solve the system of linear equations
using the Gauss–Jordan elimination method.
2x-3y=8
4x+y=-2

- UsukiDoll

solve the system of linear equations
using the Gauss–Jordan elimination method.
2x-3y=8
4x+y=-2

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- UsukiDoll

|dw:1361067093589:dw|

- UsukiDoll

|dw:1361067118822:dw|

- UsukiDoll

ugh I missed that one on my paper! that subtraction because -16-(-2) is - 14

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## More answers

- UsukiDoll

anyway|dw:1361067211535:dw|

- UsukiDoll

|dw:1361067247174:dw|

- UsukiDoll

|dw:1361067302376:dw|

- Callisto

|dw:1361067336310:dw|

- UsukiDoll

|dw:1361067340708:dw|

- Callisto

|dw:1361067371987:dw|

- UsukiDoll

I'm using elementary row operations.

- UsukiDoll

oh crud

- UsukiDoll

spotted anotehr error. not my day today

- UsukiDoll

|dw:1361067436003:dw|

- UsukiDoll

arghhh...

- UsukiDoll

ok really...I put this in reduced row echleon form. got my solution, but geez doesn't even equal back into the equation

- Callisto

Hmm... Try to do it again...

- Callisto

Since you've made a mistake in the first operation, it's hard to get the right answer :|

- UsukiDoll

something is wrong with the y

- UsukiDoll

ok let's redo this

- UsukiDoll

|dw:1361067604121:dw|

- UsukiDoll

augmented matrix [a b]

- UsukiDoll

|dw:1361067630686:dw| triangles indicate main diagonal

- UsukiDoll

so I have to get rid of the 4 and the -3.

- UsukiDoll

2row 1 - row 2 --> row2

- Callisto

Hmm.. My usual practice is -2R1 + R2 -> R2

- UnkleRhaukus

do you mean \[R_2\to R_2-2R_1\] @UsukiDoll

- UsukiDoll

yes

- UsukiDoll

|dw:1361067762579:dw|

- UsukiDoll

|dw:1361067803078:dw|

- UsukiDoll

so now... to get rid of the -3... 7row1 -3row2 --->row1

- Callisto

Why not do this: -1/7 R2 -> R2

- UsukiDoll

achhhhhhhhhhhh

- UsukiDoll

I don't wanna deal with fractions

- Callisto

But the answers are fractions :P

- UnkleRhaukus

|dw:1361067930957:dw|

- UsukiDoll

arghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I feel like Rarity Belle now

- Callisto

|dw:1361067978329:dw|

- UsukiDoll

oh wait. sign errors everywhere. Is that why I am getting frustrateD?

- UsukiDoll

wait a sec something is wrong already with row 2

- UsukiDoll

|dw:1361068119755:dw|

- UsukiDoll

|dw:1361068152547:dw|

- Callisto

Relax~~~
|dw:1361068054013:dw|

- UsukiDoll

|dw:1361068165204:dw|

- UsukiDoll

argh there could be so many combinations with this thing!

- Callisto

|dw:1361068179917:dw|

- UsukiDoll

|dw:1361068217096:dw|

- UsukiDoll

|dw:1361068228254:dw|

- Callisto

Just stop for a while, if you don't mind?

- UsukiDoll

arghhhhhhhhhhhhh doing that got me x = 7

- Callisto

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|

- UsukiDoll

big numbers -_-

- UsukiDoll

that means to get rid of the 6 I have to do 7r1 - 6r2

- Callisto

No?!

- UsukiDoll

or -7r1+6r2

- UsukiDoll

-.-

- UsukiDoll

|dw:1361068464895:dw|

- UsukiDoll

|dw:1361068472375:dw| main diagonal in triangles

- UsukiDoll

|dw:1361068495592:dw|

- UsukiDoll

|dw:1361068548208:dw|

- Callisto

Yup..Do you want to deal with big numbers or fractions?

- UsukiDoll

|dw:1361068571208:dw|

- UsukiDoll

I don't want to deal with fractions

- UsukiDoll

|dw:1361068639980:dw|

- Callisto

Yes!!!

- UsukiDoll

|dw:1361068650303:dw|

- UsukiDoll

|dw:1361068693152:dw|

- UsukiDoll

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

- UsukiDoll

just earlier I got x = -1 and y = -2

- Callisto

Hmm.. Careless mistakes perhaps?!

- UsukiDoll

now x = 1/7 y = -18/7

- UsukiDoll

what!

- Callisto

The one x=1/7; y=-18/7 is right :\

- UsukiDoll

-.- is there more than one answer to this? There could be many many things that I could have done here.

- Callisto

No.

- UnkleRhaukus

x = 1/7
y = -18/7 \(\LARGE\checkmark\)

- Callisto

There are only 3 cases for system of linear equations..
1. No solution
2. Exactly one solution
3. Infinitely many solution.

