How Do You Calculate the Inverse of a 2x2 Matrix?

[lakitu]

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]

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

 

[Gale]

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

halfway down is how i usually find matrices. always works, so that's what i use.

 

[lakitu]

thank you for your comments :)

 

[HallsofIvy]

If your book is telling you "add, subtract, and multiply" (hey, that's how you solve any mathematics problem! ) 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]

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

<br /> \begin{align*}<br /> AA^{-1}&amp;=I\\<br /> \left( \begin{array}{cc} a &amp; b \\<br /> c &amp; d \end{array} \right)<br /> \left( \begin{array}{cc} p &amp; q \\<br /> r &amp; s \end{array} \right)<br /> &amp;=<br /> \left( \begin{array}{cc} 1 &amp; 0 \\<br /> 0 &amp; 1\end{array} \right)<br /> \end{align*}<br />
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&amp;=I as well.)

 

[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.

 

[MathematicalPhysicist]

there is a trick for finding the inverse of a 2x2 matrix
A=\left( \begin{array}{cc} a &amp; b \\<br /> c &amp; d \end{array} \right)
then
A^{-1}=\frac{1}{ad-bc}\left( \begin{array}{cc} d &amp; -b \\<br /> -c &amp; 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]

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]

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.