We are given two matrices $P = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $T = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$. We are also given that the linear transformation $T \circ P$ maps to the vector $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}$. We need to find the values of $x$ and $y$. Note that in this case, $T \circ P$ refers to the matrix multiplication of $T$ and $P$ in that order.

1. Problem Description

We are given two matrices $P = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $T = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$. We are also given that the linear transformation $T \circ P$ maps to the vector $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}$. We need to find the values of $x$ and $y$. Note that in this case, $T \circ P$ refers to the matrix multiplication of $T$ and $P$ in that order.

2. Solution Steps

The expression $T \circ P$ represents the matrix product $TP$. Since $T = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ is a $2 \times 1$ matrix and $P = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ is also a $2 \times 1$ matrix, the expression should probably be interpreted as $PT$, or $P^T$. Assuming the intended operation is actually $P^T$, the problem is unsolvable. I will assume that $P$ and $T$ are $2\times 2$ matrices.

$P = \begin{pmatrix} 1 & -3 \\ 2 & 1 \end{pmatrix}$ and we are trying to solve for $x$ and $y$ of the transformation

$\begin{pmatrix} x \\ y \end{pmatrix} = T \circ P$, which should be equal to $\begin{pmatrix} 6 \\ -6 \end{pmatrix}$.

Let's assume that $P = \begin{pmatrix} 1 & -3 \\ 2 & 1 \end{pmatrix}$ and let's apply $P$ followed by $T$ which results in the coordinate $(6,-6)$

Since $(x,y) = (6, -6)$, then $x=6$ and $y=-6$. This can be confirmed without knowing the transformation.

3. Final Answer

x = 6

y = -6