Problem source

The matrix $\mathbf{F}$ is defined by 
\[
\mathbf{F}=\mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n},
\]
where $\mathbf{A}=\begin{pmatrix}-3 & -1\\
8 & 3
\end{pmatrix} $  and  $ t $ is a variable scalar. Evaluate $\mathbf{A}^{2}$, and show
that 
\[
\mathbf{F}=\mathbf{I}\cosh t+\mathbf{A}\sinh t.
\]
Show also that $\mathbf{F}^{-1}=\mathbf{I}\cosh t-\mathbf{A}\sinh t$, 
and that $\dfrac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}=\mathbf{FA}$.
The vector $\mathbf{r}=\begin{pmatrix}x(t)\\
y(t)
\end{pmatrix}$ satisfies the differential equation 
\[
\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}=\mathbf{0},
\]
with $x=\alpha$ and $y=\beta$ at $t=0.$ Solve this equation by means of a suitable matrix integrating factor, and hence show that
\begin{alignat*}{1}
x(t) & =\alpha\cosh t+(3\alpha+\beta)\sinh t\\
y(t) & =\beta\cosh t-(8\alpha+3\beta)\sinh t.
\end{alignat*}

Solution source

\begin{align*}
\begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}^2 &= \begin{pmatrix} 9-8 & 3-3 \\ -24+24 & -8+9 \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\
&= \mathbf{I}
\end{align*}

Therefore:

\begin{align*}
\mathbf{F} &= \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n} \\
&= \mathbf{I} + \sum_{n=1}^{\infty} \frac{1}{(2n)!}t^{2n} \mathbf{I} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}t^{2n+1} \mathbf{A} \\
&= \cosh t \mathbf{I} + \sinh t \mathbf{A}
\end{align*}

Notice that 
\begin{align*}
\mathbf{F} (\mathbf{I}\cosh t-\mathbf{A}\sinh t) &= (\mathbf{I}\cosh t+\mathbf{A}\sinh t)(\mathbf{I}\cosh t-\mathbf{A}\sinh t) \\
&= \mathbf{I}^2 \cosh^2 t+\mathbf{A}(\sinh t \cosh t - \cosh t \sinh t) - \mathbf{A}^2\sinh^2 t \\
&= \mathbf{I} \cosh^2 t - \mathbf{I} \sinh^2 t \\
&= \mathbf{I}
\end{align*}

Therefore $\mathbf{F}^{-1} = \mathbf{I}\cosh t-\mathbf{A}\sinh t$

\begin{align*}
\frac{\d \mathbf{F}}{\d t} &= \frac{\d }{\d t} \left [ \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}\right] \\
&= \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^n \\
&= \left ( \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^{n-1} \right) \mathbf{A} \\
&= \mathbf{F}\mathbf{A}
\end{align*}

\begin{align*}
&& \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}&=\mathbf{0} \\
\Rightarrow && \mathbf{F} \frac{\d \mathbf{r}}{\d t} + \mathbf{FAr} &= \mathbf{0} \\
&& \frac{\d }{\d t} \left ( \mathbf{F} \mathbf{r}\right) &= 0 \\
\Rightarrow && \mathbf{Fr} &= \mathbf{c} \\
\Rightarrow && \mathbf{r} &= \mathbf{F}^{-1}\mathbf{c} \\
&&&= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\mathbf{c} \\
t = 0: && \binom{\alpha}{\beta} &= \mathbf{c} \\
\Rightarrow && \mathbf{r} &= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\binom{\alpha}{\beta} \\
&&&= \binom{\alpha \cosh t}{\beta \cosh t} -  \binom{-3\alpha-\beta}{8\alpha + 3\beta}\sinh t \\
&&&= \binom{\alpha \cosh t + (3\alpha + \beta) \sinh t}{\beta \cosh t -(8\alpha + 3\beta)\sinh t}
\end{align*}

as required