how to do matrix A'

caseyn2008

a=
[ 6 2 6 ]
[ 5 8 1 ]
[ 9 1 10] thats what i got to do i dont know much about it i'm doing my homework without a teacher over the net

JavaBear

what do you want to do with it? to form the inverse of a sqaure matrix get the adjoint of the matrix and divide by the determinant of the original matrix

caseyn2008

well all it says is A'

JavaBear

whoops didn't notice it lol

Creamy Goodness

if i remember correctly A' means the transpose of matrix A which basically means you make the rows of the matrix be equal to the columns and the columns become the rows of the matrix.

so if
A =
[6,7,8,9]
[1,3,5,2]
[5,6,3,2]

A'=
[6,1,5]
[7,3,6]
[8,5,3]
[9,2,2]

caseyn2008

got it how do u find the cofactor?

Michael Collins

This is where you have to use your initiative. Have a go at it yourself, and by all means come back if you have any problem

caseyn2008

its a test i have to get it right i mean the school dont know ho to do these bc its not the algebra 2 they teach.

Grimebox

Do they teach english?

/wanker

caseyn2008

its a typo i spilt a lot of stuff on my keyboard you have to press the button real hard for it so screw off, i only misspelled one word and i wasdoing home work so i didnt take the time for aposteophe btw ur real funny i forgot to laugh

Pherekydes

Grimebox wrote: »

Do they teach english?

/wanker

Banned from the forum for personal abuse.

cunnins4

Have a look at this. It's the first hit on google if you type in cofactor of a matrix. A lot of maths notes are available if you just search google.

caseyn2008

thatsgood but it dosent tell the processfor coming up with answers

LeixlipRed

caseyn2008 wrote: »

thatsgood but it dosent tell the processfor coming up with answers

Have you considered getting grinds? A lot of people on here are reluctant to just give answers to questions. Mainly because they're afraid people are taking shortcuts and not putting the effort in themselves. I think a one on one grind would suit you best.

caseyn2008

grinds? i just wanna know the process for doing it. did i ask for answers?

cunnins4

Right, Casey, use the head here. Look at that page I linked. You're looking for cofactors yeah? It shows you that the cofactor for A11 is 24, formed from a 2x2 matrix:
4 5
0 6

What is the ONLY way you can make 24 from those 4 numbers?

Take a look at A31, the cofactor is -2 and the 2x2 is:
2 3
4 5
There's only one way to come out with -2 from that matrix too, and it's the same as A11, and all the other elements of the original matrix A.

How did those 2x2 matrices come about?

The matrix A is made up of:

1 2 3
0 4 5
1 0 6
to find the matrix used to find the cofactor for A11 look at this:

* * *
* 4 5
* 0 6

Here I've removed the rows and columns associated with A11. To find the matrix to determine the cofactor for, let's say, A22:

1 * 3
* * *
1 * 6

Is that any help?

Casey, I ask guys for help with my college work on here a lot, and they're generally helpful, but you can't come on looking for a quick answer. You need to do some work, not just ask for the "process" which will just give you the answer.

caseyn2008

i still dont know how you came up with the number i know how to elimate the rows and stuff
i tried addin subtracting i found one for a11which was cross multiply and add that worked only for one not the other so there must be another way
wait i think i got it i just dont know which one you subtract i looked at that page and it was random it seemed like

ok i'm guessing this
[x y]
[y x] in A11 where cross multiply then subtract y and in A12 you subtract x right? it alternates between positive and negative in the matrix?

LeixlipRed

One word: determinant

caseyn2008

but that dosent explain why half of them you subtract one side and the other half you subtract the other side

eoinalmighty

detA=ad - bc

where A=

a c
b d

cunnins4

As Leixlipred said, the determinant is they key to this.

The determinant of the 2x2 matrix

a b
c d

is (a*d)-(b*c), always. That's how I got the cofactors of that matrix in my post above.

