Use Gauss Jordan elimination to solve the system

[sigma]

I need some help to figure out this whole reduced row echelon form (RREF) thing. Heres the question with some of my work (what I tried).

\(\displaystyle \
\L\
\begin{array}{l}
x_1 + x_2 + x_3 + x_4 = 4 \\
2x_1 + x_2 - x_3 + 2x_4 = 4 \\
x_1 + x_2 + 2x_3 + x_4 = 5 \\
3x_1 + 2x_2 + x_3 + 3x_4 = 9 \\
\end{array}
\\)

The augmented matrix:
\(\displaystyle \L\mbox{
\left[
\begin{array}{cccc|c}
1& 1 & 1 & 1 & 4\\
2& 1 & -1 & 2 & 4\\
1& 1 & 2 & 1 & 5\\
3& 2 & 1 & 3 & 9\\
\end{array}
\right]
}\)

\(\displaystyle \
\L\
\begin{array}{l}
R1 \to R1 - R3 \\
R2 \to R2 - R1{\rm } \to \\
R3 \to R3 - R1 \\
R4 \to R4 - 3R1 \\
\end{array}
\
\L\mbox{
\left[
\begin{array}{cccc|c}
0& 0 & -1 & 0 & -1\\
1& 0 & -2 & 1 & 0\\
0& 0 & 1 & 0 & 1\\
0& -1 & -2 & 0 & -3\\
\end{array}
\right]
}\)

\(\displaystyle \
\L\
\begin{array}{l}
R3 \to R3 + R1{\rm } \to \\
R1 \to R1 + R3{\rm } \\
\end{array}
\
\L\mbox{
\left[
\begin{array}{cccc|c}
0& 0 & 0 & 0 & 0\\
1& 0 & -2 & 1 & 0\\
0& 0 & 0 & 0 & 0\\
0& -1 & -2 & 0 & -3\\
\end{array}
\right]
}\)

And I get stuck from there. I know this is not in RREF and I think I already screwed up on that part. The answer to this question is (2-t, 1, 1, t). I somewhat know how RREF works (you need that pattern with the 1's and 0's) but I can't completely grasp the concept of it and how you go about solving these systems. It seems more trial and error for me and I don't know what I'm suppose to be looking for in the pattern of the augmented matrix when you have a final solution. If somebody can help, it would be greatly appreciated. SIG

 

[pka]

Here is what MathCad found.

 

[stapel]

The complete solution does indeed have a row of zeroes. This just means that there is no unique solution. In this case, the solution will turn out to be dependent. As long as your final answer looks something like the following, you're fine:

. . . . .(x₁, x₂, x₃, x₄) = (2 - x₄, 1, 1, x₄)

Eliz.

 

[sigma]

But how do you get to that answer? How do you know completely that your matrix is in RREF? That is what I don't understand. Where do you usually begin with questions like this? Do you just try to get as many 0's as possible?

 

[pka]

But how do you get to that answer? How do you know completely that your matrix is in RREF? That is what I don't understand. Where do you usually begin with questions like this? Do you just try to get as many 0's as possible?

That is why I use a CAS, computer algebra syatem.
I find ludicrous to not use one for this problem.
Would you try to find the square root of 13 ‘by hand’?
No of course not, any sane person would use a calculator!
Row reduction is just a means to find the solution to a system of equations.
So use the tools available to you.
I have given the same talk to many in-service teachers.
Its title is “What is a solution?”
Well a solution is what makes the problem work!

 

[sigma]

No I have to do it all by hand. I'm not allowed any calculators (not for the exam anyways). Its not like I need a calculator for this anyways its just a bunch of addition and subtraction of simple numbers but its extremely tedious, and the problem is I don't know what row to add to another row or what to subtract what from. I know that your suppose to get 1's and 0's in a certain pattern in your matrix for RREF but in the end, I really don't know what I'm suppose to be working towards. All I want to know is a handy tip to know that your matrix is in RREF, that's it.

 

[stapel]

I really don't know what I'm suppose to be working towards.

You should be working toward a matrix that looks like the answers you've been given: 1's and 0's for all the entries of the first few columns (as many as you can get), with the 1's as "triangular" as possible.

Eliz.