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?

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.