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UsukiDoll
solve the system of linear equations using the Gauss–Jordan elimination method. 2x-3y=8 4x+y=-2
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ugh I missed that one on my paper! that subtraction because -16-(-2) is - 14
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I'm using elementary row operations.
spotted anotehr error. not my day today
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ok really...I put this in reduced row echleon form. got my solution, but geez doesn't even equal back into the equation
Hmm... Try to do it again...
Since you've made a mistake in the first operation, it's hard to get the right answer :|
something is wrong with the y
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augmented matrix [a b]
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so I have to get rid of the 4 and the -3.
2row 1 - row 2 --> row2
Hmm.. My usual practice is -2R1 + R2 -> R2
do you mean \[R_2\to R_2-2R_1\] @UsukiDoll
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so now... to get rid of the -3... 7row1 -3row2 --->row1
Why not do this: -1/7 R2 -> R2
I don't wanna deal with fractions
But the answers are fractions :P
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arghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I feel like Rarity Belle now
oh wait. sign errors everywhere. Is that why I am getting frustrateD?
wait a sec something is wrong already with row 2
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Relax~~~ |dw:1361068054013:dw|
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argh there could be so many combinations with this thing!
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Just stop for a while, if you don't mind?
arghhhhhhhhhhhhh doing that got me x = 7
This 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|
that means to get rid of the 6 I have to do 7r1 - 6r2
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|dw:1361068472375:dw| main diagonal in triangles
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Yup..Do you want to deal with big numbers or fractions?
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I don't want to deal with fractions
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ok why the heck am I getting different answers despite the fact that I got it into reduced row echleon form. that's just...grrrrrrrrrrrrrrrrrr
just earlier I got x = -1 and y = -2
Hmm.. Careless mistakes perhaps?!
now x = 1/7 y = -18/7
The one x=1/7; y=-18/7 is right :\
-.- is there more than one answer to this? There could be many many things that I could have done here.
x = 1/7 y = -18/7 \(\LARGE\checkmark\)
There are only 3 cases for system of linear equations.. 1. No solution 2. Exactly one solution 3. Infinitely many solution.
A quick string of row operations \[R_2\to R_2-2R_1\\R_2\to R_2/7\\ R_1\to R_1+3R_2\\R_1\to R_1/2\], 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 .
1. No solution with a row that has all zero entries on the left, but non-zero 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
but...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.
Gauss Jordan is worse than Gaussian
ohhhhh what happens if I did Gaussian and then Gauss Jordon
because the answers to the Gaussian are supposed to be the same as the Gauss Jordon
Gaussian Jordan = Gaussian then Jordan lol!!!
but sometimes when I put the matrix into echleon form I can see the steps to making it row reduced
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Guassian told you to make the left an upper triangular matrix Jordan asked you to further make the left an identity matrix
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-____________________- this is just Gaussian... WHAT THEEEEE!!!!!!! I see y = -18/7 oh I haven't solved yet oops
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uighhhhhhhhhhhhhhhh mind blown
Gaussian and then Gaussian Jordan yeahhhhh
I found the x = 1/7 just doing the Gaussian
yeahhhh...now I really should've done Gaussian first and then the Jordan version
Gaussian -> Make the left an upper triangular matrix Jordan -> Make the left an identity matrix Gaussian-Jordan => Make it an upper triangular matrix, then further make it an identity matrix. That's what Gaussan-Jordan 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.
dang that means I have to redo some practice problems...gawd not again wasting paper here.
yup works. I went way too ahead... x.x
so now I gotta correct all of these problems and write a proof...nice X___X
Callisto do you know how to write proofs? I got a partial first draft and I was wondering if you could critique it?
Are you doing high school maths or college maths?
like the one I am currently writing is a contradiction proof. contradicts a theorem at least 2-3 times
umm should I type it?
I.. didn't know what was a proof :S Please type it, if you don't mind!
okkk... 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.
There'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.
It also has something to do with the homogenous system as well.
because 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!
so, 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
If a matrix is nonsingular, an inverse exists...so Theorem 2.9 doesn't work at all
Forgive 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..
oh yeah this is just a rough draft of it
hmm 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
Hmm.. \[AA^{-1} = A^{-1}A=I\]provided that \(A^{-1}\) exists... ------------- But what about \(B \ne A^{-1}\)? And we also have to show that B is also non-singular?!
I know. What if we let A be the inverse of B?
that would be Matrix B A = B^-1 BB^-1 = identity matrix
What if \(A \ne B^{-1}\) and \(B \ne A^{-1}\) ?
I think that the only way that occurs is if A and B are singular
Hmm.. 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.
is it necessary to use that theorem ? can't we go like this way ? if AB is non-sing., \(|AB| \ne 0 \implies |A||B|\ne 0\) for that both |A|and |B| must not =0, so...
if need to use contradiction, then let any1 of the A or B be singular, say A , then |A| =0, then |A||B|=0 , then |AB|=0 which is a contardiction....hence, neither A nor B can be 0
*contradiction, * neither A nor B can be singular
omg how did u get that so fast?
because i knew the property that singularity means determinant = 0.....so something related to determinants should be there... and also, |AB|= |A||B|