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Solve the following system using Gauss Elimination

$z+4 x-y=7$
$y-x=-2$
$2 x+z=3$

$4 x-y+z=7$
$-x+y=-2$
$2 x+z=3$

$\left(\begin{array}{ccc|c}{4} & {-1} & {1} & {7} \\ {-1} & {1} & {0} & {-2} \\ {2} & {0} & {1} & {3}\end{array}\right) R_{1} \leftrightarrow R_{2}$

$\left(\begin{array}{ccc|c}{-1} & {1} & {0} & {-2} \\ {4} & {-1} & {1} & {7} \\ {2} & {0} & {1} & {3}\end{array}\right) R_{1}\rightarrow-R_{1}$

$\left(\begin{array}{ccc|c}{1} & {-1} & {0} & {2} \\ {4} & {-1} & {1} & {7} \\ {2} & {0} & {1} & {3}\end{array}\right) \begin{array}{l}{R_{2} \rightarrow R_{2}-4 R_{1}} \\ {R_{3} \rightarrow R_{3} - 2 R_{1}} \end{array}$

$\left(\begin{array}{ccc|c}{1} & {-1} & {0} & {2} \\ {0} & {3} & {1} & {-1} \\ {0} & {2} & {1} & {-1}\end{array}\right) R_{2} \rightarrow 1 / 3 R_{2}$

$\left(\begin{array}{ccc|c}{1} & {-1} & {0} & {2} \\ {0} & {1} & {1 / 3} & {-1 / 3} \\ {0} & {2} & {1} & {-1}\end{array}\right) R_{3} \rightarrow R_{3} - 2 R_{2}$

$\left(\begin{array}{ccc|c}{1} & {-1} & {0} & {2} \\ {0} & {1} & {1 / 3} & {-1 / 3} \\ {0} & {0} & {1 / 3} & {-1 / 3}\end{array}\right) R_{3} \rightarrow 3 R_{3}$

$\left(\begin{array}{ccc|c}{1} & {-1} & {0} & {2} \\ {0} & {1} & {1 / 3} & {-1 / 3} \\ {0} & {0} & {1} & {-1}\end{array}\right)$

Row echelon Form

$x-y+0=2$
$y+1 / 3 z=-1 / 3$
$z=-1$

$y+1 / 3(-1)=-1 / 3$

$y=0$

$x-0+0=2$

$x=2$

$\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{c}{2} \\ {0} \\ {-1}\end{array}\right)$

Solve the following system using Gauss Elimination

$x_{1}+x_{2}=1$
$x_{1}+x_{3}=2$
$x_{2}+x_{3}=3$

$\left(\begin{array}{lll|l}{1} & {1} & {0} & {1} \\ {1} & {0} & {1} & {2} \\ {0} & {1} & {1} & {3}\end{array}\right)$

Solve the following system using Gauss Elimination

$x_{1}+x_{2}+x_{3}+x_{4}=12$
$x_{1}+2 x_{2}+5 x_{4}=17$
$3 x_{1}+2 x_{2}+4 x_{3}-x_{4}=31$

$\left(\begin{array}{llll|l}{1} & {1} & {1} & {1} & {12} \\ {1} & {2} & {0} & {5} & {17} \\ {3} & {2} & {4} & {-1} & {31}\end{array}\right) \begin{array}{l}{R_{2} \rightarrow R_{2} - R_{1}} \\ {R_{3} \rightarrow R_{3} - 3 R_{1}} \end{array}$

$\left(\begin{array}{cccc|c}{1} & {1} & {1} & {1} & {12} \\ {0} & {1} & {-1} & {4} & {5} \\ {0} & {-1} & {1} & {-4} & {-5}\end{array}\right) {R_{3} \rightarrow R_{3}+R_{2}}$

$\left(\begin{array}{cccc|c}{1} & {1} & {1} & {1} & {12} \\ {0} & {1} & {-1} & {4} & {5} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right){R_{1} \rightarrow R_{1} - R_{2}}$

$\left(\begin{array}{cccc|c}{1} & {0} & {2} & {-3} & {7} \\ {0} & {1} & {-1} & {4} & {5} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right)$

No. of free var. = No. of var. - r(A)

$=4-2=2$

Let $x_{3}=r$
$x_{4}=5$

$x_{1}+2 r-3 s=7$
$x_{2}-r+4 s=5$

$x_{1}=7- 2r+3s$
$x_{2}=5+r-4 s$
$x_{3}=r$
$x_{4}=s$

$\left(\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right)=r\left(\begin{array}{l}{-2} \\ {1} \\ {1} \\ {0}\end{array}\right)+s\left(\begin{array}{c} {3} \\ {-4} \\ {0} \\ {1}\end{array}\right)+\left(\begin{array}{l}{7} \\ {5} \\ {0} \\ {0}\end{array}\right)$