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 one year ago
solve the system of linear equations
using the Gauss–Jordan elimination method.
2x3y=8
4x+y=2
 one year ago
solve the system of linear equations using the Gauss–Jordan elimination method. 2x3y=8 4x+y=2

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067093589:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067118822:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ugh I missed that one on my paper! that subtraction because 16(2) is  14

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0anywaydw:1361067211535:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067247174:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067302376:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3dw:1361067336310:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067340708:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3dw:1361067371987:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I'm using elementary row operations.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0spotted anotehr error. not my day today

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067436003:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok really...I put this in reduced row echleon form. got my solution, but geez doesn't even equal back into the equation

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Hmm... Try to do it again...

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Since you've made a mistake in the first operation, it's hard to get the right answer :

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0something is wrong with the y

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067604121:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0augmented matrix [a b]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067630686:dw triangles indicate main diagonal

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so I have to get rid of the 4 and the 3.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.02row 1  row 2 > row2

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Hmm.. My usual practice is 2R1 + R2 > R2

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0do you mean \[R_2\to R_22R_1\] @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067762579:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067803078:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so now... to get rid of the 3... 7row1 3row2 >row1

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Why not do this: 1/7 R2 > R2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I don't wanna deal with fractions

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3But the answers are fractions :P

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361067930957:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0arghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I feel like Rarity Belle now

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3dw:1361067978329:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh wait. sign errors everywhere. Is that why I am getting frustrateD?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0wait a sec something is wrong already with row 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068119755:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068152547:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Relax~~~ dw:1361068054013:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068165204:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0argh there could be so many combinations with this thing!

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3dw:1361068179917:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068217096:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068228254:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Just stop for a while, if you don't mind?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0arghhhhhhhhhhhhh doing that got me x = 7

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3This is basically what you do for the first row operation 2R1 + R2 > R2. I split it into two steps so that you can catch up the the arithmetic easier (I hope..) dw:1361068290374:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0that means to get rid of the 6 I have to do 7r1  6r2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068464895:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068472375:dw main diagonal in triangles

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068495592:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068548208:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Yup..Do you want to deal with big numbers or fractions?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068571208:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I don't want to deal with fractions

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068639980:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068650303:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068693152:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok why the heck am I getting different answers despite the fact that I got it into reduced row echleon form. that's just...grrrrrrrrrrrrrrrrrr

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0just earlier I got x = 1 and y = 2

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Hmm.. Careless mistakes perhaps?!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0now x = 1/7 y = 18/7

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3The one x=1/7; y=18/7 is right :\

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0. is there more than one answer to this? There could be many many things that I could have done here.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0x = 1/7 y = 18/7 \(\LARGE\checkmark\)

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3There are only 3 cases for system of linear equations.. 1. No solution 2. Exactly one solution 3. Infinitely many solution.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0A quick string of row operations \[R_2\to R_22R_1\\R_2\to R_2/7\\ R_1\to R_1+3R_2\\R_1\to R_1/2\], if you have gotten a different solution you must have a made a mistake somewhere, it is very easy to make mistakes using Gauss–Jordan elimination .

Callisto
 one year ago
Best ResponseYou've already chosen the best response.31. No solution with a row that has all zero entries on the left, but nonzero entry on right => system is inconsistent 2. Exactly one solution After elementary row operations, no. of rows = no. of unknowns (/columns) => system is consistent 3. Infinitely many solution. After elementary row operations, no. of rows < no. of unknowns => system is consistent

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0but...yeah I suppose because I have seen some matrices that had a row of zeros on the left but with a nonzero on the right that is an inconsistent system .. How is it possible to make a mistake when it's so hard to spot one? I mean...I could've used different operations on the matrix. The combinations are everywhere.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0Gauss Jordan is worse than Gaussian

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhh what happens if I did Gaussian and then Gauss Jordon

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0because the answers to the Gaussian are supposed to be the same as the Gauss Jordon

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Gaussian Jordan = Gaussian then Jordan lol!!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0but sometimes when I put the matrix into echleon form I can see the steps to making it row reduced

