A linear recurrence and matrix powers

Matrix form.
$$\begin{pmatrix} u_{n+1} \\ u_n \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} u_n \\ u_{n-1} \end{pmatrix} = A^n \begin{pmatrix} u_1 \\ u_0 \end{pmatrix}.$$

Eigenvalues. $\det(A - \lambda I) = (3-\lambda)(-\lambda) + 2 = \lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2)$, so $\lambda_1 = 1$, $\lambda_2 = 2$.

Eigenvectors. For $\lambda=1$: $v_1 = (1,1)$. For $\lambda=2$: $v_2 = (2,1)$.

Diagonalization.
$$A = P\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}P^{-1}, \qquad P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}, \quad P^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}.$$

Power.
$$A^n = P\begin{pmatrix} 1 & 0 \\ 0 & 2^n \end{pmatrix}P^{-1} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2^n \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}.$$
Compute:
$$A^n = \begin{pmatrix} -1+2^{n+1} & 2-2^{n+1} \\ -1+2^n & 2-2^n \end{pmatrix}.$$

Conclusion. $u_n = (A^n)_{21} u_1 + (A^n)_{22} u_0 = (2^n - 1)\cdot1 + 0 = 2^n - 1$.

Verification. $u_0 = 0$, $u_1 = 1$, $u_2 = 3$, $u_3 = 7$: matches $2^n - 1$.