# Matrices

Joined Jan 19, 2004
220
Hi
i need some help in matrices on topics such as finding inverses of matrices (co factor method and any other short cut method there exists)... the methods i know ae way too tedious that involve row and column transformation on an identity matrix

thx a lot

#### Dave

Joined Nov 17, 2003
6,970
Could you possibly give us an idea of a question that you want to use these simpler method on? If I personally was calculating the inverse I would use the identity matrix method you say is too tedious.

For the cofactor method, take the matrix A:

A = 2 3 5
4 1 6
1 4 0

Then look at a minor element, eg. the 2:

We multiply the 2 by the diagonals of the numbers not in the row and column of the two, i.e. 1 6
4 0

With the first diagonal minus the second diagonal:

So we get 2(0-24) = -48

We do this for all the top row to derive the determinant |A|:

So we get 2(0-24) - 3(0-6) + 5(16-1) = 45

Note that each alternate product is substracted then added and so on, so if this matrix had another row that product would be subtracted from the above calculation:

2(0-24) - 3(0-6) + 5(16-1) - .......

Now deduce the cofactors for each of the elements in the matrix, but don't multipy by the minor element. For our matrix above, the matrix of cofactors C looks like:

C = -24 6 15
20 -5 -5
13 8 -10

Remember to multiply the cofactors by the appropriate sign as per the following matrix:

M = + - + - + - ..........
- + - + - + .........
+ - + - + - ..........
.
.
.
.

Then transpose the cofactor matrix C, this is the adjoint of matrix A:

C Transpose = -24 20 13
6 -5 8
15 -5 -10

Then to find the inverse of your original matrix A, multply the transverse cofactor matrix C Transverse buy the reciprocal of the determinant:

Therefore:

Inverse A = (1/45)* -24 20 13
6 -5 8
15 -5 -10

Confusing eh! :blink: Post back if you don't understand any of that and I'll try and iron out individual problems. Good luck Edit: Apologies that the matrices are not in line (as they should be). You should be able to work them out if you write them down on paper, they are 3x3 matrices.

#### Dave

Joined Nov 17, 2003
6,970
I hope they explained it better than I did!!  Just a question d'you think that method is easier than the Identity Matrix method? I know you though that method was too tedious and this one isn't much better.