Solve system of equation with matrix.
X1 - X2 - X3 = 1
-2X1 +6X2 + 10X3 = 14
2X1 +X2 +6X3

Question

Solve system of equation with matrix.
X1 - X2 - X3 = 1
-2X1 +6X2 + 10X3 = 14
2X1 +X2 +6X3

e.mccormick · Answer

Know how to put it in matrix form?

anonymous · Answer

Yes, is there a special format for on here?

e.mccormick · Answer

You can do it in $\LaTeX$ if you know how.  I can do that too.

anonymous · Answer

1  -1  -1  1
-2  6   10 14
 2   1    6   9

anonymous · Answer

ok?

e.mccormick · Answer

$\left[ \begin{array}{ccc|c}
1 & -1 & -1 & 1\
-2 & 6 & 10 & 14\
 2 & 1 & 6 & 9\
\end{array}\right ]$

e.mccormick · Answer

OK, know your elementary row operations?

anonymous · Answer

Just learning it......can I start with R2 = R2 + R3?

e.mccormick · Answer

That is a good start.

e.mccormick · Answer

$\left[ \begin{array}{ccc|c}
1 & -1 & -1 & 1\
0 & 7 & 16 & 23\
 2 & 1 & 6 & 9\
\end{array}\right ]$

e.mccormick · Answer

If you want to know what I typed, it is:
```
\left[ \begin{array}{ccc|c}
1 & -1 & -1 & 1\
0 & 7 & 16 & 23\
 2 & 1 & 6 & 9\
\end{array}ight ]
```

Makes it all pretty.

e.mccormick · Answer

Well, that between `$` and `$`.

anonymous · Answer

ok..

e.mccormick · Answer

So, what do you think the next step would be?

anonymous · Answer

(-2)R1 + R3 = R3

e.mccormick · Answer

And what would you get?

anonymous · Answer

1   -1   -1   1
 0    7   16  23
 0    3     8    7

e.mccormick · Answer

Yep.  And now it gets fun.

e.mccormick · Answer

There are two ways to go from that point.  You can use a LCM type solution to eliminate more, ir you can work with fractions.

anonymous · Answer

Im suppose to use Guass Jordan.......Im not sure what you mean.

e.mccormick · Answer

This is Guass Jordan.  I am talking about the algebra needed to eliminate thing in the second row.

anonymous · Answer

Im not really sure where to go from here...

e.mccormick · Answer

I meant second column.  Your first cokumn is done.  You have a sole row with a number in it.  This is sometimes called a pivot.  Anyhow, you have 1 alone in the first row, first column.  Now, you need to get the second row, second column to 1 and the other rows in the second column to 0.

anonymous · Answer

R2 = R2 - 2R3   ?

e.mccormick · Answer

Oh, that works nice.  Avoids the fractions.

anonymous · Answer

so...

   1    -1    -1    1
  0       1     0     9
   0      3     8     7

e.mccormick · Answer

OK.  They set that up nice.  So now use your new R2 to eliminate the 3 in R3C2

anonymous · Answer

that would give....

1   -1   -1   1
0     1    0    9
0     0    8    -20

e.mccormick · Answer

Yep... and you need to do the same sort of thing to R1.  Get rid of C2 there.

anonymous · Answer

can I use R1 + R2 = R1 ?

e.mccormick · Answer

Yep.

anonymous · Answer

that would give...

1   0   -1   10
0   1     0    9
0   0    0      2

anonymous · Answer

sorry thats wrong..

e.mccormick · Answer

Yah.  LOL

anonymous · Answer

1   0   -1   10
0    1    0    9
0    0    2    -5

e.mccormick · Answer

OK.  Now you just need to work with the last row.  No way to avoid the fractions now.

anonymous · Answer

OK. But after that how do I take care of the -1 R1 C3?

anonymous · Answer

lol sorry I see it.

anonymous · Answer

It seems that these are always fairly tedious problems? Or something for a program to think about for large matrices?

e.mccormick · Answer

Well, there are some tools you learn later that can help, and there is software to do the work.  However, matrices allow you to do things that would be exceptionally hard without them.  Like Google.  It is a really huge matrix.  Or rotating pictures.   Or encryption.

anonymous · Answer

Google is a matrix?

e.mccormick · Answer

Yes.  They use a form of hastable lookup database, and that is based on matrixes.

e.mccormick · Answer

So, after you divide R3 by 2, and that gets you R3, and then add it to R1, what is your final matrix?

anonymous · Answer

1   0   0   25/2
0   1   0     9
0   0   1     5/2

e.mccormick · Answer

25/2?

anonymous · Answer

15/2

e.mccormick · Answer

Oh, I see.  You forgot it was -5 on the last line.  You also missed it in the answer.  So yah:

$\left[ \begin{array}{ccc|c}
1 & 0 & 0 &  \dfrac{15}{2}\
0 & 1 & 0 & 9\
 0 & 0 & 1 & - \dfrac{5}{2}\
\end{array}\right ]$



$x_1 + 0 + 0 =  \dfrac{15}{2}$

$0 + x_2 + 0 = 9$

$0 + 0 + x_3 = -\dfrac{5}{2}$

So there you are.

e.mccormick · Answer

And that is Gauss-Jordan.

e.mccormick · Answer

Here is another one I worked for someone that was interested in learning how to do these. It has a link to a PDF on the methods: http://openstudy.com/users/e.mccormick#/updates/5275c700e4b00c1700b26a41

e.mccormick · Answer

And if you end up doing any Matrix Multiplication, this was my small writeup on that: http://assets.openstudy.com/updates/attachments/515c4ceee4b0507ceba37899-e.mccormick-1369021610530-matrixmultiplication.png

anonymous · Answer

Thank you!