C31 - Page 349
For the linear transformation S compute the pre-images.
\[ S: C^3 -> C^3\ S = \left(\begin{array}{c}\begin{bmatrix} a \\ b \\ c \end{bmatrix} \end{array}\right) = \begin{bmatrix} a-2b-c \\ 3a-b+2c \\ a+b+c \end{bmatrix} \]
\[ S^-1 = \left(\begin{array}{c}\begin{bmatrix} -2 \\ 5 \\ 3 \end{bmatrix} \end{array}\right) and\ S^-1 = \left(\begin{array}{c}\begin{bmatrix} -5 \\ 5 \\ 7 \end{bmatrix} \end{array}\right) \]
Using matrix equality, we arrive at a system of two equations in three unknows x1, x2, x3 with an augmented matrix that we can row-reduce to:
\[ \begin{bmatrix} 1 & -2 & -1 & -2 \\ 3 & -1 & 2 & 5 \\ 1 & 1 & 2 & 3 \end{bmatrix} -> \begin{bmatrix} 1 & 0 & 1 & \frac{12}{5} \\ 0 & 1 & 1 & \frac{11}{5} \\ 0 & 0 & 0 & \frac{-8}{5} \end{bmatrix}\]
This system is inconsistent and we can say no vector s is a member of S^-1 (-2, 5, 3). Pre-image cannot be computed.
Using matrix equality, we arrive at a system of two equations in three unknows x1, x2, x3 with an augmented matrix that we can row-reduce to:
\[ \begin{bmatrix} 1 & -2 & -1 & -5 \\ 3 & -1 & 2 & 5 \\ 1 & 1 & 2 & 7 \end{bmatrix} -> \begin{bmatrix} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}\]
We can describe the preimage as:
\[ S^-1 = \left(\begin{array}{c}\begin{bmatrix} -2 \\ 5 \\ 3 \end{bmatrix} \end{array}\right) \\ = \begin{bmatrix} 3-x3 \\ 4-x3 \\ x3 \end{bmatrix} | x3 \Subset C^3\\ = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} + x3\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} | x3 \Subset C^3\\ \]