inverse matrices help please
hi there would anyone be able to give me a link or guidence as how i would go about finding the inverse of A? I have a book that tells me add subtract and multiply.
the question is this: find the inverse of A A = (19 81) (2 10) sorry for the crude matrices. any help would be great :) lakitu 
there is a trick for finding the inverse of a 2x2 matrix
[tex]A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)[/tex] then [tex]A^{1}=\frac{1}{adbc}\left( \begin{array}{cc} d & b \\ c & a \end{array} \right)[/tex] provided adbc is not equal to zero hope that helps 
http://www.purplemath.com/modules/mtrxinvr.htm
halfway down is how i usually find matrices. always works, so thats what i use. 
thank you for your comments :)

If your book is telling you "add, subtract, and multiply" (hey, that's how you solve any mathematics problem!:rolleyes: ) then go back and read over exactly what you add and subtract and what you multiply by. I suspect that your book is talking about "row operations" that's what Gale's website is talking about.

Before rushing off to a formula or procedure, it might be helpful to understand what you are trying to do when determining the inverse of a matrix.
Given a square matrix [tex]A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)[/tex], its inverse [if it exists] is another square matrix [tex]A^{1}=\left( \begin{array}{cc} p & q \\ r & s \end{array} \right)[/tex] such that the matrix product is the identity matrix. [tex] \begin{align*} AA^{1}&=I\\ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{cc} p & q \\ r & s \end{array} \right) &= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array} \right) \end{align*} [/tex] If you carry out the matrix multiplication, you should find a simple system of four linear equations in four unknowns... "simple" because it's really a pair of systems of two linear equations in two unknowns. You can easily solve these systems to obtain the formula given by vladimir69 above. (In addition, the inverse would satisfy [tex]A^{1}A&=I[/tex] as well.) 
Of course, if you have a 3 by 3 or 5 by 5 matrix, so that your system is 9 equation in 9 unknowns or 25 equations in 25 unknowns, you might find robphy's method a bit tedious! I think it's worth learning row reduction.

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This is how I like to think of it. If this doesn't make sense to you, feel free to forget about it so you don't get confused!
We start off with the partitioned matrix: [A : I] which has the property that: (the left matrix) = (the right matrix) * A. In particular, A = I*A. Now, if we do row operations, we will get some other partitioned matrix: [B : C] which still has the property that: (the left matrix) = (the right matrix) * A. In particular, B = C*A If we fully rowreduce the left hand side, we get the partitioned matrix: [I : V] which still has the property that (the left matrix) = (the right matrix) * A. In particular, I = V*A, and therefore V is the inverse of A. 
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