Equivalent systems

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 $$\begin{pmatrix} 2 & 3 & -1 & 1 & -2 & | & 1 \\ 1 & -3 & 2 & 1 & 1 & | & -1 \\ -2 & 1 & -2 & -1 & 2 & | & 3 \end{pmatrix} \Rightarrow col2 \leftrightarrow col3 \Rightarrow \begin{pmatrix} 2 & -1 & 3 & 1 & -2 & | & 1 \\ 1 & 2 & -3 & 1 & 1 & | & -1 \\ -2 & -2 & 1 & -1 & 2 & | & 3 \end{pmatrix} $$