# Gaussian-Jordan Elimination question?

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.

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this question needs editing! – Mathematician Feb 8 '13 at 16:24
Dont vote this down,I'm editing it please! – dgfddf Feb 8 '13 at 16:25
Can you please stop downvoting this? Please? – dgfddf Feb 8 '13 at 16:31
You can't interchange columns as you did... – David Mitra Feb 8 '13 at 16:38
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 8 '13 at 16:47

Hints:

Updated

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

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

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

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The usual collation order for unknowns goes $x, y, z, u, v, w, \ldots$. That doesn't match alphabetical order, but tradition is tradition. – Henning Makholm Feb 8 '13 at 18:01
@HenningMakholm: Thanks for that tidbit! Regards – Amzoti Feb 8 '13 at 18:11
Nice job! I keep forgetting how to format the vertical bar in an augmented matrix (when I need to use it)! – amWhy May 3 '13 at 2:20
@amWhy: To be honest, I use so many of these infrequently, I have to go to my cheat sheet to recall it! :-) Thanks for all the support! – Amzoti May 3 '13 at 2:21

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: $$\left[ \begin{array}{@{}ccccc|c@{}} 1& 0& -4/3 & 25/3 & 0&-2 \\ 0&1&-1/3 &10/3 & 0 &-1 \\ 0&0&0& 0&1&1 \\ \end{array} \right]$$

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