Math halp. Matricies, Determinants, EigenValues/Vectors Systems of equ

2009-09-07 14:09:48

#1

zeidmaan: Member; +234|6834|Vienna

I have couple of various questions.

1. When using matricies to solve a system of equasions do I need to get the "upper/lower triangular" (after endles row-operations) or is it enough that

I get one row with one value thats NOT ZERO, one row with 2 values that are NOT ZERO etc...

basically if I have a system with 4 unknowns (x, y, z, u) and 4 equasions do I need to get a matrix like:

n n n n | 6 (x) (numbers are random)
0 n n n | 5 (y)
0 0 n n | -9 (z)
0 0 0 n | 2 (u)

and from that extrapolate u, than z, than y, than x

or can I get a matrix like:

n 0 0 n | 6 (x) (numbers are random)
0 n n n | 5 (y)
n n n n | -9 (z)
0 0 0 n | 2 (u)

and from that extrapolate u, than x (because of 2 zeroes), than y, than z

2. Regarding Inverse Matricies.
As far as Ive figured:
A matrix is regular if its determinant is NOT ZERO
A Matrix is singular if its determinat is ZERO

If a matrix is singular than it DOES NOT have an inverse matrix
I think I have it right so far but does every REGULAR matrix have an inverse matrix?

SO if a determinant of matrix A != 0 doest that mean that A has to have an Inverse matrix?

More questions to come

2009-09-07 14:15:38

#2

.Sup: be nice; +2,646|6873|The Twilight Zone

dude cmon there's a thread for this
http://forums.bf2s.com/viewtopic.php?id=130201
j/k (no rly)

https://www.shrani.si/f/3H/7h/45GTw71U/untitled-1.png

2009-09-07 14:40:10

#3

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

zeidmaan wrote:
basically if I have a system with 4 unknowns (x, y, z, u) and 4 equasions do I need to get a matrix like:

n n n n | 6 (x) (numbers are random)
0 n n n | 5 (y)
0 0 n n | -9 (z)
0 0 0 n | 2 (u)

and from that extrapolate u, than z, than y, than x

Gaussian elimination (the row reduction stuff) should get the matrix down to this format. It makes back substituting a hell of a lot easier than the random on you posted after.

This format also allows you do then row reduce again (backwards), to get a diagonal matrix, which is used with eigenvectors and values to solve systems of ordinary differential equations.

Edit: Also, your x y z u should be along the top, not down the side, as each row of the matrix should represent a different equation (first column is the x coefficient, second the y coefficient and so on).

2009-09-07 14:44:59

#4

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

Blegh I hated this crap. You're still not even close to the hard stuff though

Linear Algebra FTL

2009-09-07 14:48:58

#5

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Winston_Churchill wrote:
Blegh I hatedloved this crap. You're still not even close to the hard stuff though

Linear Algebra FTLFTW

I'll get my notes on inverse matrices.

2009-09-07 14:52:00

#6

zeidmaan: Member; +234|6834|Vienna

mcminty wrote:
zeidmaan wrote:
basically if I have a system with 4 unknowns (x, y, z, u) and 4 equasions do I need to get a matrix like:

n n n n | 6 (x) (numbers are random)
0 n n n | 5 (y)
0 0 n n | -9 (z)
0 0 0 n | 2 (u)

and from that extrapolate u, than z, than y, than x
Gaussian elimination (the row reduction stuff) should get the matrix down to this format. It makes back substituting a hell of a lot easier than the random on you posted after.

This format also allows you do then row reduce again (backwards), to get a diagonal matrix, which is used with eigenvectors and values to solve systems of ordinary differential equations.

Edit: Also, your x y z u should be along the top, not down the side, as each row of the matrix should represent a different equation (first column is the x coefficient, second the y coefficient and so on).

I thought so on the first part, but are you sure about the columns representing unknowns?

I mean, If I replace every n with 5 (for example) doesnt that mean that
u=2-5=-3

If I do it your way how do I extrapolate u ?

Or did I misunderstand you?

