1 problem found
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: \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