Suppose that $\mathbf{x}(t) \in \mathbb{R}^{3}$ obeys the differential equation

$$\frac{d \mathbf{x}}{d t}=M \mathbf{x}$$

where $M$ is a constant $3 \times 3$ real matrix.

(i) Suppose that $M$ has distinct eigenvalues $\lambda_{1}, \lambda_{2}, \lambda_{3}$ with corresponding eigenvectors $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$. Explain why $\mathbf{x}$ may be expressed in the form $\sum_{i=1}^{3} a_{i}(t) \mathbf{e}_{i}$ and deduce by substitution that the general solution of $(*)$ is

$$\mathbf{x}=\sum_{i=1}^{3} A_{i} e^{\lambda_{i} t} \mathbf{e}_{i}$$

where $A_{1}, A_{2}, A_{3}$ are constants.

(ii) What is the general solution of $(*)$ if $\lambda_{2}=\lambda_{3} \neq \lambda_{1}$, but there are still three linearly independent eigenvectors?

(iii) Suppose again that $\lambda_{2}=\lambda_{3} \neq \lambda_{1}$, but now there are only two linearly independent eigenvectors: $\mathbf{e}_{1}$ corresponding to $\lambda_{1}$ and $\mathbf{e}_{2}$ corresponding to $\lambda_{2}$. Suppose that a vector $\mathbf{v}$ satisfying the equation $\left(M-\lambda_{2} I\right) \mathbf{v}=\mathbf{e}_{2}$ exists, where $I$ denotes the identity matrix. Show that $\mathbf{v}$ is linearly independent of $\mathbf{e}_{1}$ and $\mathbf{e}_{2}$, and hence or otherwise find the general solution of $(*)$.