# Thread: I simply cannot do this Gauss elimination question

1. ## I simply cannot do this Gauss elimination question

$
2x - y + 3z + t = 1
$

$
-2x + y - 2z = 1
$

$
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 : /

2. Originally Posted by thomasashcroft
$
2x - y + 3z + t = 1
$

$
-2x + y - 2z = 1
$

$
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 $t = \lambda$ and $x = \mu$ where each parameter can be any real number. Now get y and z in terms of these parameters and you have a solution.

3. Originally Posted by mr fantastic
Get the reduced row echelon form. Then it should be clear that you have to introduce two parameters .... Perhaps $t = \lambda$ and $x = \mu$ 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?

4. 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

5. 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?

6. 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?
$R_1 \longrightarrow R_1 - 3 R_2$:

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

which is equivalent to the linear system:

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

$z + t = 2$

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

Then:

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

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

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

$
-2x + y - 2z = 1
$

$
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:

$
\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:

$
\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:

$
\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:

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

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

CB