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The problem describes a linear transformation \( T : \mathbb{R}^2 \to \mathbb{R}^2 \) and provides two specific mappings of vectors:

\[
T \left( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 2 \\ 7 \end{bmatrix} \quad \text{and} \quad T \left( \begin{bmatrix} 1 \\ -2 \end{bmatrix} \right) = \begin{bmatrix} -1 \\ -5 \end{bmatrix}.
\]

We are asked to find \( T \left( \begin{bmatrix} 0 \\ 9 \end{bmatrix} \right) \).

### Solution Outline:

Since the transformation is linear, \( T(\mathbf{v}) = A\mathbf{v} \), where \( A \) is a \( 2 \times 2 \) matrix. To find \( A \), use the given vectors:

1. Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \).
   
2. From the first equation, \( T \left( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 2 \\ 7 \end{bmatrix} \), we get:
   \[
   A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix} = \begin{bmatrix} 2 \\ 7 \end{bmatrix}.
   \]
   This gives us the system:
   \[
   a + b = 2 \quad \text{and} \quad c + d = 7.
   \]

3. From the second equation, \( T \left( \begin{bmatrix} 1 \\ -2 \end{bmatrix} \right) = \begin{bmatrix} -1 \\ -5 \end{bmatrix} \), we get:
   \[
   A \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} a - 2b \\ c - 2d \end{bmatrix} = \begin{bmatrix} -1 \\ -5 \end{bmatrix}.
   \]
   This gives us another system:
   \[
   a - 2b = -1 \quad \text{and} \quad c - 2d = -5.
   \]

### Solving the system:

- From \( a + b = 2 \) and \( a - 2b = -1 \), solve for \( a \) and \( b \):
  1. Add the two equations:
     \[
     (a + b) + (a - 2b) = 2 + (-1) \quad \Rightarrow \quad 2a - b = 1.
     \]
  2. Subtract the second from the first:
     \[
     3b = 3 \quad \Rightarrow \quad b = 1.
     \]
  3. Plug \( b = 1 \) into \( a + b = 2 \):
     \[
     a + 1 = 2 \quad \Rightarrow \quad a = 1.
     \]

- From \( c + d = 7 \) and \( c - 2d = -5 \), solve for \( c \) and \( d \):
  1. Add the two equations:
     \[
     (c + d) + (c - 2d) = 7 + (-5) \quad \Rightarrow \quad 2c - d = 2.
     \]
  2. Subtract the second from the first:
     \[
     3d = 12 \quad \Rightarrow \quad d = 4.
     \]
  3. Plug \( d = 4 \) into \( c + d = 7 \):
     \[
     c + 4 = 7 \quad \Rightarrow \quad c = 3.
     \]

Thus, the transformation matrix \( A \) is:
\[
A = \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}.
\]

### Finding \( T \left( \begin{bmatrix} 0 \\ 9 \end{bmatrix} \right) \):

Now that we have \( A \), we can calculate \( T \left( \begin{bmatrix} 0 \\ 9 \end{bmatrix} \right) \):

\[
T \left( \begin{bmatrix} 0 \\ 9 \end{bmatrix} \right) = A \begin{bmatrix} 0 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 9 \end{bmatrix} = \begin{bmatrix} 1(0) + 1(9) \\ 3(0) + 4(9) \end{bmatrix} = \begin{bmatrix} 9 \\ 36 \end{bmatrix}.
\]

Thus, \( T \left( \begin{bmatrix} 0 \\ 9 \end{bmatrix} \right) = \begin{bmatrix} 9 \\ 36 \end{bmatrix} \).

Would you like more details on this, or do you have any questions?

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Here are 5 related questions to explore:

1. How can you confirm that a matrix represents a linear transformation?
2. What other properties can we deduce from knowing the matrix of a linear transformation?
3. How do eigenvalues relate to linear transformations in two dimensions?
4. Can linear transformations always be represented by matrices?
5. How would this problem change if \( T \) were a transformation in \( \mathbb{R}^3 \)?

**Tip:** When solving systems of linear equations, combining equations (addition or subtraction) is often the quickest way to isolate variables.

If T: R^2 -> R^2 is a linear transformation such that T([1,1]) = [2,7] and T([1,-2]) = [-1,-5], find T([0,9]).

Learn how to find the matrix for a linear transformation T: R^2 -> R^2 given specific mappings. This problem involves solving systems of equations and matrix multiplication. Suitable for undergraduate-level linear algebra students.

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