- UsukiDoll

I had a truckload

- UnkleRhaukus

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 .

- Callisto

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

- UsukiDoll

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.

- UsukiDoll

Gauss Jordan is worse than Gaussian

- UsukiDoll

ohhhhh what happens if I did Gaussian and then Gauss Jordon

- UsukiDoll

because the answers to the Gaussian are supposed to be the same as the Gauss Jordon

- Callisto

Gaussian Jordan = Gaussian then Jordan lol!!!

- UsukiDoll

but sometimes when I put the matrix into echleon form I can see the steps to making it row reduced

- UsukiDoll

|dw:1361069849783:dw|

- Callisto

Guassian told you to make the left an upper triangular matrix
Jordan asked you to further make the left an identity matrix

- Callisto

*Gaussian

- UsukiDoll

|dw:1361069869054:dw|

- UsukiDoll

|dw:1361069916982:dw|

- UsukiDoll

-____________________- this is just Gaussian... WHAT THEEEEE!!!!!!! I see y = -18/7 oh I haven't solved yet oops

- UsukiDoll

|dw:1361070002077:dw|

- Callisto

Just a minute..

- UsukiDoll

|dw:1361070023727:dw|

- UsukiDoll

uighhhhhhhhhhhhhhhh mind blown

- UsukiDoll

Gaussian and then Gaussian Jordan yeahhhhh

- UsukiDoll

I found the x = 1/7 just doing the Gaussian

- Callisto

|dw:1361070037568:dw|

- UsukiDoll

yeahhhh...now I really should've done Gaussian first and then the Jordan version

- Callisto

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.

- UsukiDoll

dang that means I have to redo some practice problems...gawd not again wasting paper here.

- UsukiDoll

oh yeah now I got it

- UsukiDoll

yup works. I went way too ahead... x.x

- UsukiDoll

so now I gotta correct all of these problems and write a proof...nice X___X

- UsukiDoll

Callisto do you know how to write proofs? I got a partial first draft and I was wondering if you could critique it?

- Callisto

Are you doing high school maths or college maths?

- UsukiDoll

college math

- Callisto

Proofs of??

- UsukiDoll

like the one I am currently writing is a contradiction proof. contradicts a theorem at least 2-3 times

- UsukiDoll

umm should I type it?

- Callisto

I.. didn't know what was a proof :S Please type it, if you don't mind!

- Callisto

*that instead of what

- UsukiDoll

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.

- UsukiDoll

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.

- UsukiDoll

It also has something to do with the homogenous system as well.

- UsukiDoll

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!

- UsukiDoll

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

- UsukiDoll

If a matrix is nonsingular, an inverse exists...so Theorem 2.9 doesn't work at all

- Callisto

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..

- UsukiDoll

oh yeah this is just a rough draft of it

- UsukiDoll

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

- Callisto

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?!

- UsukiDoll

I know. What if we let A be the inverse of B?

- UsukiDoll

that would be Matrix B
A = B^-1
BB^-1 = identity matrix

- Callisto

What if \(A \ne B^{-1}\) and \(B \ne A^{-1}\) ?

- UsukiDoll

I think that the only way that occurs is if A and B are singular

- Callisto

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.

- hartnn

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...

- hartnn

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

- hartnn

*contradiction, * neither A nor B can be singular

- UsukiDoll

omg how did u get that so fast?

- hartnn

because i knew the property that singularity means determinant = 0.....so something related to determinants should be there... and also, |AB|= |A||B|

- UsukiDoll

I haven't learned determinants yet. That's chapter 3 in my book

- UsukiDoll

What if I'm forced to use that Theorem? then what?

- hartnn

but can we use that if AB is non-singular, then |AB| not =0
....grrrr, i guess not ..

- UsukiDoll

nnooo...

- UsukiDoll

using advanced things aren't allowed otherwise my prof will get suspicious.

- UsukiDoll

determinants is a topic that my class didn't even learn yet, so if I put that on my proof, then I will get into trouble

- UsukiDoll

he was like DON'T USE ANYTHING THAT WASN'T INTRODUCED!!!!!!!!! IT'S A BAD IDEA

- hartnn

ok ok wait....

- hartnn

whats wrong with this logic ?
" 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

What if B =/= inverse of A?

- UsukiDoll

what if B isn't the inverse of A?

- UsukiDoll

:/

- UsukiDoll

so many contradictions everywhere. That's all I can get out of this thing.

- UsukiDoll

nontrivial solution in a homogenous system is NOT a 0, so how the heck can Ax =0?

- UsukiDoll

I gotta go...bye... x__X

- hartnn

just last small try.

- hartnn

let A and B be singular, then Ax =0 and Bx =0 [not sure about this step.]-----------------> (AB)x = 0 has a non-trivial solution,----->AB = singular.
------>a contradiction,
not sure whether it works....

- UsukiDoll

^ actually it does. thanks :D

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