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To solve this problem, we need to determine the image of the matrix \( \begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix} \) under the linear transformation \( T \). We know \( T \) is a linear transformation from \( M_{2,2} \) into \( M_{2,2} \), and it is defined by its action on the basis matrices of \( M_{2,2} \):

1. \( T\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \)
2. \( T\left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} 0 & 2 \\ 1 & 1 \end{bmatrix} \)
3. \( T\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\right) = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \)
4. \( T\left(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right) = \begin{bmatrix} 3 & -1 \\ 1 & 0 \end{bmatrix} \)

### Step-by-Step Solution

1. Express the matrix \( \begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix} \) as a linear combination of the basis matrices:
   \[
   \begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix} = a \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
   \]

   Solving for \( a \), \( b \), \( c \), and \( d \), we get:
   - \( a = 1 \)
   - \( b = -1 \)
   - \( c = 4 \)
   - \( d = 3 \)

2. Apply \( T \) to the linear combination:
   \[
   T\left(\begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix}\right) = a T\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) + b T\left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right) + c T\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\right) + d T\left(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right)
   \]

3. Substitute the values of \( a \), \( b \), \( c \), and \( d \) and the images of the basis matrices:
   \[
   T\left(\begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix}\right) = 1 \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} + (-1) \begin{bmatrix} 0 & 2 \\ 1 & 1 \end{bmatrix} + 4 \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} + 3 \begin{bmatrix} 3 & -1 \\ 1 & 0 \end{bmatrix}
   \]

4. Perform the matrix multiplications and additions:
   \[
   = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -2 \\ -1 & -1 \end{bmatrix} + \begin{bmatrix} 4 & 8 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} 9 & -3 \\ 3 & 0 \end{bmatrix}
   \]

5. Adding these matrices together:
   \[
   = \begin{bmatrix} -1 + 0 + 4 + 9 & 2 - 2 + 8 - 3 \\ 0 - 1 + 0 + 3 & 1 - 1 + 4 + 0 \end{bmatrix}
   \]

   Simplifying, we get:
   \[
   = \begin{bmatrix} 12 & 5 \\ 2 & 4 \end{bmatrix}
   \]

### Final Answer:
\[
T\left(\begin{bmatrix} 1 & -1 \\ 4 & 3 \end{bmatrix}\right) = \begin{bmatrix} 12 & 5 \\ 2 & 4 \end{bmatrix}
\]

Would you like a detailed breakdown of any specific step?

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1. What is a linear transformation?
2. How do you determine a linear transformation using basis matrices?
3. What is the purpose of expressing a matrix as a linear combination of basis matrices?
4. How do you calculate the image of a matrix under a given transformation?
5. Why is matrix addition important in transformations?

**Tip:** When working with transformations, always check if you’re using the correct basis.

Let T be a linear transformation from M_{2,2} into M_{2,2} such that T([[1, 0], [0, 0]]) = [[-1, 2], [0, 1]], T([[0, 1], [0, 0]]) = [[0, 2], [1, 1]], T([[0, 0], [1, 0]]) = [[1, 2], [0, 1]], and T([[0, 0], [0, 1]]) = [[3, -1], [1, 0]]. Find T([[1, -1], [4, 3]]).

Solve for the image of the matrix [[1, -1], [4, 3]] under a linear transformation T from M_{2,2} to M_{2,2}, given its action on basis matrices. This problem demonstrates matrix transformations using linear combinations and basis matrices, suitable for Linear Algebra students.

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