Let a linear map $T:R^2 -> R^3$ defined by
$T\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}2\\3\\-3\end{pmatrix}$, $T\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-2\\4\\0\end{pmatrix}$.
Then  $T\begin{pmatrix}1\\-4\end{pmatrix}$ is equal to

a. $\begin{pmatrix}{r}0\\4\end{pmatrix}$
b. $\begin{pmatrix}{r}0\\4\end{pmatrix}$
c. $\begin{pmatrix}{r}0\\4\end{pmatrix}$
d. $\begin{pmatrix}{r}2\\0\\2\end{pmatrix}$
e. $\begin{pmatrix}{r}2\\0\\2\end{pmatrix}$

Given the linear map $ T: \mathbb{R}^2 \to \mathbb{R}^3 $ defined by:
\[
T\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -3 \end{pmatrix}, \quad T\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix}
\]

We need to find $ T\begin{pmatrix} 1 \\ -4 \end{pmatrix} $.

First, express $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$ as a linear combination of $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$:
\[
\begin{pmatrix} 1 \\ -4 \end{pmatrix} = a \begin{pmatrix} 2 \\ 1 \end{pmatrix} + b \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

This gives us a system of linear equations:
\[
2a + b = 1
\]
\[
a + 2b = -4
\]

Solve this system by substitution or elimination. We use substitution:
First, solve for $b$ from the first equation:
\[
b = 1 - 2a
\]

Substitute $b$ into the second equation:
\[
a + 2(1 - 2a) = -4
\]
\[
a + 2 - 4a = -4
\]
\[
-3a + 2 = -4
\]
\[
-3a = -6
\]
\[
a = 2
\]

Then substitute $a = 2$ back into $b = 1 - 2a$:
\[
b = 1 - 2(2) = 1 - 4 = -3
\]

Thus:
\[
\begin{pmatrix} 1 \\ -4 \end{pmatrix} = 2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} - 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

Now apply $T$ using linearity:
\[
T\begin{pmatrix} 1 \\ -4 \end{pmatrix} = 2 T\begin{pmatrix} 2 \\ 1 \end{pmatrix} - 3 T\begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]
\[
= 2 \begin{pmatrix} 2 \\ 3 \\ -3 \end{pmatrix} - 3 \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 4 \\ 6 \\ -6 \end{pmatrix} - \begin{pmatrix} -6 \\ 12 \\ 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 4 - (-6) \\ 6 - 12 \\ -6 - 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 4 + 6 \\ 6 - 12 \\ -6 \end{pmatrix}
\]
\[
= \begin{pmatrix} 10 \\ -6 \\ -6 \end{pmatrix}
\]

Therefore, $ T\begin{pmatrix} 1 \\ -4 \end{pmatrix} $ is $\begin{pmatrix} 10 \\ -6 \\ -6 \end{pmatrix} $.

However, none of the provided answer choices match this result. There might be an error in the provided options or in the problem statement. Nonetheless, the correct solution from our calculation is:
\[
T\begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 10 \\ -6 \\ -6 \end{pmatrix}
\]

Do you have any questions or need further details?

Here are some more questions you can ask:
1. How do you solve a system of linear equations?
2. What is the definition of a linear transformation?
3. How do you find the matrix representation of a linear map?
4. What are the properties of linear transformations?
5. How do you determine if a given transformation is linear?

**Tip:** Always check if you can express vectors in terms of given basis vectors when dealing with linear transformations.

Explore the solution to a linear map problem where a transformation T: R^2 to R^3 is defined by specific mappings. Learn how to apply linear transformation principles and solve for a given vector using linear combinations and linearity.

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