If M = a, b; c, d is a 2-by-2 matrix with a d minus b c nonzero then the matrix M inverse equals d, minus b; minus c, a, all divided by (a d minus b c) satisfies M inverse M equals M M inverse equals the identity matrix. We say that M inverse is the inverse of M.
In the next few videos, we're going to answer the question: can you divide by a matrix?
This is the analogue of the reciprocal x reciprocal equals one over x of a number: the equation M inverse M equals the identity is the analogue of x reciprocal times x equals 1.
Let's just check M inverse M equals the identity (check the other equality for yourself). one over (a d minus b c) times the matrix d, minus b; minus c, a times the matrix a, b; c, d equals one over (a d minus b c) times the matrix d a minus b c, d b minus b d; minus c a + a c, minus c b + a d, which equals 1, 0; 0, 1.
We can use this to "divide" by a matrix: if we have a matrix equation A B = C then we can multiply both sides on the left by A inverse to get A inverse A B equals A inverse C, and since A inverse A equals the identity this means B equals A inverse C.
With great power comes great responsibility: you should never write B over A for matrices A, B! It's not clear if you're doing A inverse times B or B times A inverse (and these are different because A inverse and B might not commute).
We can use inverses to solve simultaneous equations. For example: x minus y equals minus 1, x + y = 3 is equivalent to A v = b where A = 1, minus 1; 1, 1 and b = minus 1, 3. We can compute A inverse equals 1, 1; minus 1, 1, all divided by 2, so v equals A inverse b equals a half times the matrix 1, 1; minus 1, 1 times the vector minus 1, 3, which equals the vector 1, 2 and this coincides with the solution x = 1, y = 2 we found earlier.
We'd like to generalise the notion of inverse to bigger matrices.
Let A be an n-by-n matrix. We say that A is invertible if there exists a matrix B such that A B = B A = 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 inverse.
To see that the inverse is unique when it exists, suppose you had two inverses B,C for A: A B = A C = I and B A = C A = I. Now evaluate B A C in two ways:
B A C equals (the product B A) times C, which equals I C, which equals C,
B A C equals B times the product A C, which equals B I which equals B,
so C = B A C = B.
If A, B are invertible matrices then A B is invertible with inverse B inverse A inverse.
we have A B times B inverse times A inverse equals A times B B inverse times A inverse, which equals A I A inverse, or A A inverse, which equals I and B inverse A inverse A B equals B inverse I B, which equals B inverse B, or I so B inverse A inverse is an inverse for A B.
Had we tried to use A inverse B inverse instead, we would have obtained A B A inverse B inverse, 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 a d minus b c is nonzero. Even more basically, a 1-by-1 matrix (x) is invertible if and only if x is nonzero. 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.