Example of finding matrix inverse

In the last video, we stumbled upon a way to figure out the inverse for an invertible matrix. So, let's actually use that method in this video right here. I'm going to use the same matrix that we started off with in the last video. It seems like a fairly good matrix. We know that it's reduced row echelon form is the identity matrix, so we know it's invertable. So, let's find its inverse. The technique is pretty straightforward. You literally just apply the same transformations you would apply to this guy to get you to the identity matrix, and you would apply those same transformations to the identity matrix. That's because the collection of those transformations, if you represent them as matrixes, are really just the inverse of this guy. Let's just do it. So I'll create an augmented matrix here. Maybe I'll do it right here. Let me make it a little bit neater. First, I'll write a. It's 1, minus 1, 1. And then minus 1, 2, 1. Minus 1, 3, 4. And then I'll augment it with the identity matrix, with 1, 0, 0, 0, 1, 0, 0, 0, 1. Now, if I want to get a into reduced row echelon form, maybe I'll replace the second row. I'll keep the first row the same for now. Let me just draw it like this. The entire first row: 1, minus 1, minus 1. It's going to be augmented with 1, 0, 0. Keep the whole first row the same. Let's replace the second row with the second row plus the first row. Minus 1 plus 1 is 0. 2 plus minus 1 is 1. 3 plus minus 1 is 2. 0 plus 1 is 0. 1 plus-- oh, sorry. That was a tricky one. 0 plus 1 is 1. 1 plus 0 is 1. 0 plus 0 is 0. All I did is I added these two rows up. Now, this third row. Let me replace-- I want to get a zero here. Let me replace the third row with the third row minus the first row. 1 minus 1 is 0. 1 minus minus 1 is 2. 4 minus minus 1 is 5. 0 minus 1 is minus 1. 0 minus 0 is 0. And then 1 minus 0 is 1. Just like that. Now, what do we want to do? Well, we've gotten this far. We want to zero out that guy and that guy. Let's keep our second row the same. Let me write it down here. It's 0, 1, 2, and then you augmented it with 1, 1, 0. Just like that. And let's replace my first row with the first row plus the second row. 1 plus 0 is 1. Minus 1 plus 1 is 0. That's why I did that, to get a zero there. Minus 1 plus 2 is 1. 1 plus 1 is 2. 0 plus 1 is 1. 0 plus 0 is 0. And now, I also want to zero out this guy right here. Let's replace the third row with the third row minus 2 times the second row. 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 5 minus 2 times 2 is 5 minus 4, that's 1. Minus 1 minus 2 times 1-- that's minus 1 minus 2-- is minus 3. 0 minus 2 times 1, that's minus 2. And then, 1 minus 2 times 0 is just 1 again. All right, home stretch. Now, I just want to zero out these guys right here. All right, so just let me keep my third row the same. Let me switch colors, keep things colorful. It's 0, 0, 1. We're going to augment it with minus 3, minus 2, and 1. Now, let's replace our first row with the first row minus the third row. 1 minus 0 is 1. 0 minus 0 is 0. 1 minus 1 is 0. 2 minus minus 3, that's 5. 1 minus minus 2 is 3. 0 minus 1 is minus 1. Now, let's replace the second row with the second row minus 2 times the third row. 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 0. 2 minus 2 times 1 is-- I'm sorry, I just-- oh, whoops. Let me-- we have to be very careful not to make any careless mistakes. 0 minus 2 times 0 is 0. 1 minus 2 times 0 is 1. It's not 0. 2 minus 2 times 1 is 0. 1 minus 2 times minus 3-- that is 1 plus 2 times 3-- that is 7. 1 minus 2 times minus 2, that's 1 plus 4, which is 5. And then, 0 minus 2 times 1, so that's minus 2. And just like that, we've gotten the A part of our augmented matrix into reduced row echelon form. This is the reduced row echelon form of A. And when you apply those exact same transformations-- because if you think about it, that series of matrix products that got you from this to the identity matrix-- that, by definition, is the identity matrix. So you apply those same transformations to the identity matrix, you're going to get the inverse of A. This right here is A inverse. And we have solved for the inverse, and it actually wasn't too painful.