Have you ever made the limit of a sequence of matrices?

$\boxed{1}$ Tricky solution.- $\displaystyle \text{ det (A) = 1/2} \Rightarrow \text{ det (} \lim_{n \to \infty} A^n ) = 0 \Rightarrow \lim_{n \to \infty} A^n = \boxed{4} \text{ or } \boxed{6}$ . Due to you have 3 tries you'll get it at most in 2 tries.

$\boxed{2}$

If you want to study a square matrix, you have to start knowing its charasteristic polynomial, Jordan matrix and its minimal polynomial. The charasterisc polynomial of $A$ is $p_{A} (x) = (x - 1)(x - 1/2) \Rightarrow A = SDS^{- 1}$ , with $D$ being the diagonal matrix $D = \begin{pmatrix} 1 & 0 \\ 0 & 1/2 \end{pmatrix} \Rightarrow A^n = SD^{n} S^{- 1}$ $\text{ Ker (A - I) =} \langle (1, 3/2) \rangle ,\quad \quad \text{ and} \quad \text{ Ker ( A - (1/2)I) = } \langle (1, 2) \rangle \Rightarrow \quad A =\begin{pmatrix} 1 & 1 \\ 3/2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1/2 \end{pmatrix} \cdot \begin{pmatrix} 4 & -2 \\ -3 & 2 \end{pmatrix} \Rightarrow$ $\displaystyle \lim_{n \to \infty} A^n = \begin{pmatrix} 1 & 1 \\ 3/2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 4 & -2 \\ -3 & 2 \end{pmatrix} = \boxed{4}$