3. Considere el sistema dado por: $$ 4 x_{1}+3 x_{2}-x_{3}=4 $$ $$ \begin{aligned} 3 x+9 y-3 z & =0 \ y+z & =1 \ -2 x-5 y+4 z & =4 \ -2 x-6 y+3 z & =4 \end{aligned} \quad \text { C. } \quad \begin{aligned} -2 x+6 y-4 z & =0 \ -x+3 y-z & =0 \ 5 x-15 y+10 z & =0 \ x-3 y & =0 \end{aligned} $$ A. $$ 8 x_{1}-6 x_{2}-2=0 $$ B. $$ x_{1}-x_{2}-x_{3}=3 $$ i. Determine el orden del sistema, tipo de solución e indique las variables libre y las variables pivotales. ii. Resuelva el sistema por el método de Gauss o de eliminación de Gauss-Jordan.

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UpStudy AI · Accepted Answer

Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}8x_{1}-6x_{2}-2=0\x_{1}-x_{2}-x_{3}=3\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}8x_{1}-6x_{2}-2=0\x_{1}=3+x_{2}+x_{3}\end{array}\right.$$
- step2: Substitute the value of $$x_{1}:$$
  $$8\left(3+x_{2}+x_{3}\right)-6x_{2}-2=0$$
- step3: Simplify:
  $$22+2x_{2}+8x_{3}=0$$
- step4: Evaluate:
  $$22+8x_{3}+2x_{2}=0$$
- step5: Move the expression to the right side:
  $$2x_{2}=0-\left(22+8x_{3}\right)$$
- step6: Subtract the terms:
  $$2x_{2}=-22-8x_{3}$$
- step7: Divide both sides:
  $$\frac{2x_{2}}{2}=\frac{-22-8x_{3}}{2}$$
- step8: Divide the numbers:
  $$x_{2}=-11-4x_{3}$$
- step9: Substitute the value of $$x_{2}:$$
  $$x_{1}=3-11-4x_{3}+x_{3}$$
- step10: Simplify:
  $$x_{1}=-8-3x_{3}$$
- step11: Calculate:
  $$\left(x_{1},x_{2},x_{3}\right) = \left(-8-3x_{3},-11-4x_{3},x_{3}\right),x_{3} \in \mathbb{R}$$
- step12: Alternative Form:
  $$\textrm{Infinitely many solutions}$$
Solve the system of equations $$ 4 x_{1}+3 x_{2}-x_{3}=4;3 x+9 y-3 z=0;y+z=1;-2 x-5 y+4 z=4;-2 x-6 y+3 z=4;-2 x+6 y-4 z=0;-x+3 y-z=0;5 x-15 y+10 z=0;x-3 y=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}4x_{1}+3x_{2}-x_{3}=4\3x+9y-3z=0\y+z=1\-2x-5y+4z=4\-2x-6y+3z=4\-2x+6y-4z=0\-x+3y-z=0\5x-15y+10z=0\x-3y=0\end{array}\right.$$
- step1: Rearrange the terms:
  $$\left\{ \begin{array}{l}4x_{1}+3x_{2}-x_{3}=4\3x+9y-3z=0\y+z=1\-2x-5y+4z=4\-2x-6y+3z=4\-2x+6y-4z=0\-x+3y-z=0\x-3y=0\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}x_{3}=-4+4x_{1}+3x_{2}\3x+9y-3z=0\y+z=1\-2x-5y+4z=4\-2x-6y+3z=4\-2x+6y-4z=0\-x+3y-z=0\x-3y=0\end{array}\right.$$
- step3: Substitute the value of $$x_{3}:$$
  $$\left\{ \begin{array}{l}3x+9y-3z=0\y+z=1\-2x-5y+4z=4\-2x-6y+3z=4\-2x+6y-4z=0\-x+3y-z=0\x-3y=0\end{array}\right.$$
- step4: Solve the equation:
  $$\left\{ \begin{array}{l}3x+9y-3z=0\y=1-z\-2x-5y+4z=4\-2x-6y+3z=4\-2x+6y-4z=0\-x+3y-z=0\x-3y=0\end{array}\right.$$
- step5: Substitute the value of $$y:$$
  $$\left\{ \begin{array}{l}3x+9\left(1-z\right)-3z=0\-2x-5\left(1-z\right)+4z=4\-2x-6\left(1-z\right)+3z=4\-2x+6\left(1-z\right)-4z=0\-x+3\left(1-z\right)-z=0\x-3\left(1-z\right)=0\end{array}\right.$$
- step6: Simplify:
  $$\left\{ \begin{array}{l}3x+9-12z=0\-2x-5+9z=4\-2x-6+9z=4\-2x+6-10z=0\-x+3-4z=0\x-3+3z=0\end{array}\right.$$
- step7: Solve the equation:
  $$\left\{ \begin{array}{l}3x+9-12z=0\-2x-5+9z=4\-2x-6+9z=4\-2x+6-10z=0\x=3-4z\x-3+3z=0\end{array}\right.$$
- step8: Substitute the value of $$x:$$
  $$\left\{ \begin{array}{l}3\left(3-4z\right)+9-12z=0\-2\left(3-4z\right)-5+9z=4\-2\left(3-4z\right)-6+9z=4\-2\left(3-4z\right)+6-10z=0\3-4z-3+3z=0\end{array}\right.$$
- step9: Simplify:
  $$\left\{ \begin{array}{l}18-24z=0\-11+17z=4\-12+17z=4\-2z=0\-z=0\end{array}\right.$$
- step10: Simplify the expression:
  $$\left\{ \begin{array}{l}18-24z=0\-11+17z=4\-12+17z=4\-2z=0\end{array}\right.$$
- step11: Solve the equation:
  $$\left\{ \begin{array}{l}z=\frac{3}{4}\-11+17z=4\-12+17z=4\-2z=0\end{array}\right.$$
- step12: Substitute the value of $$z:$$
  $$\left\{ \begin{array}{l}-11+17\times \frac{3}{4}=4\-12+17\times \frac{3}{4}=4\-2\times \frac{3}{4}=0\end{array}\right.$$
- step13: Simplify:
  $$\left\{ \begin{array}{l}\frac{7}{4}=4\\frac{3}{4}=4\-\frac{3}{2}=0\end{array}\right.$$
- step14: Calculate:
  $$\left\{ \begin{array}{l}x \in \varnothing \x_{1} \in \varnothing \x_{2} \in \varnothing \x_{3} \in \varnothing \y \in \varnothing \z \in \varnothing \end{array}\right.$$
- step15: Rewrite:
  $$(x, x_{1}, x_{2}, x_{3}, y, z) \in \varnothing$$
To solve the given problem, we will analyze the systems of equations step by step.

