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I try to get the variables for this equation:

$$\begin{cases} 6x_1 + 4x_2 + 8x_3 + 17x_4 &= -20\\ 3x_1 + 2x_2 + 5x_3 + 8x_4 &= -8\\ 3x_1 + 2x_2 + 7x_3 + 7x_4 &= -4\\ 0x_1 + 0x_2 + 2x_3 -1x_4 &= 4 \end{cases}$$

So i started with:

$$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ 3 & 2 & 5 & 8 & -8 \\ 3 & 2 & 7 & 7 & -4\\ 0 & 0 & 2 & -1 &4\\ \end{pmatrix}$$

Then I continued: $$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ -6 & -4 & -10 & -16 & 16 \\ -6 & -4 & -14 & -14 & 8\\ 0 & 0 & 2 & -1 &4\\ \end{pmatrix} $$

And began to count:

$$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ 0 & 0 & -2 & -1 & -4 \\ 0 & 0 & -6 & 3 & -12\\ 0 & 0 & 2 & -1 &4\\ \end{pmatrix} $$ And finally I transformed it to:$$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ 0 & 0 & -2 & -1 & -4 \\ 0 & 0 & -6 & 3 & -12\\ 0 & 0 & 0 & 0 &0\\ \end{pmatrix} $$

What did i wrong? I think the last row cannot be correct because:

$$(0, 0 ,0 , 0) \neq 0x_1 + 0x_2 + 2x_3 -1x_4 = 4$$

How should i solve it? Thanks

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But the last row is $(0\ 0\ 0\ 0\ 0)$, and not $(0\ 0\ 0\ 0\ 4)$. So the equation is $0x_1 + 0x_2 + 0x_3 + 0x_4 = 0$, which is perfectly alright. –  M. Vinay Jun 2 at 11:26
1  
You've done $17+(-16)=-1$ in the third matrix, but this seems to be a typo since otherwise you can't get the last row just with $0$'s. In any case your question is not related to these typos. To answer it: there is nothing wrong (except for the typos). –  Git Gud Jun 2 at 11:28
    
@GitGud but how can i get the solutions for the variables? Thanks –  John Smith Jun 2 at 11:31
1  
From the second and third rows (in the last matrix, after it is fixed) you can get $x_3$ as a function of $x_4$ or $x_4$ as a function of $x_3$. Then using the first line you can get $x_1$ as a function of $x_2, x_3$ and $x_4$ or $x_2$ as a function of $x_1, x_3$ and $x_4$, as you prefer. Two variables will act as parameters. –  Git Gud Jun 2 at 11:35

2 Answers 2

up vote 2 down vote accepted

As Git Gud points out, it should be 1, not -1. Now if you correct that and proceed further, the third row also will become zero, and you'll get

$\left(\begin{matrix} 6 & 4 & 8 & 17 & -20\\ 0 & 0 & -2 & 1 & -4\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right)$

so that

$6x_1 + 4x_2 + 8x_3 + 17x_4 = -20\\ -2x_3 + x_4 = -4$

which can be solved by back-substitution.

From the second equation, you'll get $x_4 = 2x_3 - 4$. Substitute this in the first to get $x_1$ in terms of $x_2$ and $x_3$.

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Should mean i begin with removing the $-20$ in the matrix? –  John Smith Jun 2 at 11:40
    
I've edited the answer, please check out the last part. –  M. Vinay Jun 2 at 11:50

$$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ 0 & 0 & -2 & \color{red}{-1} & -4 \\ 0 & 0 & -6 & 3 & -12\\ 0 & 0 & 2 & -1 &4\\ \end{pmatrix}$$

should be $$ \begin{pmatrix} 6 & 4 & 8 & 17 & -20 \\ 0 & 0 & -2 & \color{green}{+1} & -4 \\ 0 & 0 & -6 & 3 & -12\\ 0 & 0 & 2 & -1 &4\\ \end{pmatrix}$$

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And how should i continue to get the solution? –  John Smith Jun 2 at 11:33
1  
Solve it, edit the question with the new solution and I will have a look. I am not going to solve it for you. –  5xum Jun 2 at 11:34

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