# The Inverse of a Square Matrix - Problem 2

Let's find the inverse of another 2 by 2 matrix. Here I have matrix A which is -2 3, -4 6. I want to find the inverse. Remember the first step is to create an augmented matrix that starts with matrix A -2 3, -4 6. Then you augment it with the identity matrix 1 0, 0 1.

The idea here is, you want to do row operations on this augmented matrix, until you can get the identity matrix on the left. Then whatever is on the right is your inverse. So you want to change this to a 1. Change this to a 0. This to a 0. This to a 1 through row operations. So my idea here is I'm going to try to change this entry to 0. I'm going to do that by multiplying the top row by -2, and adding it to the bottom row. The notation for that is -2 times row 1 plus row 2.

The row number you end with, that's the row that's going to get changed. So I'm going to leave the top row alone -2 3, 1, 0. Then -2 times -2 is 4, plus 4 -4, 0. -2 times 3 is -6 plus 6 is 0. -2 times 1 is -2 plus 0, -2. -2 times 0 is 0, plus 1 is 1.

Now we've got a problem. I mean I did manage to get this first entry in the bottom row to be 0, but I didn't want a 0 here. I wanted to end up with a 1 here. Now that I have a 0, there is no way to actually get a 1 there.

So this means that this matrix is not invertible. The matrix I started with A is not invertible. Now if you look at it, the bottom row is actually 2 times the top row. Whenever one row is a constant multiple of the other, the matrix is not going to be invertible. So A has no inverse. Later on we'll talk about a quicker way finding out if a matrix is invertible using determinants.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete