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Minor Math Emergency.
Posted: Thu Oct 03, 2002 7:48 pm
by Obsidian
Heres the deal folks, I'm sure someone out there must be good at math, so heres what Im asking.
Does anyone out there know how to find the inverse of a 3x3 Matrix?
Posted: Thu Oct 03, 2002 9:33 pm
by ruined letters
It's kind of hard to explain without drawing it out, but if you have been studying matrices then you should probably understand what i'm trying to say.
It's actually incredibly simple. The inverse of any square matrix is the matrix that you can multiply it by to get the Identity martix. Get it? The identity matrix is the one where it's a bunch of diagonal 1's and 0's everywhere else. So it's just the the recipricol of a number, and is denoted, in fact, in the same exact way. It's just like how A X A^-1 = 1 in intergers, it's the exact same for matrices, except it's A X A^-1 = I (identity matrix)
Posted: Thu Oct 03, 2002 9:33 pm
by ruined letters
By the way, what math class are you taking?
Posted: Thu Oct 03, 2002 9:48 pm
by Obsidian
well ya, thanks ruined, but I've gone that far,
so, say you have a matrix of
1 2 3
3 4 6
1 0 3
I need to multiply that to get the identity matrix of
1 0 0
0 1 0
0 0 1
How can I find the [A] value, that when multiplied by the given matrix, will give me my identity matrix.
I'm guessing it has something to do with the determinant.
Sweet, 999 posts.
I'm currently taking math levels B and C, which are high level highschool classes.
My teacher promises 10 free marks to the first person to figure this out.
Posted: Thu Oct 03, 2002 9:58 pm
by ruined letters
Do you know how to multiply and divide matrices? If you do then it should be incredibly simple. You have the idea completely right, but unless I completely forgot how to multiply and divide matrices within the last year (which is very possible

) it doesn't really have much to do with the determinant. It should be in your math book in the next chapter or something if you're allready studying matrices.
It's been a year since i've done anything with matrices, so be careful cuz my memory seems to go bad over the summer.
Posted: Sat Oct 05, 2002 5:00 am
by garazdawi
The inverse of the matrix
A=
[1 2 3]
[3 4 6]
[1 0 3]
is
A-1=
[-2 1 0]
[.5 0 -.5]
[2/3 -1/3 1/3]
So A*A-1=I
Posted: Sat Oct 05, 2002 9:15 pm
by Obsidian
Oh gara, if only your timely assistance came but 2 days earlier.
How do you find those numbers though?
Posted: Sun Oct 06, 2002 2:48 am
by garazdawi
N my calculator

I have teh furmulae somewhere though, it DOES use the determinent IIRC
Posted: Sun Oct 06, 2002 3:46 am
by Osiris
Calculating by hand is tedious stuff. Been a long time since I've done this, so no guarantees this is correct.....
If A =
a b c
d e f
h i j
and inverse A =
k l m
n o p
q r s
the two must multiply to equal I
1 0 0
0 1 0
0 0 1
Hence:
ak + bn + cq = 1
dk + en + fq = 0
hk + in + jq = 0 etc.
Fill in the values you know (for a, b etc), and solve each set of three equations simultaneously to find k, l, m etc etc.
The determinant appears only indirectly.
No guarantees on the above and there may be easier ways....

Posted: Sun Oct 06, 2002 3:52 am
by garazdawi
There is a formula to do this. For a 2*2 matrix it is 1/|A|*A where |A| is the determinant.
There si something similar with a 3*3 matrix not sure how though