Matrix exercise asked by Mohamed Bella in Ol_Math group

Exercise:

Consider a matrix $A=\left[ \begin{matrix}
   0 & -r & 0  \\
   r & 0 & 0  \\
   0 & 0 & 0  \\
\end{matrix} \right]$ for every $r\in \mathbb{R}$

Show that ${{e}^{A}}=\left[ \begin{matrix}
   \cos r & -\sin r & 0  \\
   \sin r & \cos r & 0  \\
   0 & 0 & 1  \\
\end{matrix} \right]$

Solution: let’s compute, $\left( \lambda I-A \right)=\left( \begin{matrix}
   \lambda  & 0 & 0  \\
   0 & \lambda  & 0  \\
   0 & 0 & \lambda   \\
\end{matrix} \right)-\left( \begin{matrix}
   0 & -r & 0  \\
   r & 0 & 0  \\
   0 & 0 & 0  \\
\end{matrix} \right)=\left[ \begin{matrix}
   \lambda  & r & 0  \\
   -r & \lambda  & 0  \\
   0 & 0 & \lambda   \\
\end{matrix} \right]$

but ${{\left( \lambda I-A \right)}^{-1}}=\frac{1}{\det \left( \lambda I-A \right)}adj\left( \lambda I-A \right)$

Observe that, $\left| \lambda I-A \right|=\lambda \left( {{\lambda }^{2}} \right)-r\left( -r\lambda  \right)={{\lambda }^{3}}+{{r}^{2}}\lambda =\lambda \left( {{\lambda }^{2}}+{{r}^{2}} \right)$

Let’s compute the minor matrix for $\lambda I-A$ as follows

\[\left[ \begin{matrix}
   \left| \begin{matrix}
   \lambda  & 0  \\
   0 & \lambda   \\
\end{matrix} \right| & \left| \begin{matrix}
   -r & 0  \\
   0 & \lambda   \\
\end{matrix} \right| & \left| \begin{matrix}
   -r & \lambda   \\
   0 & 0  \\
\end{matrix} \right|  \\
   \left| \begin{matrix}
   r & 0  \\
   0 & \lambda   \\
\end{matrix} \right| & \left| \begin{matrix}
   \lambda  & 0  \\
   0 & \lambda   \\
\end{matrix} \right| & \left| \begin{matrix}
   \lambda  & r  \\
   0 & 0  \\
\end{matrix} \right|  \\
   \left| \begin{matrix}
   r & 0  \\
   \lambda  & 0  \\
\end{matrix} \right| & \left| \begin{matrix}
   \lambda  & 0  \\
   -r & 0  \\
\end{matrix} \right| & \left| \begin{matrix}
   \lambda  & r  \\
   -r & \lambda   \\
\end{matrix} \right|  \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{\lambda }^{2}} & -r\lambda  & 0  \\
   r\lambda  & {{\lambda }^{2}} & 0  \\
   0 & 0 & {{\lambda }^{2}}+{{r}^{2}}  \\
\end{matrix} \right]\]

So the cofactor matrix is $\left[ \begin{matrix}
   {{\lambda }^{2}} & r\lambda  & 0  \\
   -r\lambda  & {{\lambda }^{2}} & 0  \\
   0 & 0 & {{\lambda }^{2}}+{{r}^{2}}  \\
\end{matrix} \right]$

so $adj\left( \lambda I-A \right)=\left[ \begin{matrix}
   {{\lambda }^{2}} & -r\lambda  & 0  \\
   r\lambda  & {{\lambda }^{2}} & 0  \\
   0 & 0 & {{\lambda }^{2}}+{{r}^{2}}  \\
\end{matrix} \right]$ Thus ${{\left( \lambda I-A \right)}^{-1}}=\left[ \begin{matrix}
   \frac{\lambda }{{{\lambda }^{2}}+{{r}^{2}}} & \frac{-r}{{{\lambda }^{2}}+{{r}^{2}}} & 0  \\
   \frac{r}{{{\lambda }^{2}}+{{r}^{2}}} & \frac{\lambda }{{{\lambda }^{2}}+{{r}^{2}}} & 0  \\
   0 & 0 & \frac{1}{\lambda }  \\
\end{matrix} \right]$

So ${{e}^{At}}={{\mathcal{L}}^{-1}}\{{{\left( \lambda I-A \right)}^{-1}}\}=\left[ \begin{matrix}
   {{\mathcal{L}}^{-1}}\{\frac{\lambda }{{{\lambda }^{2}}+{{r}^{2}}}\} & {{\mathcal{L}}^{-1}}\{\frac{-r}{{{\lambda }^{2}}+{{r}^{2}}}\} & 0  \\
   {{\mathcal{L}}^{-1}}\{\frac{r}{{{\lambda }^{2}}+{{r}^{2}}}\} & {{\mathcal{L}}^{-1}}\{\frac{\lambda }{{{\lambda }^{2}}+{{r}^{2}}}\} & 0  \\
   0 & 0 & {{\mathcal{L}}^{-1}}\{\frac{1}{\lambda }\}  \\
\end{matrix} \right]=\left[ \begin{matrix}
   \cos r & -\sin r & 0  \\
   \sin r & \cos r & 0  \\
   0 & 0 & 1  \\
\end{matrix} \right]$