## Elementary Linear Algebra 7th Edition

This system has infinitely many solutions and they are determined by $$x_1 = 6-13t;\quad x_2 = 2-4t;\quad x_3 = t$$ where $t$ can be any real number.
To transform this system to Row-Echelon form we will follow the steps below: Step 1: To eliminate $x_1$ from the second equation subtract from it the first one multiplied by $3$ term by term: $$3x_1-3x_1+2x_2-3\times(-2x_2)-x_3-3\times5x_3 = -2-3\times2$$ which becomes $$-4x_2-16x_3 = -8.$$ We can simplify it dividing by $-4$: $$x_2 + 4x_3 = 2.$$ Now that we successfully eliminated $x_1$ our system reads \begin{align*} x_1-2x_2+5x_3 = &2\\ x_2+4x_3=&2 \end{align*} Step 2: Notice that we have only two equations with three unknowns. This means that we will have infinitely many solutions. To find them equate one of the unknowns with $t$ - a parameter that can take any real value. Let it be $x_3$: \begin{align*} x_1-2x_2+5x_3= &2 x_2+4x_3 =&2 x_3= &t \end{align*} Step 3: We will now express $x_2$ and $x_1$ in terms of $t$. Firstly put $t$ instead of $x_3$ in the second equation $$x_2+4t= 2\Rightarrow x_2 = 2-4t.$$ Now put $t$ instead of $x_3$ and $2-4t$ instead of $x_2$ in the first equation: $$x_1 - 2(2-4t) + 5t =2\Rightarrow x_1 - 4 + 8t +5t = 2\Rightarrow x_1 = 6-13t$$ Step 4: We can now rewrite the solution of this system: $$x_1 = 6-13t;\quad x_2 = 2-4t;\quad x_3 = t.$$