$\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 : /

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- Aug 15th 2009, 06:19 PMthomasashcroftI 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 PMmr 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.

- Aug 15th 2009, 06:52 PMthomasashcroft
- Aug 15th 2009, 07:02 PMmr fantastic
Why do you think that? Of course it is.

PlanetMath: reduced row echelon form - Aug 15th 2009, 07:06 PMthomasashcroft
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 PMmr fantastic
$\displaystyle R_1 \longrightarrow R_1 - 3 R_2$:

$\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 AMCaptainBlack

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