Linear Algebra ****ing SUCKS

[semtex]

this shyt.


can someone Gaussian Elimination algorithm in a clear manner please?

 

[Megatron Island]

Trust me breh, once you get passed the first month or so of that course and get used to matrices, you'll get through that class easy. I'm willing to bet that's your main problem right now.

Do you need help with deriving the operator matrix?

Just remember, your final goal for Gaussian Elimination is getting the matrix in row-echelon form.

That means maybe having to move around rows. Then you can start multiplying a row by a scalar, then adding that scalar multiple of a row to another row in order to make the first element of that row a zero. Rinse and repeat. When you need a leading one, do the samething except combine rows in a way to get that leading 1 instead of a zero. Sometimes you'll have to multiply by a fraction though and it can get messy.

Example:
2x+y-z=3
X-y+2=1
-4x+6y-7z=1
2x+0y+z=3

you want this 4x4 matrix in a form:
1 a b c
0 1 d e
0 0 1 f
0 0 0 0
Which is row echelon form. (a,b,c,d,e,f are just some #'s).

The matrix as it stands right now is:

2 1 -1 3
1 -1 2 1
-4 6 -7 1
2 0 1 3

So first thing is you should exchange rows 1 and 2 so you'll have a leading one in a11

After flipping the rows, you have:
1 -1 2 1
2 1 -1 3
-4 6 -7 1
2 0 1 3

Now you have to get the rest of the elements under that column to equal zero like:
1
0
0
0

You do this by:
Multiplying the first row row by -2 and adding it to row 2
Multiplying the 1st row by 4 and adding it to row 3
Multiplying the first row by -2 and adding it to row 4.

So first you changed rows 1 and 2 described by: P12
Then you did all that mutiplying and adding described by: A12(-2); A13(4); A14(-2)

This will leave you with:
1 -1 2 1
0 3 -5 1
0 2 1 5
0 2 -3 1

Now you need a leading 1 in the 2nd row to replace that 3. Which means you can multiply row 3 by -1 and add that to the 2nd row because you're trying to get row-echelon form. Then you need to have zeros under that 1 in the 2nd column, and you do that similarly to how we did that earlier. Afterward you'll find that you'll have to multiply the third row by 1/13 though to get a leading 1.

Eventually you'll have row-echelon form.

Eventually the bottom equation will all be zero and you'll be left with:
1 -1 2 1
0 1 -6 -4
0 0 1 1
0 0 0 0

So now the new system of linear equations is:
X-y+2z=1
0x+y-6z=-4
0x+0y+z=1

Which means from the 3rd row, z=1. Throw that value back into the system of linear equations to solve for x and y.

It's crazy because all we pretty much did was do the samething we've done to previous systems of linear equations in lower math classes, but using matrix notation.

Hope that helped.

 

[Julius Skrrvin]

the worst shyt

 

[Serious]

sh*t looks like Alegbreh II

 

[Dafunkdoc_Unlimited]

Thanks to the Internet, Linear Algebra is EZ-Mode.....

Linear Algebra | Khan Academy

 

[Megatron Island]

sh*t looks like Alegbreh II

What happened technically was Algebra II, but in matrix notation. That's how the course works, the first month doing matrix operations and determinants, then it goes on into vector spaces, eigen vectors, transformations, then you learn how to tie it all into solving systems of differential equations.

 

[O.iatlhawksfan]

If yu need help with yo homework try mathway.com

 

[tru_m.a.c]

the power of the internet

where were y'all 6 yrs ago!?!?!?!?

 

[Serious]

What happened technically was Algebra II, but in matrix notation. That's how the course works, the first month doing matrix operations and determinants, then it goes on into vector spaces, eigen vectors, transformations, then you learn how to tie it all into solving systems of differential equations.

Makes sense to refresh

Come to think of it, I used love doing Matrices.

*looks up Matrix problems*

 

[Habit]

sh*t looks like Alegbreh II

It is, I guess they revisit it in Linear Algebra. I was pretty good at getting matrix into row echelon form.

I think linear algebra is a junior level class in college.

He must be a Computer Science major.

I hate math, theorems, postulates and theory, so I avoided any math and calculation intensive majors. Buildings and roads are not supposed to move, so I can those calculations by counting my ten fingers.

 

[Dat Migo]

Graph Theory also fukking sucks, appreciated if anyone has some good tutorials/resources for this class...

 

[Type Username Here]

semtex, post a specific problem and we'll show you how it's possible to take shortcuts. Megatron did a pretty good job explaining it.

 

[semtex]

Trust me breh, once you get passed the first month or so of that course and get used to matrices, you'll get through that class easy. I'm willing to bet that's your main problem right now.

Do you need help with deriving the operator matrix?

Just remember, your final goal for Gaussian Elimination is getting the matrix in row-echelon form.

That means maybe having to move around rows. Then you can start multiplying a row by a scalar, then adding that scalar multiple of a row to another row in order to make the first element of that row a zero. Rinse and repeat. When you need a leading one, do the samething except combine rows in a way to get that leading 1 instead of a zero. Sometimes you'll have to multiply by a fraction though and it can get messy.

Example:
2x+y-z=3
X-y+2=1
-4x+6y-7z=1
2x+0y+z=3

you want this 4x4 matrix in a form:
1 a b c
0 1 d e
0 0 1 f
0 0 0 0
Which is row echelon form. (a,b,c,d,e,f are just some #'s).

The matrix as it stands right now is:

2 1 -1 3
1 -1 2 1
-4 6 -7 1
2 0 1 3

So first thing is you should exchange rows 1 and 2 so you'll have a leading one in a11

After flipping the rows, you have:
1 -1 2 1
2 1 -1 3
-4 6 -7 1
2 0 1 3

Now you have to get the rest of the elements under that column to equal zero like:
1
0
0
0

You do this by:
Multiplying the first row row by -2 and adding it to row 2
Multiplying the 1st row by 4 and adding it to row 3
Multiplying the first row by -2 and adding it to row 4.

So first you changed rows 1 and 2 described by: P12
Then you did all that mutiplying and adding described by: A12(-2); A13(4); A14(-2)

This will leave you with:
1 -1 2 1
0 3 -5 1
0 2 1 5
0 2 -3 1

Now you need a leading 1 in the 2nd row to replace that 3. Which means you can multiply row 3 by -1 and add that to the 2nd row because you're trying to get row-echelon form. Then you need to have zeros under that 1 in the 2nd column, and you do that similarly to how we did that earlier. Afterward you'll find that you'll have to multiply the third row by 1/13 though to get a leading 1.

Eventually you'll have row-echelon form.

Eventually the bottom equation will all be zero and you'll be left with:
1 -1 2 1
0 1 -6 -4
0 0 1 1
0 0 0 0

So now the new system of linear equations is:
X-y+2z=1
0x+y-6z=-4
0x+0y+z=1

Which means from the 3rd row, z=1. Throw that value back into the system of linear equations to solve for x and y.

It's crazy because all we pretty much did was do the samething we've done to previous systems of linear equations in lower math classes, but using matrix notation.

Hope that helped.

 

[semtex]

Makes sense to refresh

Come to think of it, I used love doing Matrices.

*looks up Matrix problems*

shyt was cake in high school it was just basic operations. The fanciest thing u did was determinants

 

[semtex]

semtex, post a specific problem and we'll show you how it's possible to take shortcuts. Megatron did a pretty good job explaining it.

I'll look but it's the concept in general not just a problem. How do u do this on a TI-84?