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Let's solve the given problems by finding the inverses of the linear transformations provided.

### Problem (a)
The linear transformation \( T_1: \mathbb{R}^2 \to \mathbb{R}^2 \) is given by:
\[
T_1(x, y) = (2x + 5y, 6x + 16y)
\]

To find the inverse transformation \( T_1^{-1}(x, y) \), we need to express the variables \( x \) and \( y \) in terms of the output coordinates.

1. Represent the transformation in matrix form:
   \[
   T_1(x, y) = \begin{bmatrix} 2 & 5 \\ 6 & 16 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}
   \]

2. Find the inverse of the matrix \( A = \begin{bmatrix} 2 & 5 \\ 6 & 16 \end{bmatrix} \). 

The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is:
\[
A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]

3. Compute the determinant of \( A \):
\[
\text{det}(A) = (2)(16) - (5)(6) = 32 - 30 = 2
\]

4. The inverse matrix \( A^{-1} \) is:
\[
A^{-1} = \frac{1}{2} \begin{bmatrix} 16 & -5 \\ -6 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -\frac{5}{2} \\ -3 & 1 \end{bmatrix}
\]

5. Thus, the inverse transformation \( T_1^{-1}(x, y) \) is:
\[
T_1^{-1}(x, y) = \left( 8x - \frac{5}{2}y, -3x + y \right)
\]

### Problem (b)
The linear transformation \( T_2: \mathbb{R}^3 \to \mathbb{R}^3 \) is given by:
\[
T_2(x, y, z) = (x + 2z, 2x + y, 2y + z)
\]

To find the inverse transformation \( T_2^{-1}(x, y, z) \), we again need to find the inverse of the corresponding matrix.

1. Represent the transformation in matrix form:
   \[
   T_2(x, y, z) = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}
   \]

2. Find the inverse of matrix \( B = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \).

3. Compute the determinant of \( B \):
\[
\text{det}(B) = 1(1 \cdot 1 - 0 \cdot 2) - 0(2 \cdot 1 - 0 \cdot 0) + 2(2 \cdot 2 - 1 \cdot 0) = 1 - 0 + 8 = 9
\]

4. Find the adjugate of matrix \( B \) and multiply by \( \frac{1}{\text{det}(B)} \) to get \( B^{-1} \).

5. The inverse matrix \( B^{-1} \) is calculated as:
\[
B^{-1} = \frac{1}{9} \begin{bmatrix} 1 & 2 & -4 \\ 2 & 4 & -2 \\ 2 & 1 & 1 \end{bmatrix}
\]

6. Therefore, the inverse transformation \( T_2^{-1}(x, y, z) \) is:
\[
T_2^{-1}(x, y, z) = \left( \frac{1}{9}x + \frac{2}{9}y - \frac{4}{9}z, \frac{2}{9}x + \frac{4}{9}y - \frac{2}{9}z, \frac{2}{9}x + \frac{1}{9}y + \frac{1}{9}z \right)
\]

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

### Related Questions:
1. What is the geometric interpretation of a linear transformation and its inverse?
2. How do you find the inverse of a 3x3 matrix using cofactor expansion?
3. What properties must a matrix have to be invertible?
4. How can linear transformations be used in computer graphics?
5. What is the significance of the determinant in linear transformations?

**Tip:** The determinant of a matrix not only helps in finding its inverse but also provides information about the scaling factor of the transformation.

Find the inverse of two linear transformations: T1: R^2 -> R^2 and T2: R^3 -> R^3.

Step-by-step solution to finding the inverse of the linear transformations T1: R^2 -> R^2 and T2: R^3 -> R^3. This problem involves matrix inversion, determinants, and linear algebra concepts, suitable for undergraduate math students.

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