How do I compute a determinant using Leibniz's (big and inefficient) formula?
I'm talking about this ( http://en.wikipedia.org/wiki/Leibniz_for … ) but, the Wikipedia link is too "proofy" and, there is no regular hands-on problem so, I'm having trouble getting it. Any input would be greatly appreciated!
Leibniezs formula eh what do you know of it so far?
http://answers.yahoo.com/question/index?qid=20130325071913AApgrux seems to have been another resource you used? maybe?
Yes, there isn't much useful (for me) stuff that I can find about this. All I know is that it involves products, I think, (n-1)! times and that it works only for square matrices. All I think I need is super trivial examples with 2x2 and maybe 3x3 matrices or whatever you think is best.
so the formula looks a little something like this correct? \[\Large det(A)=\sum_{\sigma \in S_n} sgn(\sigma)\prod_{i=1}^{n}A_{i,\sigma_i}\]
It seems to be correct but, I'm not sure. Wikipedia has the stuff inside/on the right of the product symbol as a_sigma(i), i. While doing a trivial example, could you please explain what the notation is doing?
from what i can tell, there are permutation that can be had to account for sigma. but i got no idea where they are getting it from. they say of n=3, then sigma = [1,2,3], or [3,2,1], or [1,3,2] etc ... and they assign an even or odd to a permutation based on some unknown concept. the sign of the sigma is either a + or a - depending on if its an even or odd parity maybe? the product is then just the matrix element of the ith row and the column of the ith sigma. i assume from this that the n is the number corresponding to an nxn matrix say a 2x2 is:\[\Large det(A)=\sum_{\sigma \in S_n} sgn([1,2])\prod_{i=1}^{2}A_{i,\sigma_i}\] \[\Large det(A)=sgn([1,2])(A_{1,1}*A_{2,2})+sgn([2,1])(A_{1,2}*A_{2,1})\]
Sorry for the potentially stupid question but, what is sgn([1,2])?
You want the determinant correct?
Yes (using this technique - not any technique).
im not so sure what the innards are, they have to do with permuations dealing with an entry inside of the matrix; and the sgn indicates "sign" as either negative of postive.
i think it equates to the standard:\[(-1)^{i+j}\text{ for a given} ~a_{i,j}~\text{entry in A}\]
if i+j is even, sgn is + if i+j is odd, sgn is - this conforms the subdeterminants into proper values without haveing to adjust the i,j to 1,1 each time
Is the sigma a subscript to the A as well?
\[A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\] subdeterminant for a1,1 is +d subdeterminant for a1,2 is -c subdeterminant for a2,1 is -b subdeterminant for a2,2 is +a
the sigma notation is shorthand since the subdeterminant of nxn matrix can grow rather large
(I was asking because the subscript has its own subscript.) Also, sorry if I ask dumber questions than usual but, my brain is worn out from thinking too much.: I didn't understand what you said here: "subdeterminant for a1,1 is +d subdeterminant for a1,2 is -c subdeterminant for a2,1 is -b subdeterminant for a2,2 is +a"
the determinant of a matrix is composed of subdeterminants .... according to some rule that i cant quite recall
i remember ....
take the value at Ai,j, and multiply it by its subdeterminant value then add up all the Ai,j subdeterminant values ... perhaps
Use matrix([1,2],[3,4]) as an example.
please
\[A=\begin{pmatrix}1&2\\3&4\end{pmatrix}\] A1,1 = 1 is subdeterminant is the value of the determinant obtained from corssing out the row and column that contain A1,1; in this case we are left with the determinant of |4| = 4, then move across the columns or rows taking a sub determinant along the way
so det(A) = 1|4| - 2|3| - 3|2| + 4|1| is what i read the setup as, but thats 2 times as much
Don't take this in a bad way but, I need to go now. I'll be back as soon as I can to read what you said. (Don't wait for me unless you'll be here anyways.)
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