Write a system of linear equations with 5 unknowns and 3 equations, and using the 5 given rules separately create equivalent systems.
Development:
First of all we write a system of $$5$$ unknowns and only $$3$$ equations: $$$\left\{ \begin{array} {rcl} 2x+3y-z+u-2v & = & 1 \\ x-3y+2z-8+2v &=& -1 \\ -2x+y-2z-u+2v &=& 3 \end{array}\right.$$$
or we can also write it in the matricial form: $$$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix}$$$
1) It is unimportant: $$$2x+3y-z+u-2v=1$$$ $$$(2x+3y-z+u-2v)+\fbox{$$3x-2$$}=1+\fbox{$$3x-2$$}$$$
2)We multiply a line by a number other than zero: $$$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} \Rightarrow 5\cdot row1 \Rightarrow \begin{pmatrix} 10 & 15 & -5 & 5 & -10 & | & 5 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} $$$ and we obtain an equivalent system
3) Now we add up row2 to row1 to obtain an equivalent system: $$$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} \Rightarrow row1+row2 \Rightarrow \begin{pmatrix} 3 & 0 & -1 & 2 & -1 & | & 0 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} $$$
4) Now we add a linear combination of the lines 2 and 3 to row1 to obtain an equivalent system: $$$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} \Rightarrow row1-2\cdot row2+row3 \Rightarrow \begin{pmatrix} -2 & 10 & -7 & -2 & -2 & | & 6 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} $$$
5) Finally a simple change of order also gives us an equivalent system $$$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} \Rightarrow col2 \leftrightarrow col3 \Rightarrow \begin{pmatrix} 2 & -1 & 3 & 1 & -2 & | & 1 \\ 1 & 2 & -3 & 1 & 1 & | & -1 \\ -2 & -2 & 1 & -1 & 2 & | & 3 \end{pmatrix} $$$
Solution:
1) Trivial
2) $$\begin{pmatrix} 10 & 15 & -5 & 5 & -10 & | & 5 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix}$$
3) $$\begin{pmatrix} 3 & 0 & -1 & 2 & -1 & | & 0 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} $$
4) $$\begin{pmatrix} -2 & 10 & -7 & -2 & -2 & | & 6 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix}$$
5) $$\begin{pmatrix} 2 & -1 & 3 & 1 & -2 & | & 1 \\ 1 & 2 & -3 & 1 & 1 & | & -1 \\ -2 & -2 & 1 & -1 & 2 & | & 3 \end{pmatrix} $$