Find a linear transformation $T\begin{pmatrix}-1\\-2\end{pmatrix}$

Question

$T:\mathbb{R}^2 \Rightarrow\mathbb{R}^3, T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\4\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}4\\-1\\1\end{pmatrix}$. Find $T\begin{pmatrix}-1\\-2\end{pmatrix}$ and $T\begin{pmatrix}-3\\-1\end{pmatrix}$.

For $T\begin{pmatrix}-1\\-2\end{pmatrix}=-1T\begin{pmatrix}1\\0\end{pmatrix}-2T\begin{pmatrix}0\\1\end{pmatrix}=-\begin{pmatrix}-1\\4\\3\end{pmatrix}-2\begin{pmatrix}4\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-4\\-3\end{pmatrix}+\begin{pmatrix}-8\\2\\-2\end{pmatrix}=\begin{pmatrix}-7\\-2\\-5\end{pmatrix}$.

Is this correct? I would like to know if my procedure is correct by finding T.

Answer 1

It's easy to see that the matrix representation of $T$ is $A=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1}$

so applying this transformation to a vector $(x,y)$ you get $$T_A(x,y)=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1} \pmatrix{x \\ y}=\pmatrix{-x+4y \\ 4x-y \\ 3x+y }$$

so now just plug the values you want $$T_A(-1,-2)=\pmatrix{-7 \\-2 \\-5}$$

Answer 2

Note also that $$T \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} \implies T= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} I^{-1}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix}.$$ Next $$T\begin{pmatrix} -1 \\ -2 \end{pmatrix}=\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}$$