1. **State the problem:** Solve the system of differential equations given by $$\mathbf{x}' = A\mathbf{x}$$ where $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$ and $$A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$$. 2. **Find eigenvalues:** Solve the characteristic equation $$\det(A - \lambda I) = 0$$. $$\det\left(\begin{pmatrix} 1-\lambda & 0 \\ 2 & 1-\lambda \end{pmatrix}\right) = (1-\lambda)(1-\lambda) - 0 = (1-\lambda)^2 = 0$$ So, the eigenvalue is $$\lambda = 1$$ with algebraic multiplicity 2. 3. **Find eigenvectors:** Solve $$(A - I)\mathbf{v} = 0$$. $$\left(\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\right) \mathbf{v} = \begin{pmatrix} 0 & 0 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ This gives the system: $$0 \cdot v_1 + 0 \cdot v_2 = 0$$ $$2 v_1 + 0 \cdot v_2 = 0 \implies 2 v_1 = 0 \implies v_1 = 0$$ So eigenvectors satisfy $$v_1 = 0$$ and $$v_2$$ free. Choose $$v_2 = 1$$, eigenvector is $$\mathbf{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$. 4. **Find generalized eigenvector:** Since algebraic multiplicity is 2 but only one eigenvector, find $$\mathbf{w}$$ such that $$(A - I)\mathbf{w} = \mathbf{v}$$ Solve: $$\begin{pmatrix} 0 & 0 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ This gives: $$0 \cdot w_1 + 0 \cdot w_2 = 0$$ $$2 w_1 + 0 \cdot w_2 = 1 \implies 2 w_1 = 1 \implies w_1 = \frac{1}{2}$$ $$w_2$$ is free, choose $$w_2 = 0$$. So generalized eigenvector is $$\mathbf{w} = \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}$$. 5. **Write general solution:** For repeated eigenvalue $$\lambda = 1$$ with eigenvector $$\mathbf{v}$$ and generalized eigenvector $$\mathbf{w}$$, solution is $$\mathbf{x}(t) = c_1 e^{t} \mathbf{v} + c_2 e^{t} (t \mathbf{v} + \mathbf{w})$$ Substitute vectors: $$\mathbf{x}(t) = c_1 e^{t} \begin{pmatrix} 0 \\ 1 \end{pmatrix} + c_2 e^{t} \left(t \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\right) = e^{t} \begin{pmatrix} \frac{c_2}{2} \\ c_1 + c_2 t \end{pmatrix}$$ **Final answer:** $$\boxed{\mathbf{x}(t) = e^{t} \begin{pmatrix} \frac{c_2}{2} \\ c_1 + c_2 t \end{pmatrix}}$$ This is the complete solution to the system.