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Transform the following system of linear equations in a row echelon form, discuss and solve if having solutions

$$ \left\{ \begin{array}{rrrrr} x_1 & -2x_2 & + x_3 & = & 7\\ 2x_1 & -5x_2 & +2x_3 & = & 6\\ 3x_1 & 2x_2 & - x_3 & = & 1\\ \end{array}\right.$$

The system has the following augmented matrix $$(A|b)= \left( \begin{array}{ccc|c} 1 & -2 & 1 & 7 \\ 2 & -5 & 2 & 6 \\ 3 & 2 & -1 & 1 \\ \end{array}\right)$$ on which we perform elementary row transformations $$ \left( \begin{array}{ccc|c} 1 & -2 & 1 & 7 \\ 2 & -5 & 2 & 6 \\ 3 & 2 & -1 & 1 \\ \end{array}\right)\simeq_f \left( \begin{array}{ccc|c} 1 & -2 & 1 & 7 \\ 0 & -1 & 0 & -8 \\ 0 & 8 & -4 & -20 \\ \end{array}\right)\simeq_f \left( \begin{array}{ccc|c} 1 & -2 & 1 & 7 \\ 0 & -1 & 0 & -8 \\ 0 & 0 & -4 & -84 \\ \end{array}\right)\simeq_f$$ $$\simeq_f \left( \begin{array}{ccc|c} 1 & -2 & 1 & 7 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 21 \\ \end{array}\right)\simeq_f \left( \begin{array}{ccc|c} 1 & 0 & 1 & 23 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 21 \\ \end{array}\right)\simeq_f \left( \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 21 \\ \end{array}\right) $$ So our system is equivalent to the following one $$ \left\{ \begin{array}{rrrrr} x_1 & & & = & 2\\ & x_2 & & = & 8\\ & & x_3 & = & 21\\ \end{array}\right.$$ that is determinate compatible with solution $x_1=2$, $x_2=8$ and $x_3=21$. Lastly, let's check that these values indeed constitute a solution $$ \left\{ \begin{array}{rrrrrrr} 2 & -2\cdot 8 & + 21 & = & 2-16+21 & = & 7\\ 2\cdot 2 & -5\cdot 8 & +2\cdot 21 & = & 4-40+42 & = & 6\\ 3\cdot 2 & 2\cdot 8 & - 21 & = & 6+16-21 & = & 1\\ \end{array}\right.$$