\[ S: \mathbb{C}^3 \rightarrow \mathbb{C}^3, S (\begin{bmatrix}a\\b\\c\end{bmatrix}) = \begin{bmatrix}a - 2b -c \\3a-b+2c\\a+b+2c\end{bmatrix} \]
\[ S^{-1}(\begin{bmatrix}-2\\5\\3\end{bmatrix}) \]
\[ S^{-1}(\begin{bmatrix}-5\\5\\7\end{bmatrix}) \] Find rref for each vectors to see if pre-image exists.
\[ \begin{bmatrix} 1&-2&-1&-2\\ 3&-1&2&5\\ 1&1&2&3 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1&0&1&0\\ 0&1&1&0\\ 0&0&0&1 \end{bmatrix} \] The row \(\begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}\) essentially represents an equation \(0a + 0b + 0c = 1\), which is impossible (a contradiction) and thus signifies that no set of values for \(a\), \(b\), and \(c\) can satisfy the transformation equation for the given vector.
\[ S^{-1}(v) = \emptyset \]
\[ \begin{bmatrix} 1&-2&-1&-5\\ 3&-1&2&5\\ 1&1&2&7 \end{bmatrix} \xrightarrow{RREF} \begin{bmatrix} 1&0&1&3\\ 0&1&1&4\\ 0&0&0&0 \end{bmatrix} \] \[ a + c = 3 \rightarrow c = 3 - a\\ b + c = 4 \rightarrow c = 4 - b \\ S^{-1}(v)= \begin{bmatrix}3\\4\\0\end{bmatrix} + c\begin{bmatrix}-1\\-1\\0\end{bmatrix} \] Thus, for any value of c, you can find a corresponding pre-image by substituting c into these equations.
below is the R code to compute rref.
pracma::rref(matrix(c(1,-2,-1,-2
,3,-1,2,5,
1,1,2,3),byrow=TRUE, nrow = 3))## [,1] [,2] [,3] [,4]
## [1,] 1 0 1 0
## [2,] 0 1 1 0
## [3,] 0 0 0 1pracma::rref(matrix(c(1,-2,-1,-5,3,-1,2,5,1,1,2,7),byrow=TRUE, nrow = 3))## [,1] [,2] [,3] [,4]
## [1,] 1 0 1 3
## [2,] 0 1 1 4
## [3,] 0 0 0 0First vector has no pre-image under S, while the second vector has an infinite set of pre-images parameterized by c