@kelliegirl33 matrix

Question

e.mccormick · Answer

And the matrix method, which is the same stuff but without the x, y, and z in the way:

$\left[\begin{array}{ccc|c}
3 & 2 & 1 & 7 \
5 & 5 & 4 & 3 \
3 & 2 & 3 & 1 
\end{array}\right]

\begin{matrix}
\leftarrow & \text{ R1}\
\leftarrow & \text{ R2}\
\leftarrow & \text{ R3}
\end{matrix}$

R3 + (-1) R1 $\rightarrow $ new R3

$\left[\begin{array}{ccc|c}
3 & 2 & 1 & 7 \
5 & 5 & 4 & 3 \
0 & 0 & 2 & -6 
\end{array}\right]$

R3 $\times \dfrac{1}{2}$

$\left[\begin{array}{ccc|c}
3 & 2 & 1 & 7 \
5 & 5 & 4 & 3 \
0 & 0 & 1 & -3 
\end{array}\right]$

R1 + (-1) R3 $\rightarrow $ new R1
R2 + (-4) R3 $\rightarrow $ new R2

$\left[\begin{array}{ccc|c}
3 & 2 & 0 & 10 \
5 & 5 & 0 & 15 \
0 & 0 & 1 & -3 
\end{array}\right]$

R2 $\times \dfrac{1}{5}$

$\left[\begin{array}{ccc|c}
3 & 2 & 0 & 10 \
1 & 1 & 0 & 3 \
0 & 0 & 1 & -3 
\end{array}\right]$
R1 + (-2) R2 $\rightarrow $ new R1

$\left[\begin{array}{ccc|c}
1 & 0 & 0 & 4 \
1 & 1 & 0 & 3 \
0 & 0 & 1 & -3 
\end{array}\right]$

R2 + (-1) R1 $\rightarrow $ new R2

$\left[\begin{array}{ccc|c}
1 & 0 & 0 & 4 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -3 
\end{array}\right]$

1x + 0y + 0z = 4
0x + 1y + 0z = -1
0x + 0y + 1z = -3 

x = 4
y = -1
z = -3

e.mccormick · Answer

This is shorthand.  I can explain it a bit more and how to do the side work, and you will be able to use it.

anonymous · Answer

I really appreciate this.....elimination is a pain in the rear :)

e.mccormick · Answer

In each column and each row on the left of the | you want only one thing.  The thing needs to end as 1 because then the right is the value for that column.

anonymous · Answer

got it

e.mccormick · Answer

So when I kill off the x and y at the same time it does not matter.  That gives me z.  Once I know z, I can use the 3rd equation to eliminate the z column out of the first two.

anonymous · Answer

oh...I see

e.mccormick · Answer

The fraction on row 2 then lets me clean it up some.  Then I used the 2nd row to get rid of y in the first row.  It happens to solve for x at the same time, which is not always true.  If it had not, I would have multiplied by a fraction there too.

anonymous · Answer

I will probably have to go over this a few times to get it down...it just seems easier this way. Easier then elimination. Thank you sooooooooo much :)

e.mccormick · Answer

It is actually a form of elimination.  Gauss/Jordan.  But it lets you see all the processes as you do them, which makes it easier to keep track.

e.mccormick · Answer

4 pages that go over it pretty well: http://www.epcc.edu/tutorialservices/valleverde/Documents/Gauss-Jordan_Method.pdf

anonymous · Answer

and with elimination, I messed up early on, and that mistake carries through to the end. I am going to practice on this. Thank you again . I would give you 100 medals for this if I could :)

anonymous · Answer

I will definitely read those 4 pages :)

e.mccormick · Answer

That also shows the side work, but really that is just wrting out the equations I did in short hand:
R3 + (-1) R1 $\rightarrow $ new R3

becomes:
(3  2  1  7 ) -1
3  2  3  1 

-3 -2 -1 -7 
 3  2  3  1 
----------
0 0 2 -6

Then you rewrite the matrix with the new row 3.

anonymous · Answer

ok...I see the steps.....I am understanding better now

e.mccormick · Answer

Also, when you do the side work, you can sometimes see the next operation, like multiplying through by half, and do that then.  Then write it out again.

anonymous · Answer

makes sense.....its not really that different from elimination, in a way

e.mccormick · Answer

And that is really it.  Oh, and it took 10 minutes... not 5.  LOL.

Exactly, it is elimination, but without writing x, y, and z, but with keeping track of all the equations so you can see what is happening the entire time.

e.mccormick · Answer

cutegirl seems to have gone silent on the other question....  /sigh  And there was not much left to do!

anonymous · Answer

I am so glad you tagged me to teach me this. In my opinion, you ARE one of the best on OS. I really mean that. You take the time to explain and not just give answers. Thumbs up to you :)

e.mccormick · Answer

Thanks.  Yah, well, the goal is learning.  I take it from the peer tutoring standpoint.  Discuss.  Work together.  Explain.  It takes more time, but it results in learning.

anonymous · Answer

cutegirl sent me a message saying she was back and I told her to come here...I copied and pasted the question. Did she ever show...did you notice ?

e.mccormick · Answer

Hehe.  Never noticed... but if all she needed was the answer, it is there...

anonymous · Answer

its a shame some people just want the answer instead of not learning how to do it. You should be a teacher. You would make an excellent one :)

e.mccormick · Answer

Lot of people tell me that.  Hehe.

anonymous · Answer

because it is the truth. Thanks again....teacher mccormick...hehe

e.mccormick · Answer

np.  have fun!

anonymous · Answer

you too :)