# Matrix Multiplication
### Matrix Multiplication
The product of two matrices is defined *only* when the number of columns of the first matrix is the same as the number of rows of the second; in other words, it is only possible to multiply m x n and n x p size matrices. The reason for this becomes clear upon defining the product:
Another way to see is this:
and
Another way to look at multiplication through columns
4th way to multiply
Also, the **multiplication of matrices need not be commutative**. Therefore AB =/= BA generally.
When AB = BA then AB are said to commute. This is the case of the identity matrix.
### Elimination
Say you have this matrix you want to solve:
Find the matrix you need to multiply it by to solve it. Or to get U (upper triangle which allows you to easily solve the system.
Step 1 : You want a 0 in position (2, 1) and (3,1) - but in this example there is already a 0 in (3,1).
Step 2: You want a 0 in position (3,2). So in this example you subtract 2 x row 2 from row 3 to get the 0.
So now, z=5, and you can easily solve the rest.
You can change the order in which you do multiplications with matrix multiplication.
Let's finish the topic of elimination.
Ex: 2 x 2 elimnation. Let A be a matrix as below where I can do elimination, but no pivots. I want to get from A to U to solve A. But then I want to know how is A related to U where there is a matrix L where A = LU. How do you get there?
First, to solve A and get U (that is then easy to solve), I multiply by my elementary matrix at the position 2, 1 (E(2,1)) because that is how I get a 0 in position 2,1. Therefore we get:
Then to get A = L U: you need to multiply E, (2,1) by the inverse which becomes L (-4 becomes 4).
Where L is the Lower triangle, and U is the Upper triangle.
Now let's try to do this is a 3x3 matrix. What are the steps to producing elimination?
Now, suppose we want all the E's on the right hand side of the equation. We get:
So L is the product of inverses. Why do we do this? Because when you multiply the non inverses, you don't get a good matrix. The multipliers go directly into L (see the number 10 in the matrix below). However, when you multiply the inverses, you get a clean L with no 0 in position 3,1. See:
### Cost of the operations
Say you have a matrix nxn where n=100. How many operations will we have to do with elimination? Turns out that the operations for A and b (Ax = b) there will be:
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