\(S:\mathbb{C}^3\) \(\rightarrow\) \(\mathbb{C}^3\)
\(S\begin{bmatrix}a \\ b \\ c\end{bmatrix}\) = \(\begin{bmatrix}a-2b-c \\ 3a-b+2c \\ a+b+2c\end{bmatrix}\)
\({S}^{-1}\)\(\begin{bmatrix}-2 \\ 5 \\ 3\end{bmatrix}\) \({S}^{-1}\)\(\begin{bmatrix}-5 \\ 5 \\ 7\end{bmatrix}\)
Working from the definition of the pre-image, Definition PI. Setting \({S}\)\(\begin{bmatrix}a \\ b \\ c\end{bmatrix}\) = \(\begin{bmatrix}-2 \\ 5 \\ 3\end{bmatrix}\)
A <- matrix(c(1,-2,-1,3,-1,2,1,1,2),nrow = 3,byrow = TRUE)
S1 <- matrix(c(-2,5,3),byrow = TRUE)
Aug_matrix <- cbind(A,S1)
Aug_matrix## [,1] [,2] [,3] [,4]
## [1,] 1 -2 -1 -2
## [2,] 3 -1 2 5
## [3,] 1 1 2 3We got the system of 3 equations in 3 variables with an augmented matrix now we reduce it to get solutions
rref(Aug_matrix)## [,1] [,2] [,3] [,4]
## [1,] 1 0 1 0
## [2,] 0 1 1 0
## [3,] 0 0 0 1refering to reduced form we see 1 in the last column, so this system is inconsistent, and there are no valuesof a, b and c that will create an element of the pre-image. So the preimage is the empty set.
Now working for the other set \({S}\)\(\begin{bmatrix}a \\ b \\ c\end{bmatrix}\) = \(\begin{bmatrix}-5 \\ 5 \\ 7\end{bmatrix}\)
A <- matrix(c(1,-2,-1,3,-1,2,1,1,2),nrow = 3,byrow = TRUE)
S2 <- matrix(c(-5,5,7),byrow = TRUE)
Aug_matrix1 <- cbind(A,S2)
Aug_matrix1## [,1] [,2] [,3] [,4]
## [1,] 1 -2 -1 -5
## [2,] 3 -1 2 5
## [3,] 1 1 2 7We got the system of 3 equations in 3 variables with an augmented matrix now we reduce it to get solutions
rref(Aug_matrix1)## [,1] [,2] [,3] [,4]
## [1,] 1 0 1 3
## [2,] 0 1 1 4
## [3,] 0 0 0 0by solving the above setting c free and substitute we get (-1,-1,1) From here we can write the solution set in vector form
\(S^{-1} \bigg(\begin{bmatrix}-5 \\5 \\7 \end{bmatrix} \bigg)\) = \(\bigg(\begin{bmatrix}3 \\4 \\0 \end{bmatrix} + c\begin{bmatrix}-1 \\-1 \\1 \end{bmatrix} \bigg)\)