# I simply cannot do this Gauss elimination question

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• Aug 15th 2009, 06:19 PM
thomasashcroft
I simply cannot do this Gauss elimination question
$\displaystyle 2x - y + 3z + t = 1$
$\displaystyle -2x + y - 2z = 1$
$\displaystyle 4x - 2y + 7z + 3t = 4$

I got it to

(2 -1 3 1 1)
(0 0 1 1 2)

and I don't know what to do from there : /
• Aug 15th 2009, 06:51 PM
mr fantastic
Quote:

Originally Posted by thomasashcroft
$\displaystyle 2x - y + 3z + t = 1$
$\displaystyle -2x + y - 2z = 1$
$\displaystyle 4x - 2y + 7z + 3t = 4$

I got it to

(2 -1 3 1 1)
(0 0 1 1 2)

and I don't know what to do from there : /

Get the reduced row echelon form. Then it should be clear that you have to introduce two parameters .... Perhaps $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where each parameter can be any real number. Now get y and z in terms of these parameters and you have a solution.
• Aug 15th 2009, 06:52 PM
thomasashcroft
Quote:

Originally Posted by mr fantastic
Get the reduced row echelon form. Then it should be clear that you have to introduce two parameters .... Perhaps $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where each parameter can be any real number. Now get y and z in terms of these parameters and you have a solution.

It's not possible to get the reduced row echelon form is it?
• Aug 15th 2009, 07:02 PM
mr fantastic
Quote:

Originally Posted by thomasashcroft
It's not possible to get the reduced row echelon form is it?

Why do you think that? Of course it is.

PlanetMath: reduced row echelon form
• Aug 15th 2009, 07:06 PM
thomasashcroft
Well all I would have to do is multiply the first row by 1/2 which would leave me with 1 - 0.5 1.5 0.5 0.5 in the first row.. and then it'd be in r.r.e.f. But how does that help me see that y should be arbitrary?
• Aug 15th 2009, 10:30 PM
mr fantastic
Quote:

Originally Posted by thomasashcroft
Well all I would have to do is multiply the first row by 1/2 which would leave me with 1 - 0.5 1.5 0.5 0.5 in the first row.. and then it'd be in r.r.e.f. But how does that help me see that y should be arbitrary?

$\displaystyle R_1 \longrightarrow R_1 - 3 R_2$:

$\displaystyle \left( \begin{array}{cccc} 2 & -1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{array} \right|$ $\displaystyle \left. \begin{array}{c} -5 \\ 2 \end{array} \right)$

which is equivalent to the linear system:

$\displaystyle 2x - y - 2t = -5$

$\displaystyle z + t = 2$

Now there is choice. I chose $\displaystyle t = \lambda$ and $\displaystyle x = \mu$ where both parameters can be any real number.

Then:

$\displaystyle z + \lambda = 2 \Rightarrow z = 2 - \lambda$.

$\displaystyle 2 \mu - y - 2 \lambda = -5 \Rightarrow y = 2 \mu + 3$.

Other choices leading to equivalent solutions can obviously be made.
• Aug 16th 2009, 10:43 AM
CaptainBlack
Quote:

Originally Posted by thomasashcroft
$\displaystyle 2x - y + 3z + t = 1$
$\displaystyle -2x + y - 2z = 1$
$\displaystyle 4x - 2y + 7z + 3t = 4$

I got it to

(2 -1 3 1 1)
(0 0 1 1 2)

and I don't know what to do from there : /

The initial tablau is:

$\displaystyle \begin{array}{rrrr|r} 2&-1&3&1&1\\ -2&1&-2&0&1\\ 4&-2&7&3&4 \end{array}$

add the first row to the second:

$\displaystyle \begin{array}{rrrr|r} 2&-1&3&1&1\\ 0&0&1&1&2\\ 4&-2&7&3&4 \end{array}$

Now subtract twice the first row from the third:

$\displaystyle \begin{array}{rrrr|r} 2&-1&3&1&1\\ 0&0&1&1&2\\ 0&0&1&1&2 \end{array}$

Removing the duplicate leaves you with:

$\displaystyle \begin{array}{rrrr|r} 2&-1&3&1&1\\ 0&0&1&1&2 \end{array}$

NOw let $\displaystyle t=\alpha$ and $\displaystyle y=\beta$ and solve for $\displaystyle x$ and $\displaystyle z$ in terms of $\displaystyle \alpha$ and $\displaystyle \beta$.

CB