1. **State the problem:** We are given a matrix $$A = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}$$ and need to find the sum $$S = I + A + A^2 + A^3 + \cdots + A^{11} + A^{12}$$ where $$I$$ is the identity matrix. 2. **Recognize the sum as a geometric series:** The sum of powers of a matrix $$A$$ from 0 to $$n$$ is a geometric series: $$ S = \sum_{k=0}^{12} A^k = I + A + A^2 + \cdots + A^{12}. $$ 3. **Formula for sum of geometric series of matrices:** If $$A - I$$ is invertible, then $$ S = (A^{13} - I)(A - I)^{-1}. $$ 4. **Check invertibility of $$A - I$$:** $$ A - I = \begin{pmatrix} 2-1 & -1 \\ 3 & -2-1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix}. $$ The determinant is $$1 \times (-3) - (-1) \times 3 = -3 + 3 = 0$$, so $$A - I$$ is not invertible. 5. **Since $$A - I$$ is not invertible, the formula $$S = (A^{13} - I)(A - I)^{-1}$$ does not apply.** 6. **Try to find a pattern or use the Cayley-Hamilton theorem:** The characteristic polynomial of $$A$$ is $$ \det(\lambda I - A) = \det\begin{pmatrix} \lambda - 2 & 1 \\ -3 & \lambda + 2 \end{pmatrix} = (\lambda - 2)(\lambda + 2) - (-3)(1) = \lambda^2 - 4 + 3 = \lambda^2 - 1. $$ So the characteristic polynomial is $$\lambda^2 - 1 = 0$$. 7. **By Cayley-Hamilton theorem, $$A$$ satisfies its characteristic polynomial:** $$ A^2 - I = 0 \implies A^2 = I. $$ 8. **Use this to simplify powers of $$A$$:** $$ A^0 = I, \quad A^1 = A, \quad A^2 = I, \quad A^3 = A, \quad A^4 = I, \quad \ldots $$ So powers of $$A$$ alternate between $$I$$ and $$A$$: - Even powers: $$A^{2k} = I$$ - Odd powers: $$A^{2k+1} = A$$ 9. **Sum the series:** There are 13 terms from $$k=0$$ to $$k=12$$. - Number of even powers: $$7$$ (0,2,4,6,8,10,12) - Number of odd powers: $$6$$ (1,3,5,7,9,11) Sum: $$ S = 7I + 6A. $$ 10. **Answer:** The sum is $$7I + 6A$$, which corresponds to option 2. **Final answer:** $$\boxed{7I + 6A}$$