1. Selesaiakan SPL berikut dengan metode Eliminasi Gauss, Gauss Jordan, dan Matriks Balikan! $$ x_{1}+3 x_{2}+2 x_{3}=4 $$ $$ 2 x_{1}+x_{2}+6 x_{3}=-11 $$ $$ x_{1}-4 x_{2}=-13 $$

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Solve the system of equations $$ x_{1}+3x_{2}+2x_{3}=4;2x_{1}+x_{2}+6x_{3}=-11;x_{1}-4x_{2}=-13 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x_{1}+3x_{2}+2x_{3}=4\2x_{1}+x_{2}+6x_{3}=-11\x_{1}-4x_{2}=-13\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x_{1}=4-3x_{2}-2x_{3}\2x_{1}+x_{2}+6x_{3}=-11\x_{1}-4x_{2}=-13\end{array}ight.$$
- step2: Substitute the value of $$x_{1}:$$
  $$\left\{ \begin{array}{l}2\left(4-3x_{2}-2x_{3}ight)+x_{2}+6x_{3}=-11\4-3x_{2}-2x_{3}-4x_{2}=-13\end{array}ight.$$
- step3: Simplify:
  $$\left\{ \begin{array}{l}8-5x_{2}+2x_{3}=-11\4-7x_{2}-2x_{3}=-13\end{array}ight.$$
- step4: Solve the equation:
  $$\left\{ \begin{array}{l}x_{2}=\frac{19+2x_{3}}{5}\4-7x_{2}-2x_{3}=-13\end{array}ight.$$
- step5: Substitute the value of $$x_{2}:$$
  $$4-7	imes \frac{19+2x_{3}}{5}-2x_{3}=-13$$
- step6: Simplify:
  $$4-\frac{7\left(19+2x_{3}ight)}{5}-2x_{3}=-13$$
- step7: Multiply both sides of the equation by LCD:
  $$\left(4-\frac{7\left(19+2x_{3}ight)}{5}-2x_{3}ight)	imes 5=-13	imes 5$$
- step8: Simplify the equation:
  $$-113-24x_{3}=-65$$
- step9: Move the constant to the right side:
  $$-24x_{3}=-65+113$$
- step10: Add the numbers:
  $$-24x_{3}=48$$
- step11: Change the signs:
  $$24x_{3}=-48$$
- step12: Divide both sides:
  $$\frac{24x_{3}}{24}=\frac{-48}{24}$$
- step13: Divide the numbers:
  $$x_{3}=-2$$
- step14: Substitute the value of $$x_{3}:$$
  $$x_{2}=\frac{19+2\left(-2ight)}{5}$$
- step15: Calculate:
  $$x_{2}=3$$
- step16: Substitute the values of $$x_{2},x_{3}:$$
  $$x_{1}=4-3	imes 3-2\left(-2ight)$$
- step17: Calculate:
  $$x_{1}=-1$$
- step18: Calculate:
  $$\left\{ \begin{array}{l}x_{1}=-1\x_{2}=3\x_{3}=-2\end{array}ight.$$
- step19: Check the solution:
  $$\left\{ \begin{array}{l}x_{1}=-1\x_{2}=3\x_{3}=-2\end{array}ight.$$
- step20: Rewrite:
  $$\left(x_{1},x_{2},x_{3}ight) = \left(-1,3,-2ight)$$
The solution to the system of equations using the method of Elimination Gauss, Gauss Jordan, and Inverse Matrix is:

$$ x_{1} = -1 $$
$$ x_{2} = 3 $$
$$ x_{3} = -2 $$
$$ x_{1} = -1 $$, $$ x_{2} = 3 $$, $$ x_{3} = -2 $$.

1. Selesaiakan SPL berikut dengan metode Eliminasi Gauss, Gauss Jordan, dan Matriks Balikan! \( x_{1}+3 x_{2}+2 x_{3}=4 \) \( 2 x_{1}+x_{2}+6 x_{3}=-11 \) \( x_{1}-4 x_{2}=-13 \)

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