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361069849783:dw

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Guassian told you to make the left an upper triangular matrix Jordan asked you to further make the left an identity matrix

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361069869054:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361069916982:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0____________________ this is just Gaussian... WHAT THEEEEE!!!!!!! I see y = 18/7 oh I haven't solved yet oops

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361070002077:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361070023727:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0uighhhhhhhhhhhhhhhh mind blown

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0Gaussian and then Gaussian Jordan yeahhhhh

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I found the x = 1/7 just doing the Gaussian

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3dw:1361070037568:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yeahhhh...now I really should've done Gaussian first and then the Jordan version

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Gaussian > Make the left an upper triangular matrix Jordan > Make the left an identity matrix GaussianJordan => Make it an upper triangular matrix, then further make it an identity matrix. That's what GaussanJordan is! Jordon helps you to get the answer by matching the left and the right That is for the row [1 0  1/7], you can tell immediately that x = 1/7 It's just the same as doing Gaussian elimination then back substitution.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dang that means I have to redo some practice problems...gawd not again wasting paper here.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah now I got it

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yup works. I went way too ahead... x.x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so now I gotta correct all of these problems and write a proof...nice X___X

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0Callisto do you know how to write proofs? I got a partial first draft and I was wondering if you could critique it?

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Are you doing high school maths or college maths?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0like the one I am currently writing is a contradiction proof. contradicts a theorem at least 23 times

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0umm should I type it?

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3I.. didn't know what was a proof :S Please type it, if you don't mind!

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3*that instead of what

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0okkk... Let A and B be n x n matrices. Show that if AB is nonsingular then A and B must be nonsingular.(Hint Use Theorem 2.9) Theorem 2.9 states that the homogeneous system of n linear equations in n unknowns Ax = 0 has a nontrivial solution if and only if A is singular.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0There's a contradiction....namely because if Matrix A is singular, then the inverse of Matrix A does NOT exist which means that AB doesn't exist as well.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0It also has something to do with the homogenous system as well.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0because basically a homogeneous system is always consistent. However, a homogeneous system's solution is always 0 and it's a trivial solution. Nontrivial solution in a homogenous system means that the solution IS NOT 0 at all!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so, what is given before proving is that A and B are n x n matrices A and B must be nonsingular AB is also nonsingular Theorem 2.9

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0If a matrix is nonsingular, an inverse exists...so Theorem 2.9 doesn't work at all

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Forgive me, I don't understand the part "... which means that AB doesn't exist as well" I understand if A is singular, A^(1) doesn't exist, but not the which means part..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah this is just a rough draft of it

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0hmm if A is nonsingular, an inverse exists. I remember back from Theorem 1.5 that it was AB=BA=In. B is the inverse of A. hmmm..if A is singular, yes the inverse doesn't exist. Let's have B = the inverse of A. Matrix A is singular Inverse doesn't exist so B doesn't exist AB is impossible to achieve unless A is nonsingular

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Hmm.. \[AA^{1} = A^{1}A=I\]provided that \(A^{1}\) exists...  But what about \(B \ne A^{1}\)? And we also have to show that B is also nonsingular?!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I know. What if we let A be the inverse of B?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0that would be Matrix B A = B^1 BB^1 = identity matrix

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3What if \(A \ne B^{1}\) and \(B \ne A^{1}\) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I think that the only way that occurs is if A and B are singular

Callisto
 one year ago
Best ResponseYou've already chosen the best response.3Hmm.. I'm sorry that I don't know how to do it :( @hartnn Would you mind giving a hand here? The problem we are discussing now is: Let A and B be n x n matrices. Show that if AB is nonsingular then A and B must be nonsingular.(Hint Use Theorem 2.9) Theorem 2.9 states that the homogeneous system of n linear equations in n unknowns Ax = 0 has a nontrivial solution if and only if A is singular.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1is it necessary to use that theorem ? can't we go like this way ? if AB is nonsing., \(AB \ne 0 \implies AB\ne 0\) for that both Aand B must not =0, so...