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

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

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

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

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