1. The problem involves a square OABC with vertices O(0,0), A(3,0), B(3,3), and C(0,3). 2. We first consider the translation of the square by the vector $\begin{pmatrix}2 \\ 2\end{pmatrix}$. 3. Translation moves every point of the shape by adding the vector components to the coordinates. 4. For example, point O(0,0) translates to O'(0+2,0+2) = (2,2). 5. Since translation moves all points by the same vector, no point on the perimeter remains fixed (invariant). 6. Therefore, the number of invariant points on the perimeter after translation by $\begin{pmatrix}2 \\ 2\end{pmatrix}$ is 0. 7. Next, translate shape A by the vector $\begin{pmatrix}6 \\ -2\end{pmatrix}$ to get shape B. 8. Point A(3,0) translates to A'(3+6,0-2) = (9,-2). 9. Since this translation moves the shape outside the original grid and all points move, no points on the perimeter remain invariant. 10. Hence, the number of invariant points on the perimeter after this translation is also 0. Final answers: - Number of invariant points after translation by $\begin{pmatrix}2 \\ 2\end{pmatrix}$: 0 - Number of invariant points after translation by $\begin{pmatrix}6 \\ -2\end{pmatrix}$: 0