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Multivariable Linear Systems

+1
157
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+5220

Solve by any method:

4x-y+z=-5

2x+2y+3z=10

5x-2y+6z=1

I tried using Gaussian Elimination but I got kinda messed up, I'd appreciate some help. Thanks!

rarinstraw1195  Dec 19, 2017
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#1
+85757
+3

4x - y + z = -5    ⇒   8x - 2y + 2z  =  -10    (1)

2x + 2y + 3z =10       (2)

5x - 2y  + 6z = 1        (3)

10x + 5z  = 0        ⇒    2x  + z  =  0   ⇒   z  =  - 2x     (4)

7x + 9z  = 11    (5)

Sub (4)  into (5)

7x + 9 (-2x)  =  11

7x - 18x = 11

-11x  =  11

x  =  -1

So.......z   =  -2(-1)  =  2

And using  4x - y + z = -5   to find y, we have

4(-1) - y +  2  =  -5

-4   -  y   +  2   = - 5

-2 - y   =  - 5

-y  =  - 3

y  =  3

So    {x, y, z }  =   ( -1, 3,  2 }

CPhill  Dec 19, 2017
#2
+19207
+2

Solve by any method:

4x-y+z=-5

2x+2y+3z=10

5x-2y+6z=1

I tried using Gaussian Elimination

Gaussian Elimination:

$$\left( \begin{matrix} 4 & -1 & 1 \\ 2 & 2 & 3 \\ 5 & -2 & 6 \end{matrix} \left| \begin{matrix} -5 \\ 10 \\ 1 \end{matrix} \right. \right) \\ \overset{II^{*}= 2\cdot II - I}{\curvearrowright} \left( \begin{matrix} 4 & -1 & 1 \\ 0 & 5 & 5 \\ 5 & -2 & 6 \end{matrix} \left| \begin{matrix} -5 \\ 25 \\ 1 \end{matrix} \right. \right) \\ \overset{III^{*}= \frac45 \cdot III - I}{\curvearrowright} \left( \begin{matrix} 4 & -1 & 1 \\ 0 & 5 & 5 \\ 0 & -\frac35 & \frac{19}{5} \end{matrix} \left| \begin{matrix} -5 \\ 25 \\ \frac{29}{5} \end{matrix} \right. \right) \\ \overset{III^{*}= \frac{25}{3} \cdot III + II}{\curvearrowright} \left( \begin{matrix} 4 & -1 & 1 \\ 0 & 5 & 5 \\ 0 & 0 & \frac{110}{3} \end{matrix} \left| \begin{matrix} -5 \\ 25 \\ \frac{220}{3} \end{matrix} \right. \right) \\$$

$$\begin{array}{|rcll|} \hline \frac{110}{3}z &=& \frac{220}{3}\\ z &=& \frac{220}{110} \\ \mathbf{z} &\mathbf{=}& \mathbf{2} \\\\ 5y+5z&=& 25 \\ 5y + 10 &=& 25 \\ 5y &=& 15 \\ \mathbf{y} &\mathbf{=}& \mathbf{3} \\\\ 4x-y+z &=& -5 \\ 4x - 3 + 2 &=& -5 \\ 4x &=& -4 \\ \mathbf{x} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array}$$

heureka  Dec 20, 2017

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