Define the linear transformation
\(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\) , \(T \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{c} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{array} \right]\)
A.) compute the preimage \(T^{-1}\left(\begin{array}{c} \left[\begin{array}{c} 2 \\ 3 \end{array} \right]\end{array} \right) and\) \(T^{-1}\left(\begin{array}{c} \left[\begin{array}{c} 4 \\ -8 \end{array} \right]\end{array} \right)\)
\[ T \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 3 \end{array} \right] \]
Given \(T\), we have:
\[ \begin{cases} 2x_1 - x_2 + 5x_3 = 2 \\ -4x_1 + 2x_2 - 10x_3 = 3 \end{cases} \]
first solve the system equation
From the first equation solve for \(x_2\):
\[ 2x_1 - x_2 + 5x_3 = 2 \]
\[ x_2 = 2x_1 + 5x_3 - 2 \]
Then, Substitute this expression for \(x_2\) into the second equation:
\[ -4x_1 + 2(2x_1 + 5x_3 - 2) - 10x_3 = 3 \]
Simplify:
\[ -4x_1 + 4x_1 + 10x_3 - 4 - 10x_3 = 3 \]
\[ -4 = 3 \]
This equation has no solutions. so, there are no preimages for the vector \([2,3]\) under the transformation \(T\).
B.) compute the preimage \(T^{-1}\left(\begin{array}{c} \left[\begin{array}{c} 4 \\ -8 \end{array} \right]\end{array} \right)\)
\[ T \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{c} 4 \\ -8 \end{array} \right] \]
Given \(T\), we have:
\[ \begin{cases} 2x_1 - x_2 + 5x_3 = 4 \\ -4x_1 + 2x_2 - 10x_3 = -8 \end{cases} \]
solve this system of equations to find the preimage.
From the first equation solve for \(x_2\):
\[ 2x_1 - x_2 + 5x_3 = 4 \]
\[ x_2 = 2x_1 + 5x_3 - 4 \]
Substitute this expression for \(x_2\) into the second equation:
\[ -4x_1 + 2(2x_1 + 5x_3 - 4) - 10x_3 = -8 \]
Simplify:
\[ -4x_1 + 4x_1 + 10x_3 - 8 - 10x_3 = -8 \]
\[ -8 = -8 \]
I would assume that this equation holds true for \(x_1\), \(x_2\), and \(x_3\). so, the preimage \(T^{-1}([4,-8])\) would be the space \(\mathbb{C}^3\).