Define the linear transformation
\[T:\mathbb{C}^3 \rightarrow \mathbb{C}^2, T \left(\left[\begin{array} {r} x_1 \\ x_2 \\ x_3 \end{array}\right]\right) = \begin{bmatrix} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{bmatrix} \]
Compute the preimages, \(T^-1\begin{pmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix}\end{pmatrix}\) and \(T^-1\begin{pmatrix}\begin{bmatrix} 4 \\ -8 \end{bmatrix}\end{pmatrix}\)
We’ll start with the first pre-image: \[T^-1 \begin{pmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix}\end{pmatrix} = T(x) = T \begin{pmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{bmatrix} \]
## [,1] [,2] [,3] [,4]
## [1,] 1 -0.5 2.5 0
## [2,] 0 0.0 0.0 1
We augment the matrix and row reduce: \[ \begin{bmatrix} 2 & 1 & 5 & 2\\ -4 & 2 & 10 & 3 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1 & -.5 & 2.5 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
This system is inconsistent, meaning: \[T^-1\begin{pmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix}\end{pmatrix} = \emptyset\]
We’ll proceed onto the next pre-image: \[T^-1 \begin{pmatrix}\begin{bmatrix} 4 \\ -8 \end{bmatrix}\end{pmatrix} = T(x) = T \begin{pmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{bmatrix} \]
## [,1] [,2] [,3] [,4]
## [1,] 1 -0.5 2.5 2
## [2,] 0 0.0 0.0 0
We augment the matrix and row reduce: \[ \begin{bmatrix} 2 & 1 & 5 & 4\\ -4 & 2 & 10 & -8 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1 & -.5 & 2.5 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
\(x_2\) and \(x_3\) are free variables, meaning the system is undetermined and consistent, with infinite solutions.
\[T^-1\begin{pmatrix}\begin{bmatrix} 4 \\ -8 \end{bmatrix}\end{pmatrix} = \begin{Bmatrix} x_1\begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} -x_2\begin{bmatrix} .5 \\ 1 \\ 0 \end{bmatrix} -x_3\begin{bmatrix} -2.5 \\ 0 \\ 1 \end{bmatrix} | x_2, x_3 \in \mathbb{C} \end{Bmatrix} \]