### Step 1: Analyze the Systems

#### System 1:
1. $$4 x_{1}+3 x_{2}-x_{3}=4$$
2. $$3 x+9 y-3 z=0$$
3. $$y+z=1$$
4. $$-2 x-5 y+4 z=4$$
5. $$-2 x-6 y+3 z=4$$
6. $$-2 x+6 y-4 z=0$$
7. $$-x+3 y-z=0$$
8. $$5 x-15 y+10 z=0$$
9. $$x-3 y=0$$

#### System 2:
1. $$8 x_{1}-6 x_{2}-2=0$$
2. $$x_{1}-x_{2}-x_{3}=3$$

### Step 2: Determine the Order and Type of Solution

#### System 1:
- The first system has 8 equations and 6 variables $$(x, x_{1}, x_{2}, x_{3}, y, z)$$.
- The result indicates that there is **no solution** for this system.

#### System 2:
- The second system has 2 equations and 3 variables $$(x_{1}, x_{2}, x_{3})$$.
- The result indicates that there are **infinitely many solutions** for this system.

### Step 3: Identify Free and Pivotal Variables

#### System 1:
- Since there is no solution, we cannot identify free or pivotal variables.

#### System 2:
- In the second system, we have 2 equations and 3 variables, which implies that one variable will be free. 
- We can express one variable in terms of the others, leading to infinitely many solutions.

### Step 4: Solve System 2

Now, let's solve the second system using the equations:
1. $$8 x_{1}-6 x_{2}-2=0$$
2. $$x_{1}-x_{2}-x_{3}=3$$

We can express $$x_{1}$$ in terms of $$x_{2}$$ from the first equation:
$$
8 x_{1} = 6 x_{2} + 2 \implies x_{1} = \frac{6 x_{2} + 2}{8} = \frac{3 x_{2}}{4} + \frac{1}{4}
$$

Substituting $$x_{1}$$ into the second equation:
$$
\frac{3 x_{2}}{4} + \frac{1}{4} - x_{2} - x_{3} = 3
$$
$$
\frac{3 x_{2}}{4} - \frac{4 x_{2}}{4} - x_{3} + \frac{1}{4} = 3
$$
$$
-\frac{x_{2}}{4} - x_{3} + \frac{1}{4} = 3
$$
$$
-x_{3} = 3 + \frac{x_{2}}{4} - \frac{1}{4}
$$
$$
-x_{3} = \frac{12}{4} + \frac{x_{2}}{4} - \frac{1}{4} = \frac{11 + x_{2}}{4}
$$
$$
x_{3} = -\frac{11 + x_{2}}{4}
$$

### Final Solution for System 2
The solutions can be expressed as:
- $$x_{1} = \frac{3 x_{2}}{4} + \frac{1}{4}$$
- $$x_{3} = -\frac{11 + x_{2}}{4}$$

Where $$x_{2}$$ is a free variable. Thus, the system has infinitely many solutions depending on the value of $$x_{2}$$.
The first system has no solution, while the second system has infinitely many solutions where $$x_{1} = \frac{3x_{2}}{4} + \frac{1}{4}$$ and $$x_{3} = -\frac{11 + x_{2}}{4}$$, with $$x_{2}$$ being a free variable.

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