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 lakitu Apr26-06 03:51 PM

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

 vladimir69 Apr26-06 06:42 PM

there is a trick for finding the inverse of a 2x2 matrix
$$A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
then
$$A^{-1}=\frac{1}{ad-bc}\left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$$
provided ad-bc is not equal to zero
hope that helps

 Gale Apr26-06 08:40 PM

http://www.purplemath.com/modules/mtrxinvr.htm

halfway down is how i usually find matrices. always works, so thats what i use.

 lakitu Apr27-06 06:48 AM

 HallsofIvy Apr27-06 01:45 PM

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.

 robphy Apr27-06 09:22 PM

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 $$A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$,
its inverse [if it exists] is another square matrix $$A^{-1}=\left( \begin{array}{cc} p & q \\ r & s \end{array} \right)$$ such that the matrix product is the identity matrix.

\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*}
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 $$A^{-1}A&=I$$ as well.)

 HallsofIvy Apr28-06 05:59 AM

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.

 MathematicalPhysicist Apr28-06 06:05 AM

Quote:
 Quote by vladimir69 there is a trick for finding the inverse of a 2x2 matrix $$A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$ then $$A^{-1}=\frac{1}{ad-bc}\left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$$ provided ad-bc is not equal to zero hope that helps
i think it's a bit more trickier than just plain trick, it's actually a theorem.

 robphy Apr28-06 06:26 AM

Quote:
 Quote by HallsofIvy 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.
Agreed!... assuming one first understands what it means to take the inverse of a matrix.

 Hurkyl Apr28-06 07:42 AM

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 row-reduce 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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