What is the minimum generating set of a matrix group? Any finite group $G$ can be generated $G=\langle A\rangle$ by a finite set $A$. A minimal generating set for $G$ is a generating set $A$ of minimal size.
For example, any cyclic group has a minimal generating set of size 1; and a minimal generating set for the dihedral group $D_{2n}$ is $\{r,s\}$.
What would be the minimal generating set for the group $\mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$? 
 A: It is difficult to know how to answer this question most helpfully without knowing more about your background, and where this problem comes from.
In fact ${\rm GL}(3,3)$ is generated by the two matrices
$$ a = \left(\begin{array}{ccc}1&2&0\\0&1&0\\0&0&2\end{array}\right),\ \ \ \ \ \
b = \left(\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right).$$
Note first that $a \not\in {\rm SL}(3,3)$, so it is enough to prove that ${\rm SL}(3,3) = \langle a^2,b\rangle$ and that $ a^2 = \left(\begin{array}{ccc}1&1&0\\0&1&0\\0&0&1\end{array}\right)$ is a so-called transvection. Now $b$ is a permutation matrix, and it is easily seen that
$$b^{-1}a^2b= \left(\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right),\ \ \ 
[a^2,b^{-1}a^2b]= \left(\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right),\ \ \ b^{-1}[a^2,b^{-1}a^2b]b= \left(\begin{array}{ccc}1&0&0\\1&1&0\\0&0&1\end{array}\right),$$
and by further conjugations by $b$ we get all of the upper and lower unitriangular matrices that represent transvections.
Showing that ${\rm SL}(3,3)$ is generated by these transvection matrices is just basic linear algebra, and is essntially equivalent to row and column reducing an invertible matrix. (You can make life easier by using the diagonal matrix $a^3$ and conjugates under $b$ to handle diagonal entries equal to $-1$.)
