in \mathbb{C}\)
We can separate it into \(\begin{bmatrix}3 \\ 4 \\ 0 \end{bmatrix} + c\begin{bmatrix}-1\\ -1\\ 1\\ \end{bmatrix}\Bigg|c \in \mathbb{C}\)
Until we have \(\begin{bmatrix}3 \\ 4\\ 0\end{bmatrix} + \Bigg\langle\Bigg\{\begin{bmatrix}-1 \\ -1\\ 1 \end{bmatrix}\Bigg\}\Bigg\rangle\)