To solve this problem, we need to find the inverse of the composition $T \circ S$ in matrix form, where:

• $T(x, y, z) = (3x - z, 5x - y + z)$
• $S(x, y) = (2x - y, x - 3y, -3x + y)$

Step 1: Represent $T$ and $S$ as matrices

Matrix for $S(x, y)$

The transformation $S(x, y) = (2x - y, x - 3y, -3x + y)$ can be written as a matrix multiplying the vector $\begin{pmatrix} x \\ y \end{pmatrix}$:

$S(x, y) = \begin{pmatrix} 2 & -1 \\ 1 & -3 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

So, the matrix representation of $S$ is:

$S = \begin{pmatrix} 2 & -1 \\ 1 & -3 \\ -3 & 1 \end{pmatrix}$

Matrix for $T(x, y, z)$

The transformation $T(x, y, z) = (3x - z, 5x - y + z)$ can be written as:

$T(x, y, z) = \begin{pmatrix} 3 & 0 & -1 \\ 5 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

So, the matrix representation of $T$ is:

$T = \begin{pmatrix} 3 & 0 & -1 \\ 5 & -1 & 1 \end{pmatrix}$

Step 2: Multiply $T$ and $S$

Now, we compute the composition $T \circ S$, which is the matrix multiplication $T \times S$:

$T \circ S = \begin{pmatrix} 3 & 0 & -1 \\ 5 & -1 & 1 \end{pmatrix} \times \begin{pmatrix} 2 & -1 \\ 1 & -3 \\ -3 & 1 \end{pmatrix}$

Carrying out the matrix multiplication:

$T \circ S = \begin{pmatrix} (3)(2) + (0)(1) + (-1)(-3) & (3)(-1) + (0)(-3) + (-1)(1) \\ (5)(2) + (-1)(1) + (1)(-3) & (5)(-1) + (-1)(-3) + (1)(1) \end{pmatrix}$

Simplifying the elements:

$T \circ S = \begin{pmatrix} 6 + 3 & -3 - 1 \\ 10 - 1 - 3 & -5 + 3 + 1 \end{pmatrix} = \begin{pmatrix} 9 & -4 \\ 6 & -1 \end{pmatrix}$

Step 3: Find the inverse of $T \circ S$

Now, we need to find the inverse of the matrix $\begin{pmatrix} 9 & -4 \\ 6 & -1 \end{pmatrix}$.

The formula for the inverse of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is:

$A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

For the matrix $\begin{pmatrix} 9 & -4 \\ 6 & -1 \end{pmatrix}$:

• $a = 9$, $b = -4$, $c = 6$, $d = -1$
• Determinant $\text{det}(A) = (9)(-1) - (6)(-4) = -9 + 24 = 15$

So, the inverse is:

$A^{-1} = \frac{1}{15} \begin{pmatrix} -1 & 4 \\ -6 & 9 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} \frac{-1}{15} & \frac{4}{15} \\ \frac{-6}{15} & \frac{9}{15} \end{pmatrix} = \begin{pmatrix} \frac{-1}{15} & \frac{4}{15} \\ \frac{-2}{5} & \frac{3}{5} \end{pmatrix}$

Step 4: Compare with the options

This matches option C, which is:

$\begin{pmatrix} \frac{-1}{15} & \frac{4}{15} \\ \frac{-2}{5} & \frac{3}{5} \end{pmatrix}$

Thus, the correct answer is C.

Would you like more details or have any questions? Here are some related questions to extend this:

1. How do you find the inverse of a 3x3 matrix?
2. What is the significance of the determinant when calculating matrix inverses?
3. Can the composition of transformations be interpreted geometrically?
4. What happens if the determinant of a transformation matrix is zero?
5. How do transformations apply in practical fields such as computer graphics?

Tip: Always check if the determinant is non-zero before computing the inverse of a matrix.