## Elementary Linear Algebra 7th Edition

This system has infinitely many solutions and they are determined by $$x_1 = 13-4t;\quad x_2 = \frac{15}{2}(3-t);\quad x_3=t.$$ where $t$ can be any real number.
We will transform this system to Row-Echelon form by following the steps bellow. Step 1: Replace the 1st and the 3rd equations and rewrite them so that the terms appear in order $x_2, x_1$ and $x_3$: \begin{align*} -2x_2+2x_1-7x_3=&-19\\ -2x_2+4x_1+x_3=&7\\ x_1+4x_3=&13 \end{align*} Step 2: Subtract the first equation term by term from the second to eliminate $x_2$ which will be step by step: $$-2x_2-(-2x_2)+4x_1-2x_1+x_3-(-7x_3) = 7-(-19)$$ which is then $$2x_1+8x_3=26.$$ To make it simpler we divide it by $2$: $$x_1+4x_3=13.$$ Our system now reads: \begin{align*} -2x_2+2x_1-7x_3 =&-19\\ x_1+4x_3=&13\\ x_1+4x_3=&13 \end{align*} Step 3: We see that the last two equations say exactly the same so we actually have two equations with three unknowns. That means that we have infinitely many solutions. To find them we will equate one of the unknowns, let's say $x_3$ with real parameter $t$ and then express the other two in terms of $t$: \begin{align*} -2x_2+2x_1-7x_3 =&-19\\ x_1+4x_3=&13\\ x_3=&t \end{align*} Now substitute $x_3$ for $t$ to the second equation and express $x_1$ in terms of $t$: $$x_1+4t=13\Rightarrow x_1 = 13-4t.$$ Finally substitute $x_3$ for $t$ and $x_1$ for $13-4t$ in the first equation to express $x_2$ in terms of $t$: $$-2x_2+2(13-4t)-7t=-19\Rightarrow -2x_2-15t + 26 = -19$$ and finally $$x_2 = \frac{1}{2}(45-15 t) = \frac{15}{2}(3-t)$$ Step 4: Just rewriting the solutions we have that they are determined by $$x_1 = 13-4t;\quad x_2 = \frac{15}{2}(3-t);\quad x_3=t.$$