For an $n \times n$ matrix $A$, define the matrix exponential by

$$\exp (A)=\sum_{m=0}^{\infty} \frac{A^{m}}{m !}$$

where $A^{0} \equiv I$, with $I$ being the $n \times n$ identity matrix. [You may assume that $\exp ((s+t) A)=\exp (s A) \exp (t A)$ for real numbers $s, t$ and you do not need to consider issues of convergence.] Show that

$$\frac{\mathrm{d}}{\mathrm{d} t} \exp (t A)=A \exp (t A)$$

Deduce that the unique solution to the initial value problem

$$\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} t}=A \mathbf{y}, \quad \mathbf{y}(0)=\mathbf{y}_{0}, \quad \text { where } \mathbf{y}(t)=\left(\begin{array}{c} y_{1}(t) \\ \vdots \\ y_{n}(t) \end{array}\right)$$

is $\mathbf{y}(t)=\exp (t A) \mathbf{y}_{0}$.

Let $\mathbf{x}=\mathbf{x}(t)$ and $\mathbf{f}=\mathbf{f}(t)$ be vectors of length $n$ and $A$ a real $n \times n$ matrix. By considering a suitable integrating factor, show that the unique solution to

$$\frac{\mathrm{d} \mathbf{x}}{\mathrm{d} t}-A \mathbf{x}=\mathbf{f}, \quad \mathbf{x}(0)=\mathbf{x}_{0}$$

is given by

$$\mathbf{x}(t)=\exp (t A) \mathbf{x}_{0}+\int_{0}^{t} \exp [(t-s) A] \mathbf{f}(s) \mathrm{d} s$$

Hence, or otherwise, solve the system of differential equations $(*)$ when

$$A=\left(\begin{array}{ccc} 2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3 \end{array}\right), \quad \mathbf{f}(t)=\left(\begin{array}{c} \sin t \\ 3 \sin t \\ 0 \end{array}\right), \quad \mathbf{x}_{0}=\left(\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right)$$

[Hint: Compute $A^{2}$ and show that $\left.A^{3}=0 .\right]$