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I have the linear system

$$ \begin{align*} 2x-y-z+v&=0 \\ x-2y-z+5u-v&=1 \\ 2x-z+v&=1 \end{align*}$$

Very well. I form the matrix

$$ \left[ \begin{array}{@{}ccccc|c@{}} 2&-1&-1 & 0 & 1 &0 \\ 1&-2&-1 & 5 & -1 &1 \\ 2&0&-1 & 0&1&1 \\ \end{array} \right] $$

So I thought about exchanging the first row with the second one,and the first column with the last column. The first row is $a1$,the second $a$2 and the third $a3$. So what I do is find $a3-a2$ ,then multiply $a1$ by $-2$, then do $a2+a1$ , then $a3-a1$, this way I will have a $scaled$ $matrix$ and it will look like

$$ \left[ \begin{array}{@{}ccccc|c@{}} -2&2&2 & -10 & -4 &-2 \\ 0&-3&1 &-10 & -5 &-2 \\ 0&0&0& 0&1&1 \\ \end{array} \right] $$

When I proceed with the Gaussian method, I don't know what to multiply or divide to get some of the numbers in the matrix zero. Help me ? Please.

bryan.blackbee
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dgfddf
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  • this question needs editing! – Mathematician Feb 08 '13 at 16:24
  • Dont vote this down,I'm editing it please! – dgfddf Feb 08 '13 at 16:25
  • Can you please stop downvoting this? Please? – dgfddf Feb 08 '13 at 16:31
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    You can't interchange columns as you did... – David Mitra Feb 08 '13 at 16:38
  • :/....what do I do then? – dgfddf Feb 08 '13 at 16:39
  • Take a deep breath, relax, and don't worry about the downvotes. If you stick around (we hope you do), you'll accumulate reputation enough that downvotes won't matter. You've already learned a useful fact, that it's better to edit your question before you post it. – Rick Decker Feb 08 '13 at 16:44
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    Look at the original augmented matrix. You want to eliminate (that is "make $0$") the "$1$" in the second row, first column. How can this be done by using row operations? One way would be to take twice row two and subtract row one; this gives you a new row two (notation $a_2\rightarrow 2a_2-a_1$). This gives $[0\ -3\ -1\ 10\ -3\ | 2]$ as the new row two. Now eliminate the "$2$" in row three, column one (by using row one). Write down the new matrix obtained. There will be only one more elimination needed (eliminate the entry in the third row, second column using the new row two). – David Mitra Feb 08 '13 at 16:47

2 Answers2

2

Hints:

Updated

Variables arranged using $(x, y, z, u, v)$.

From your system, we have:

$$ \left[ \begin{array}{@{}ccccc|c@{}} 2&-1&-1&0&1 &0 \\ 1&-2&-1 &5&-1& 1 \\ 2&0&-1 &0&1& 1 \\ \end{array} \right] $$

The row-reduced-echelon-form should be:

$$ \left[ \begin{array}{@{}ccccc|c@{}} 1&0& 0 & -5 & 2 & -2\\ 0&1&0 & 0 & 0 &1 \\ 0&0&1 & -10&3&-5 \\ \end{array} \right] $$

Regards

Amzoti
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Assuming that you've done everything right so far

$$ \left[ \begin{array}{@{}ccccc|c@{}} -2&2&2 & -10 & -4 &-2 \\ 0&-3&1 &-10 & -5 &-2 \\ 0&0&0& 0&1&1 \\ \end{array} \right] $$

Now, let the first row be L1, the second L2, the third L3.

L1$\times$(-0.5) and then, L1-2L2. You got
$$ \left[ \begin{array}{@{}ccccc|c@{}} 1&-1&-1 & 5 & 0&-1 \\ 0&-3&1 &-10 & -5 &-2 \\ 0&0&0& 0&1&1 \\ \end{array} \right] $$

next, you do L2$\times$(-1/3)-(5/3)L3. You got

$$ \left[ \begin{array}{@{}ccccc|c@{}} 1&-1&-1 & 5 & 0&-1 \\ 0&1&-1/3 &10/3 & 0 &-2/3 \\ 0&0&0& 0&1&1 \\ \end{array} \right] $$

Finally, L1+L2, you have: $$ \left[ \begin{array}{@{}ccccc|c@{}} 1& 0& -4/3 & 25/3 & 0&-2 \\ 0&1&-1/3 &10/3 & 0 &-1 \\ 0&0&0& 0&1&1 \\ \end{array} \right] $$