20. Inverses

20. Inverses

In the next few videos, we're going to answer the question: can you divide by a matrix?

Theorem:

If M=(abcd) is a 2-by-2 matrix with ad-bc0 then the matrix M-1=1ad-bc(d-b-ca) satisfies M-1M=MM-1=I . We say that M-1 is the inverse of M .

This is the analogue of the reciprocal x-1=1/x of a number: the equation M-1M=I is the analogue of x-1x=1 .

Let's just check M-1M=I (check the other equality for yourself). 1ad-bc(d-b-ca)(abcd)=1ad-bc(da-bcdb-bd-ca+ac-cb+ad)=(1001)=I.

We can use this to "divide" by a matrix: if we have a matrix equation AB=C then we can multiply both sides on the left by A-1 to get A-1AB=A-1C , and since A-1A=I this means B=A-1C .

With great power comes great responsibility: you should never write BA for matrices A,B ! It's not clear if you're doing A-1B or BA-1 (and these are different because A-1 and B might not commute).

We can use inverses to solve simultaneous equations. For example: x-y=-1,x+y=3

is equivalent to Av=b where A=(1-111) and b=(-13) . We can compute A-1=12(11-11) , so v=A-1b=12(11-11)(-13)=(12),
and this coincides with the solution x=1 , y=2 we found earlier.

Bigger matrices

We'd like to generalise the notion of inverse to bigger matrices.

Definition:

Let A be an n -by-n matrix. We say that A is invertible if there exists a matrix B such that AB=BA=I . (Here I is the n -by-n identity matrix). If such a B exists, then it's unique, so we're justified in calling it the inverse of A and writing it as A-1 .

To see that the inverse is unique when it exists, suppose you had two inverses B,C for A : AB=AC=I,BA=CA=I.

Now evaluate BAC in two ways:

  • BAC=(BA)C=IC=C,

  • BAC=B(AC)=BI=B,

so C=BAC=B .

Lemma:

If A,B are invertible matrices then AB is invertible with inverse (AB)-1=B-1A-1 .

we have (AB)(B-1A-1)=A(BB-1)A-1=AIA-1=AA-1=I

and (B-1A-1)(AB)=B-1(A-1A)B=B-1IB=B-1B=I,
so B-1A-1 is an inverse for AB .

Had we tried to use A-1B-1 instead, we would have obtained ABA-1B-1 , and we couldn't have cancelled anything because the various terms don't commute.

We saw for 2-by-2 matrices that a matrix is invertible if ad-bc0 . Even more basically, a 1-by-1 matrix (x) is invertible if and only if x0 . We'll see in a few videos' time that there is an analogous quantity (the determinant) det(A) which is nonzero precisely when A is invertible. First though, we're going to explain how to calculate inverses of n -by-n matrices.