$\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 : /
$\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.
Why do you think that? Of course it is.
PlanetMath: reduced row echelon form
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.
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