## Elementary Linear Algebra 7th Edition

The solution to the system is $$x_1=-26\qquad x_2=13\qquad x_3=-7\qquad x_4=4.$$
The system associated to this matrix is \begin{align*} x_1+2x_2\qquad+x_3=&4\\ x_2+2x_3+x_4=&3\\ x_3+2x_4=&1\\ x_4=&4 \end{align*} To solve it follow the steps below: Step 1: We already have the solution for $x_4$. Put it into the third equation to find $x_3$: $$x_3+2\times4=1$$ which gives $$x_3 = -7.$$ Step 2: Put $x_4=4$ and $x_3=-7$ to the second equation and find $x_2$: $$x_2+2\times(-7)+4=3\Rightarrow x_2-14+4=3$$ and this gives $$x_2 = 13.$$ Step 3: Put all of the calculated values into the first equation and find $x_1$: $$x_1+2\times13+4=4$$ and this gives $$x_1=-26.$$