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To solve this problem, we need to find the image of the vector \(\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix}\) under the linear transformation \( T: \mathbb{R}^4 \to \mathbb{R}^3 \). We are given that

\[
T\begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \quad \text{and} \quad T\begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix}.
\]

### Step 1: Setting Up the Problem
Since \( T \) is a linear transformation, any vector in \(\mathbb{R}^4\) can be expressed as a linear combination of basis vectors. Let's express \(\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix}\) as a linear combination of \(\begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix}\) and \(\begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix}\), if possible.

Suppose:
\[
\begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} = a \begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} + b \begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix}
\]

This gives us a system of equations for \(a\) and \(b\):
\[
\begin{cases}
a = 1 \\
a - b = 3 \\
b = -2 \\
-2a + b = -4
\end{cases}
\]

### Step 2: Solving for \(a\) and \(b\)
From the equations:
1. From \(a = 1\), we substitute into the second equation:
   \[
   1 - b = 3 \implies b = -2
   \]

2. The third equation confirms \(b = -2\).
3. The fourth equation is also consistent:
   \[
   -2(1) + (-2) = -4
   \]

So, we have \(a = 1\) and \(b = -2\).

### Step 3: Applying the Linear Transformation
Now, we use linearity to find \( T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) \):
\[
T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = a T\left( \begin{bmatrix} 1 \\ 1 \\ 0 \\ -2 \end{bmatrix} \right) + b T\left( \begin{bmatrix} 0 \\ -1 \\ 1 \\ 1 \end{bmatrix} \right)
\]

Substituting \(a = 1\) and \(b = -2\), we get:
\[
T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = 1 \cdot \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} + (-2) \cdot \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix}
\]

Calculating each component:
\[
= \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} + \begin{bmatrix} -10 \\ 0 \\ -2 \end{bmatrix}
= \begin{bmatrix} 2 - 10 \\ 3 + 0 \\ -1 - 2 \end{bmatrix}
= \begin{bmatrix} -8 \\ 3 \\ -3 \end{bmatrix}
\]

### Final Answer
\[
T\left( \begin{bmatrix} 1 \\ 3 \\ -2 \\ -4 \end{bmatrix} \right) = \begin{bmatrix} -8 \\ 3 \\ -3 \end{bmatrix}
\]

Would you like more details on any step or have any questions?

### Related Questions

1. How do we verify if a transformation is linear?
2. What are other methods for finding the image of a vector under a linear transformation?
3. How can we represent linear transformations using matrices?
4. How do we determine if a set of vectors spans a vector space?
5. What is the significance of basis vectors in linear transformations?

**Tip:** When solving systems of equations, always check if each equation is consistent with your solution to avoid errors in linear combinations.

Let T: R^4 -> R^3 be a linear transformation such that T([1, 1, 0, -2]) = [2, 3, -1] and T([0, -1, 1, 1]) = [5, 0, 1]. Find T([1, 3, -2, -4]).

Solve the problem of finding the image of the vector [1, 3, -2, -4] under a linear transformation T: R^4 -> R^3. Detailed solution includes steps to express the vector as a linear combination of basis vectors, solving a system of equations, and applying linearity of transformations.

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