Elementary Linear Algebra 7th Edition

$x_{1}=1$ $x_{2}=1$ $x_{3}=1$ $x_{5}=1$
$x_{1}+3x_{4}=4 \rightarrow x_{1}=4-3x_{4}$ $2x_{2}-x_{3}-x_{4}=0$ $3x_{2}-2x_{4}=1$ $2x_{1}-x_{2}+4x_{3}=5$ Substitute $x_{1}$ from the first equation to the fourth $2(4-3x_{4})-x_{2}+4x_{3}=5$ Simplify and express $x_{2}$ from the equation above. $x_{2}=3-6x_{4}+4x_{3}$ Substitute $x_{2}$ to the second and the third original equation. $2(3-6x_{4}+4x_{3})-x_{3}-x_{4}=0$ $3(3-6x_{4}+4x_{3})-2x_{4}=1$ Simplify. $-13x_{4}+7x_{3}+6=0$ $-20x_{4}+12x_{3}+9=1$ Express $x_{3}$ from the first equation and substitute it to the second. $x_{3}=\frac{13x_{4}-6}{7}$ $-20x_{4}+12(\frac{13x_{4}-6}{7}) +9=1 \rightarrow$ Simplify. $x_{4}=1$ Use the value of $x_{4}$ to calculate $x_{3}$, $x_{2}$ and $x_{1}$. $x_{3}=\frac{13x_{4}-6}{7}=1$ $x_{2}=3-6x_{4}+4x_{3}=1$ $x_{1}=4-3x_{4}=1$