The signs before the cofactors of a 3x3 matrix are

+ - +
- + -
+ - + for a bigger matrix it just follows the same pattern, just alternating signs before the cofactors.

That's one of the many properties of matrices, you just have to learn it off. Look at cofactor expansion on wikipedia. Cofactors are a convenient way of finding the determinant of a large matrix, ie bigger than 2x2.

caseyn2008

so i had the pattern right?

cunnins4

^^Try it and see if you got the answer right. If you don't have the answer, post up the problem and your proposed answer and we'll tell you if it's correct or not, and if not, help you get the answer out.

caseyn2008

kay ill do that,
matrix a cofactor 4,1
[4 3 1 2]
[3 -1 1 -1]
[0 2 -3 0]
[5 -2 2 1]
comes to
[3 1 2]
[-1 1 -1]
[2 -3 0]
detriment, 9 im not sure of the polarity
+-+-
-+-+
+-+-
-+-+

a13

4 3 1 -2
3 -1 1 -1
0 2 -3 0
5 -2 2 1

3 1 -2
-1 1 -1
-2 2 1
detriment = -2

a33 = -3
4 3 1 2
3 -1 1 -1
0 2 -3 0
5 -2 2 1

4 3 2
3 -1 -1
-5 -2 1

detriment = 6
coreect me wherever my polarity is wrong

Tar.Aldarion

http://tutorial.math.lamar.edu/Classes/LinAlg/Matrices.aspx
That site taught me any college maths I would need in two days. Pretty dang good.

cunnins4

caseyn2008 wrote: »

kay ill do that,
matrix a cofactor 4,1
[4 3 1 2]
[3 -1 1 -1]
[0 2 -3 0]
[5 -2 2 1]
comes to
[3 1 2]
[-1 1 -1]
[2 -3 0]
detriment, 9 im not sure of the polarity
+-+-
-+-+
+-+-
-+-+

a13

4 3 1 -2
3 -1 1 -1
0 2 -3 0
5 -2 2 1

3 1 -2
-1 1 -1
-2 2 1
detriment = -2

a33 = -3
4 3 1 2
3 -1 1 -1
0 2 -3 0
5 -2 2 1

4 3 2
3 -1 -1
-5 -2 1

detriment = 6
coreect me wherever my polarity is wrong

Okay, good stuff, you're trying it yourself. For the first one you have the cofactor correct - The determinant of that 3x3 matrix is 9. Well done!

The second one, unfortunately you made up the wrong matrix, it should be:
3 -1 -1
0 2 0
5 -2 1

The determinant of that matrix is 16.

You just mixed up the rows to cancel out. The determinant of the 3x3 you posted is actually 12 so I'm afraid you got that one wrong.

I'm afraid you got the determinant for the last one wrong, it should be -38. Mind you, you used -5 instead of 5, so that's where you may have gone wrong there.

Is this college maths yeah? Maybe start out finding the cofactors of a 3x3 matrix first-it's A LOT easier.

I've attached a PDF with a list of all of these matrices with their determinants calculated and also the original 4x4 matrix and all of its cofactors. I used Matlab to calculate these-this was an EXCELLENT tool for me in first year when I was studying matrix algebra. If your college has Matlab on the computers it's well worth a look!

caseyn2008

thanks man im on heavy painkillers bc i had a full spinal fusion that wby i messed up

caseyn2008

thank u for all your help wheres matlab?

caseyn2008

am i right that adding and subtracting matrices they have be the same shape? and if i have say
[ x x ] [ x x x]
[ x x ] [ x x x]
[ x x ] + only the second matrix lets call it b, is b' where it is like shown on the first page of this thread would that work? also to find a detriment i do x + x + x - x + x + x right?

edit if that b' would it be
[ x x ]
[ x x ]
[ x x ]

caseyn2008

I'd like to say I went to prom last night and I'm starting my internship at the hospital tomorrow.

Tar.Aldarion

Random enough...