2009-09-07 14:55:50

#7

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

zeidmaan wrote:
2. Regarding Inverse Matricies.
As far as Ive figured:
A matrix is regular if its determinant is NOT ZERO
A Matrix is singular if its determinat is ZERO

If a matrix is singular than it DOES NOT have an inverse matrix
I think I have it right so far but does every REGULAR matrix have an inverse matrix?

SO if a determinant of matrix A != 0 doest that mean that A has to have an Inverse matrix?

More questions to come

My notes say:

If a matrix has an inverse, then it is a non-singular matrix.
If a matrix has no inverse, then it is a singular matrix.

Existence of a matrix: For an n x n matrix A the inverse exists if and only if the rank of A is equal to n. This is equivalent to saying that the determinant of A is non-zero.

2009-09-07 14:58:44

#8

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

zeidmaan wrote:
mcminty wrote:
zeidmaan wrote:
basically if I have a system with 4 unknowns (x, y, z, u) and 4 equasions do I need to get a matrix like:

n n n n | 6 (x) (numbers are random)
0 n n n | 5 (y)
0 0 n n | -9 (z)
0 0 0 n | 2 (u)

and from that extrapolate u, than z, than y, than x
Gaussian elimination (the row reduction stuff) should get the matrix down to this format. It makes back substituting a hell of a lot easier than the random on you posted after.

This format also allows you do then row reduce again (backwards), to get a diagonal matrix, which is used with eigenvectors and values to solve systems of ordinary differential equations.

Edit: Also, your x y z u should be along the top, not down the side, as each row of the matrix should represent a different equation (first column is the x coefficient, second the y coefficient and so on).
I thought so on the first part, but are you sure about the columns representing unknowns?

I mean, If I replace every n with 5 (for example) doesnt that mean that
u=2-5=-3

If I do it your way how do I extrapolate u ?

Or did I misunderstand you?

err.. Yes, I'm sure.

Write out the system of equations you are trying to solve.

2009-09-07 15:02:02

#9

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S

2009-09-07 15:03:21

#10

zeidmaan: Member; +234|6834|Vienna

mcminty wrote:
zeidmaan wrote:
2. Regarding Inverse Matricies.
As far as Ive figured:
A matrix is regular if its determinant is NOT ZERO
A Matrix is singular if its determinat is ZERO

If a matrix is singular than it DOES NOT have an inverse matrix
I think I have it right so far but does every REGULAR matrix have an inverse matrix?

SO if a determinant of matrix A != 0 doest that mean that A has to have an Inverse matrix?

More questions to come
My notes say:

If a matrix has an inverse, then it is a non-singular matrix.
If a matrix has no inverse, then it is a singular matrix.

Existence of a matrix: For an n x n matrix A the inverse exists if and only if the rank of A is equal to n. This is equivalent to saying that the determinant of A is non-zero.

OK so I got that part right.

Singular (det=0) has no inverse, non-singular (det !=0) has inverse.
also
AA^-1 = A^-1A=In
etc

thanks so far...

2009-09-07 15:09:26

#11

zeidmaan: Member; +234|6834|Vienna

Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S

Yes thats what Im thinking. Each row represents an equasion. If I reduce the matrix to a lower-trianguar or what its called than that means that the 4th row represents the 4th unknown. So If I do bunch of row operations and get the matrix down to something like:

1 3 -4 2 | 4
0 2 1 -3 | 2
0 0 5 6 | 5
0 0 0 3 | 6

this means that the 4th unknown: u=6-3

Than I use that to get the rest... ?

2009-09-07 15:11:55

#12

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S

Yeah, each row represents a different equation. If each column didn't represent a variable in the equations, then what would it represent?

2009-09-07 15:12:11

#13

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

IIRC I think youre supposed to have those right columns' values at 0

2009-09-07 15:13:13

#14

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

mcminty wrote:
Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S
Yeah, each row represents a different equation. If each column didn't represent a variable in the equations, then what would it represent?

The column could be filled with 0s?

2009-09-07 15:16:56

#15

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

zeidmaan wrote:
Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S
Yes thats what Im thinking. Each row represents an equasion. If I reduce the matrix to a lower-trianguar or what its called than that means that the 4th row represents the 4th unknown. So If I do bunch of row operations and get the matrix down to something like:

1 3 -4 2 | 4
0 2 1 -3 | 2
0 0 5 6 | 5
0 0 0 3 | 6

this means that the 4th unknown: u=6-3

Than I use that to get the rest... ?

