For the linear transformation T, compute the pre-images.
\(\begin{equation} T:
C^3\end{equation}\) –> \(\begin{equation}C^2,
T\end{equation}\)(\(\begin{bmatrix}x1\\x2\\x3\end{bmatrix}\)) =
\(\begin{bmatrix}2x1-x2+5x3\\-4x1+2x2-10x3\end{bmatrix}\)
\(\begin{equation} T^-1 \end{equation}\)(\(\begin{bmatrix}2\\3\end{bmatrix}\)) \(\begin{equation} T^-1 \end{equation}\)(\(\begin{bmatrix}4\\-8\end{bmatrix}\))
I took T and created a matrix with the end goal (2,3) added that to
the right. We can then do some row reduction:
\(\begin{bmatrix}2&-1&5&2\\-4&2&-10&3\end{bmatrix}\)
–> R2 + 2R1 & .5R1 –> \(\begin{bmatrix}1&-.5&2.5&1\\0&0&0&7\end{bmatrix}\)
The equation we can make from this is:
x1 - .5x2 + 2.5x3 = 1 => x1 = .5x2 - 2.5x3 + 1
using x2, x3:
\(\begin{equation} T^-1
\end{equation}\)(\(\begin{bmatrix}2\\3\end{bmatrix}\)) =
{\(\begin{bmatrix}1\\0\\0\end{bmatrix}\) +
x2\(\begin{bmatrix}.5\\1\\0\end{bmatrix}\) +
x3\(\begin{bmatrix}-2.5\\0\\1\end{bmatrix}\)} |
x2, x3 ∈ \(\begin{equation} C^3
\end{equation}\)
I took T and created a matrix with the end goal (2,3) added that to
the right. We can then do some row reduction:
\(\begin{bmatrix}2&-1&5&4\\-4&2&-10&-8\end{bmatrix}\)
–> R2 + 2R1 & .5R1 –> \(\begin{bmatrix}1&-.5&2.5&2\\0&0&0&0\end{bmatrix}\)
The equation we can make from this is:
x1 - .5x2 + 2.5x3 = 2 => x1 = .5x2 - 2.5x3 + 2
using x2, x3:
\(\begin{equation} T^-1
\end{equation}\)(\(\begin{bmatrix}4\\-8\end{bmatrix}\)) =
{\(\begin{bmatrix}2\\0\\0\end{bmatrix}\) +
x2\(\begin{bmatrix}.5\\1\\0\end{bmatrix}\) +
x3\(\begin{bmatrix}-2.5\\0\\1\end{bmatrix}\)} |
x2, x3 ∈ \(\begin{equation} C^3
\end{equation}\)