Example 2 (to item $2^{\circ}$). Find the rank of the matrix

$$
A=\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
7 & -1 & 2 & 1 & -12 \\
11 & -1 & 3 & 0 & -16 \\
10 & -2 & 3 & 3 & -20
\end{array}\right)
$$
Solution. After subtracting the first row from all the others (from the last one with a factor of 2), we obtain the equivalent matrix

$$
A \sim\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
4 & 0 & 1 & -1 & -4 \\
8 & 0 & 2 & -2 & -8 \\
4 & 0 & 1 & -1 & -4
\end{array}\right) \sim\left(\begin{array}{rrrrr}
3 & -1 & 1 & 2 & -8 \\
4 & 0 & 1 & -1 & -4
\end{array}\right)
$$

Since three rows of the intermediate matrix were proportional, we can obtain two zero rows from them, which we have discarded.

It is clear that $\operatorname{rank} A=2$, because $M_{2}=\left|\begin{array}{rr}3 & -1 \\ 4 & 0\end{array}\right|=4$.