Given the Lie group that consist from matrices like: \begin{bmatrix}1&x&z+1/2xy\\0&1&y\\0&0&1\end{bmatrix}
Find algebra Lie.
Using the next formula $\mathfrak{g}=T_I G=\left\{A^{\prime}\left(t_0\right) \in \operatorname{Mat}(n, \mathbb{R}) \mid A \in C^{\infty}((\alpha, \beta), G): A\left(t_0\right)=I\right\}$
Is next answer correct?
\begin{bmatrix}0&a&c+1/2ab\\0&0&b\\0&0&0\end{bmatrix} where a,b,c from R.
The description of $G$ is a little strange -- if $$ G = \{A \in \text{Mat}(3,\mathbb R): A=\left(\begin{array}{ccc} 1 & x & z+\frac{1}{2}xy \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array}\right), \quad x,y,z \in \mathbb R\}, $$ Then if $a=x, b=y,c= z+\frac{1}{2}xy$, then $a,b,c$ are arbitrary real numbers since $(x,y,z) \mapsto (x,y,z+\frac{1}{2}xy)$ has inverse $(a,b,c) \mapsto (a,b,c-\frac{1}{2}ab)$. But then $G=I_3+N$ where $$ N = \{A\in \text{Mat}(3,\mathbb R): A=\left(\begin{array}{ccc} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{array}\right), \quad a,b,c \in \mathbb R\}, $$ That is, if $E_{ij}$ denotes the elementary matrix with all entries $0$ except the one in the $i$-th row and $j$-th column, which is equal to $1$, then $N$ is $\text{span}\{E_{12},E_{13},E_{23}\}$.
In this case we have $\mathfrak{g} = \text{Lie}(G)=N$. Can you see how to show this using the definition you quote?