Let $$ G=\left\{ \begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} \text{with $\bar{a}$ and $\bar{c}$ in $\mathbb{F}^{*}_{7}$ and $\bar{b}$ in $\mathbb{F}_{7}$}\right\}. $$
I have to do two things. Determine the elements of $G$ which have order $2$ and compute the centre of $Z(G)$ of $G$.
First the elements with order $2$. I think I need to find elements such that the matrix times itself equals the matrix with identity elements as coefficients. That is we need $$\begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} \cdot \begin{bmatrix} \bar{a} & \bar{b} \\ \bar{0} & \bar{c} \end{bmatrix} = \begin{bmatrix} \bar{a}\bar{a} & \bar{a}\bar{b}+\bar{b}\bar{c} \\ \bar{0} & \bar{c}\bar{c} \end{bmatrix}= \begin{bmatrix} \bar{1} & \bar{0} \\ \bar{0} & \bar{1} \end{bmatrix}$$
Thus $\bar{a}\bar{a}=\bar{c}\bar{c}=1$ which is only the case for $\bar{a},\bar{c}=\{\bar{1},\bar{6}\}$. This leaves us with four possibilities. Namely, $(\bar{a}=1, \bar{c}=1)$,$(\bar{a}=1, \bar{c}=6)$,$(\bar{a}=6, \bar{c}=1)$ and $(\bar{a}=6, \bar{c}=6)$. Now for $\bar{b}$ this implies $\bar{a}\bar{b}+\bar{b}\bar{c}=\bar{b}(\bar{a}+\bar{c})=\bar{b}=\bar{0}$ for all $b \in \mathbb{F}_{7}$.
Now about the centre $Z(G)$. I know that for the centre subgroup the following holds: $Z(G)=\{z \in Z \mid zgz^{-1}=g \text{ for all } g \in G\}.$ I have absolutely no idea how to proceed so any suggestions are welcome.