## Elementary Linear Algebra 7th Edition

The solution is $$x_1=1\qquad x_2=1\qquad x_3=0$$
The system associated with this matrix is \begin{align*} 2x_1+x_2-x_3=&3\\ x_1-x_2+x_3=&0\\ x_2+2x_3=&1 \end{align*} To solve it follow the steps below: Step 1: Subtract from the first equation the second one multiplied by $2$: $$2x_1-2x_1+x_2-2(-x_2)-x_3-2x_3=3-2\times 0$$ which becomes $$3x_2-3x_3=3.$$ Divide this by $3$ for better looks $$x_2-x_3=1.$$ Now we have a system \begin{align*} x_2-x_3=&1\\ x_1-x_2+x_3=&0\\ x_2+2x_3=&1. \end{align*} Step 2: Subtract from the third equation the first one: $$x_2-x_2+2x_3-(-x_3) = 1-1$$ which becomes $$3x_3=0$$ and this gives $$x_3 =0$$ Step 3: Put this into the first equation and find $x_1$: $$x_2-0=1$$ which gives $$x_2=1.$$ Step 4: Put $x_3=0$ and $x_2=1$ into the second equation to find $x_1$: $$x_1-1+0=0$$ and this gives $$x_1 =1.$$