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. $\begin{pmatrix}{r}2\\0\\2\end{pmatrix}$

To find $ T \begin{pmatrix} 1 \\ -4 \end{pmatrix} $, we first need to determine the linear transformation $ T $ based on the given information. 

The linear map $ T $ is defined by:
\[
T \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -3 \end{pmatrix}
\]
\[
T \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix}
\]

Any vector $\begin{pmatrix} x \\ y \end{pmatrix}$ in $\mathbb{R}^2$ can be written as a linear combination of $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Thus:
\[
\begin{pmatrix} x \\ y \end{pmatrix} = a \begin{pmatrix} 2 \\ 1 \end{pmatrix} + b \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

We need to solve for $a$ and $b$ in the equation:
\[
\begin{pmatrix} 1 \\ -4 \end{pmatrix} = a \begin{pmatrix} 2 \\ 1 \end{pmatrix} + b \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

This gives us the system of linear equations:
\[
2a + b = 1 \quad \text{(1)}
\]
\[
a + 2b = -4 \quad \text{(2)}
\]

We can solve this system using substitution or elimination. Let's use elimination:

Multiply equation (1) by 2:
\[
4a + 2b = 2 \quad \text{(3)}
\]

Subtract equation (2) from equation (3):
\[
(4a + 2b) - (a + 2b) = 2 - (-4)
\]
\[
3a = 6
\]
\[
a = 2
\]

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

Thus, we have $a = 2$ and $b = -3$. Therefore:
\[
\begin{pmatrix} 1 \\ -4 \end{pmatrix} = 2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} - 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]

We now use the linearity of $T$:
\[
T \begin{pmatrix} 1 \\ -4 \end{pmatrix} = 2T \begin{pmatrix} 2 \\ 1 \end{pmatrix} - 3T \begin{pmatrix} 1 \\ 2 \end{pmatrix}
\]
\[
= 2 \begin{pmatrix} 2 \\ 3 \\ -3 \end{pmatrix} - 3 \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 4 \\ 6 \\ -6 \end{pmatrix} - \begin{pmatrix} -6 \\ 12 \\ 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 4 + 6 \\ 6 - 12 \\ -6 - 0 \end{pmatrix}
\]
\[
= \begin{pmatrix} 10 \\ -6 \\ -6 \end{pmatrix}
\]

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

None of the provided answer choices match $ \begin{pmatrix} 10 \\ -6 \\ -6 \end{pmatrix} $, so it seems there may have been an error in the provided options.

Do you want more details or have any questions?

Here are five more questions you could ask:
1. How do you find the kernel of a linear transformation?
2. Can you explain how to compute the determinant of a 3x3 matrix?
3. How do you solve a system of linear equations using matrix inversion?
4. What is the significance of eigenvalues and eigenvectors in linear algebra?
5. How do you determine if a set of vectors forms a basis for a vector space?

**Tip:** When solving systems of linear equations, it is often helpful to use matrix methods such as Gaussian elimination for efficiency and clarity.

Explore a linear transformation example where we find T(1, -4) using given mappings. Understand matrix operations and solving systems of linear equations in this advanced high school level problem.

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