Help with matrices needed - Tilted Forum Project Discussion Community

essendoubleop · 03-10-2005, 09:22 PM

I have a 3x4 matrix and I need to put it in reduced row echelon form. I plugged it into the calculator and found the answer should be (10,2,3), but I can't figure out how I got the first number wrong. Here are each step I took:

5x + 9y - 7z = 47
4x - 3y + 2z = 40
9x + 8y - 3z = 97

{5 9 -7 47
4 -3 2 40
9 8 -3 97}

{1 12 -9 7
4 -3 2 40
9 8 -3 97}

{1 12 -9 7
0 -51 38 12
0 -100 78 34}

{1 0 3 (277/3)
0 -1 -1 -5
0 -100 78 34}

{1 0 3 (277/3)
0 1 1 5
0 0 178 534}

{1 0 0 (250/3)
0 1 1 5
0 0 178 534}

{1 0 0 (250/3)
0 1 0 2
0 0 1 3}

By my answer I get (250/3), 2, 3. I can't figure out what I did wrong.

tres · 03-11-2005, 04:37 AM

Maybe I should have paid attention in school... That looks like um...a bunch of numbers.... good luck with that....

filtherton · 03-11-2005, 07:31 AM

How did you go from this:

{1 12 -9 7
0 -51 38 12
0 -100 78 34}

to this:

{1 0 3 (277/3)
0 -1 -1 -5
0 -100 78 34}
?

How did you get -1 -1 -5 for the second row?
Maybe it's just early, but i don't see it.

essendoubleop · 03-11-2005, 08:59 AM

Quote:

Originally Posted by filtherton

How did you go from this:

{1 12 -9 7
0 -51 38 12
0 -100 78 34}

to this:

{1 0 3 (277/3)
0 -1 -1 -5
0 -100 78 34}
?

How did you get -1 -1 -5 for the second row?
Maybe it's just early, but i don't see it.

(25/3) R1 + R3 --> R1
R2 - (1/2) R3 --> R2

vbuser · 03-11-2005, 05:27 PM

Can you edit your post to include each operation done the row?

vinaur · 03-11-2005, 05:46 PM

Truly, I can't even see how you got the first step: 5 9 -7 47 to 1 12 -9 7. Wouldn't you want to divide the first row by 5??? So you get 1 9/5 -7/5 47/5.

spectre · 03-11-2005, 09:18 PM

This should get you started (as long as I didn't screw anything up):

5x + 9y - 7z = 47
4x - 3y + 2z = 40
9x + 8y - 3z = 97

5 9 -7 47
4 -3 2 40
9 8 -3 97

(1/5)R1 =>
1 9/5 -7/5   47/5
4 -3    2      40
9   8  -3      97

-4R1 + R2 =>
-9R1 + R3 =>
1 9/5   -7/5   47/5
0 -51/5 38/5 12/5
0 -41/5 48/5 62/5

FloydianOne · 03-11-2005, 10:40 PM

Here is one of my professor's textbooks on the subject... It explains it a little bit.

http://www.math.gatech.edu/~bourbaki...eb-notes/3.pdf

You can read the whole text here...

http://www.math.gatech.edu/%7Ecarlen...S04/index.html

essendoubleop · 03-12-2005, 12:48 AM

Thanks for the help. I decided to download a program called MATRIX that really helped me out.

FloydianOne · 03-12-2005, 11:08 AM

Quote:

Originally Posted by essendoubleop

Thanks for the help. I decided to download a program called MATRIX that really helped me out.

Sounds interesting... give us a link

essendoubleop · 03-13-2005, 12:07 AM

Quote:

Originally Posted by FloydianOne

Sounds interesting... give us a link

http://members.shaw.ca/mtx/matrix.htm

03-10-2005, 09:22 PM	#1 (permalink)
essendoubleop Psycho	Help with matrices needed I have a 3x4 matrix and I need to put it in reduced row echelon form. I plugged it into the calculator and found the answer should be (10,2,3), but I can't figure out how I got the first number wrong. Here are each step I took: 5x + 9y - 7z = 47 4x - 3y + 2z = 40 9x + 8y - 3z = 97 {5 9 -7 47 4 -3 2 40 9 8 -3 97} {1 12 -9 7 4 -3 2 40 9 8 -3 97} {1 12 -9 7 0 -51 38 12 0 -100 78 34} {1 0 3 (277/3) 0 -1 -1 -5 0 -100 78 34} {1 0 3 (277/3) 0 1 1 5 0 0 178 534} {1 0 0 (250/3) 0 1 1 5 0 0 178 534} {1 0 0 (250/3) 0 1 0 2 0 0 1 3} By my answer I get (250/3), 2, 3. I can't figure out what I did wrong.

03-11-2005, 09:18 PM	#7 (permalink)
spectre Junkie	This should get you started (as long as I didn't screw anything up): 5x + 9y - 7z = 47 4x - 3y + 2z = 40 9x + 8y - 3z = 97 5 9 -7 47 4 -3 2 40 9 8 -3 97 (1/5)R1 => 1 9/5 -7/5   47/5 4 -3    2      40 9   8  -3      97 -4R1 + R2 => -9R1 + R3 => 1 9/5   -7/5   47/5 0 -51/5 38/5 12/5 0 -41/5 48/5 62/5 __________________ "Fuck these chains No goddamn slave I will be different" ~ Machine Head Last edited by spectre; 03-11-2005 at 09:28 PM..

03-11-2005, 10:40 PM	#8 (permalink)
FloydianOne Crazy	Here is one of my professor's textbooks on the subject... It explains it a little bit. http://www.math.gatech.edu/~bourbaki...eb-notes/3.pdf You can read the whole text here... http://www.math.gatech.edu/%7Ecarlen...S04/index.html __________________ Fight apathy! ..... or dont.

03-11-2005, 04:37 AM	#2 (permalink)
tres Insane Location: Long Island, NY	Maybe I should have paid attention in school... That looks like um...a bunch of numbers.... good luck with that....

03-11-2005, 07:31 AM	#3 (permalink)
filtherton Junkie Location: In the land of ice and snow.	How did you go from this: {1 12 -9 7 0 -51 38 12 0 -100 78 34} to this: {1 0 3 (277/3) 0 -1 -1 -5 0 -100 78 34} ? How did you get -1 -1 -5 for the second row? Maybe it's just early, but i don't see it.

03-11-2005, 05:27 PM	#5 (permalink)
vbuser Upright	Can you edit your post to include each operation done the row?

03-11-2005, 05:46 PM	#6 (permalink)
vinaur Insane	Truly, I can't even see how you got the first step: 5 9 -7 47 to 1 12 -9 7. Wouldn't you want to divide the first row by 5??? So you get 1 9/5 -7/5 47/5.

03-12-2005, 12:48 AM	#9 (permalink)
essendoubleop Psycho	Thanks for the help. I decided to download a program called MATRIX that really helped me out.