See, what you have in that matrix is:

1x + 3y - 4z + 2u = 4
0x + 2y + 1z - 3u = 2
0x + 0y +5z + 6u = 5
0z + 0y + 0z +3u = 6

Thus,
3u = 6
u = 2

5z + 6(2) = 5
5z = 5 - 12 = -7
z = -7/5

and so on and so on.

2009-09-07 15:17:50

#16

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Winston_Churchill wrote:
mcminty wrote:
Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S
Yeah, each row represents a different equation. If each column didn't represent a variable in the equations, then what would it represent?
The column could be filled with 0s?

Then it would be pretty pointless

But in that case, it would represent a variable that isn't being used in the system of equations. No?

2009-09-07 15:21:24

#17

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

mcminty wrote:
Winston_Churchill wrote:
mcminty wrote:

Yeah, each row represents a different equation. If each column didn't represent a variable in the equations, then what would it represent?
The column could be filled with 0s?
Then it would be pretty pointless

But in that case, it would represent a variable that isn't being used in the system of equations. No?

Ah, I get what you were saying now. I thought you were trying to say that that a row solves into a variable. Now that you wrote it out in the post above it makes sense

2009-09-07 15:23:12

#18

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Haha, ok. Yeah, its early here.. so my writing isn't the best.

2009-09-07 15:28:41

#19

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

What are you in to have to take this crap? I had it for engineering

/suicide

2009-09-07 15:32:03

#20

zeidmaan: Member; +234|6834|Vienna

mcminty wrote:
zeidmaan wrote:
Winston_Churchill wrote:
Im confused by what you said too mcminty, doesnt every row represent an equation? Each column doesnt necessarily have an unknown. Maybe I'm misunderstanding you too :S
Yes thats what Im thinking. Each row represents an equasion. If I reduce the matrix to a lower-trianguar or what its called than that means that the 4th row represents the 4th unknown. So If I do bunch of row operations and get the matrix down to something like:

1 3 -4 2 | 4
0 2 1 -3 | 2
0 0 5 6 | 5
0 0 0 3 | 6

this means that the 4th unknown: u=6-3

Than I use that to get the rest... ?
See, what you have in that matrix is:

1x + 3y - 4z + 2u = 4
0x + 2y + 1z - 3u = 2
0x + 0y +5z + 6u = 5
0z + 0y + 0z +3u = 6

Thus,
3u = 6
u = 2

5z + 6(2) = 5
5z = 5 - 12 = -7
z = -7/5

and so on and so on.

OH I see now what you meant.
so than I take
2y+1z-3u=2
than insert the values for u and z and calculate...

my wrong conclusion was that:
u=6-3
the right way is:
3u=6
u=2

because 4th column represents u... Thanks.

edit:
I guess the best way to avoid confusion is to convert the reduced matrix back to a full system of equations like you did.

Last edited by zeidmaan (2009-09-07 15:35:57)

2009-09-07 15:35:17

#21

ghettoperson: Member; +1,943|7069

Guys, if I post up all my coursework for next semester, will you do it all for me? Thanks.

2009-09-07 15:42:54

#22

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Winston_Churchill wrote:
What are you in to have to take this crap? I had it for engineering

/suicide

This was second year civil engineering maths. Oh god, and matrices were one of the easier topics.

I'm doing a double degree, so this semester I'm doing commerce subjects. Fuck they are easy.

2009-09-07 19:08:52

#23

Winston_Churchill: Bazinga!; +521|7158|Toronto | Canada

Heh, mine was first semester first year

2009-09-08 00:25:10

#24

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Winston_Churchill wrote:
Heh, mine was first semester first year

Oh, we did matrices in first semester, first year...

But that was from my second year set of notes, since it went into more detail/was much harder.

BF2S Forums Math halp. Matricies, Determinants, EigenValues/Vectors Systems of equ

zeidmaan wrote:

Winston_Churchill wrote:

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Winston_Churchill wrote:

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mcminty wrote:

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zeidmaan wrote:

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mcminty wrote:

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zeidmaan wrote:

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