Formula for 2x2 inverse

I've got a 2 by 2 matrix here. Let's say A is equal to a, b, c, d. So I'm going to keep it really general. So this is really any 2 by 2 matrix. What I want to do is use our technique for finding an inverse of this matrix to essentially find a formula for the inverse of a 2 by 2 matrix. So I want to essentially find a inverse, and I want to do it just using a formula that it just applies to this matrix right here. So how can I do that? Well, we know a technique. We just create an augmented matrix. So let's just create an augmented matrix right here. So we have a, b, c, d, and then we augment it with the identity in R2, so 1, 0, 0, 1. And we know if we perform a series of row operations on this augmented matrix to get the left-hand side in reduced row echelon form. The right-hand side, if the reduced row echelon form here gets to the identity, then the right-hand side is going to be the inverse. So let's do it in this general case, not dealing with particular numbers here. So the first thing I want to do, or I would like to do, is I would like to zero this guy out. What we want to do is we want to zero that out, zero that out, and then these two terms have to become equal to 1. So the best way to zero this out, let's perform a little transformation here. So if I perform the transformation on the columns, C1, so those are the entries of a column-- this would be one column right here, that would be another column right there, that's the third column, that's the fourth column. But the transformation I'm going to perform on each of these columns, and we know this is equivalent to a row operation, is going to be equal to-- since I want to zero this one out, I'm going to keep my first row the same, so it's going to be C1, and I'm going to replace my second row with a times my second row minus c times my first row. Now why am I doing that? Because a times c minus c times a is going to be 0. So this guy's going to be 0. That's the row operation I'm going to perform. And I'm doing this so we can kind of keep track, account for what we're doing because the algebra's going to get hairy in a little bit. So let me perform this operation. So if I perform that operation on our matrix, what do we have. So our first row's going to be the same. Let me start with our second row because that's a little bit more complicated. So I'm going to replace c with a times c minus c times a. That's ac-- so let me put it this way-- so that's going to be 0 right there. I'm going to replace d with d times a or a times d minus c times C1 in this column vector. So minus c times b. Let me write this as bc. And then let me augment it. And then this guy's going to be a times 0, because he's C2 minus c times C1. So it's going to be minus c. And then finally, this guy right here is going to be a times 1-- a times this 1 right here-- minus c times 0. So that's just going to be an a. And then the first row is pretty straightforward. We know that the first row or the first entries in our column vectors just stay the same through this transformation. So it's a, b, 1, 0. And just to make sure you're clear what we're doing, when you perform this transformation on this column vector right here, you got this column vector right there. When you perform the transformation on this column vector right there, you get this column vector right there. Now I just want to make that clear because I did all of the second entries of all of the column vectors at once because we all were essentially performing the same row operation, so that just helped me simplify at least my thinking a little bit. Let me stay in this mode. So let's continue to get this in reduced row echelon form. The next thing we want to do is, let's make another transformation, we'll call this T1. That was our first transformation. Let's do another transformation. T2, or another set of row operations. So if I start with the column vector C1, C2, what I want to do now is I want to keep my second row the same and I want to zero out this character right here. I want to zero him out. I know I'm going to keep my second row the same, so C2 is just going to still be C2. But in order to zero this out, what I can do is I can replace the first row with this scaling factor times the first row minus this scaling factor times the second row. So it'll be ad minus bc times your first entry in your column vector minus b times your second entry. And the whole reason why I'm doing that is so that this guy zeroes out. So if we apply that to this matrix up here-- let's do the first row first. So this first entry right here is going to be ad minus bc times a, because that's C1-- let me write that down. So it's ad minus bc times a minus b times C2 minus 0. So it's just going to be-- that second term just becomes 0. Fair enough. Now what is this guy going to be? He's going to be-- I'll write it out. He's going to be ad minus bc times b minus b times your C2 in this column vector, minus b times ad minus bc. And you can see immediately that these two guys are going to cancel out and you're going to get a 0 there. And then we've got to augment it. I want to make sure I don't run out of space, I should've started to the left a little bit more. So what's this guy going to be? Well I'm going to have this guy times ad minus bc-- I'll do it in pink-- so you're going to have ad minus bc times 1, which is just ad minus bc minus b times C2, so minus b times minus c. So that's plus bc. So 1 times ad minus bc minus b times minus c is equal to that. And you can immediately see that these two guys will cancel out. You're just [INAUDIBLE] ad. And then this guy over here, you're going to have 0 times ad minus bc, which is just a 0, minus b times a. So you have minus ab-- just squeeze it in there. And we know that our second row just stays the same. Our second row just stays the same in this transformation. So we had a 0 here, we're still going to have a 0. We had an ad minus bc. We'll still have an ad minus bc. We had a minus c. Then we had an a. Just like that. Now let me re-write this matrix just so it gets cleaned up a little bit. So let me re-write it right here. I'll do it in my orange-- well, let me do it in this yellow. So I have ad minus bc times a. And then this term right here just became a 0. This term right here is a 0. This term right here is an ad minus bc. And then our augmented part, this part was just an ad. This was a minus ab. This is a minus c. And then this is an a. Now we're almost at reduced row echelon form right here. These two things just have to be equal to 1 in order to get reduced row echelon form. So let's define a transformation that'll make both of these equal to 1. So if this was T2 let me define my transformation T3. You give it a column vector, C1, C2. And it's just going to scale each of the column vectors. So what I want to do is I want to divide my first entries by this scaling factor right here so that this becomes a 1. So I'm essentially going to multiply 1 over ad minus bc times a, so 1 over ad, bc, a times my first entry in each of my column vectors. And then my second one I want to divide by this. So that this guy becomes a 1. So I'm doing two scalar divisions in one transformation. So this one's going to be 1 over ad minus bc times C2. So I'm just scaling everything by these two scaling factors. So if you apply this transformation to that right there, what do we get? We get a matrix. And this guy, I'm going to divide him by ad minus bc times a, so I'm dividing it by itself, so that guy's going to be 1. I'm going to divide 0 by this, but 0 divided by anything is just 0. Then we're in an augmented part. ad divided by-- so let me write it like this. So ad-- I'm going to divide by this-- so it's going to be ad minus bc times a-- you immediately see that the a's cancel out. This is going to be minus ab divided by ad minus bc times a. Once again, the a's cancel out. And then in my second row are my second entries in my column vectors, 0 divided by anything is 0. So 0 divided by this thing is going to be 0, assuming we can divide by that, and we're going to talk about that in a second. This guy divided by this guy, we're just dividing by himself, so it's going to be equal to 1. Now we have minus c divided by this, or ad minus bc. And then we have an a. a divided by ad, ad minus bc. And we're done. We put the left-hand side of our augmented matrix into reduced row echelon form. And now this is going to be our inverse. So let me clean it up a little bit. So, so far we started off with a matrix-- I'll do it in purple-- we started off with a matrix a is equal to a, b, c, d. And now just using our technique we figured out that a inverse is equal to this thing right here. And just to simplify-- well let me just write it the way I have it there, because I don't want to skip any steps-- this is equal to d over ad minus bc. Right, this guy and that guy canceled out. And then we have a minus b over ad minus bc because that guy and that guy canceled out. Then you have a minus c over ad minus bc. And then finally you have an a over ad minus bc, which is our inverse. But one thing might just pop out at you immediately is that everything in our inverse is being divided by this. So maybe an easier way to write our inverse. We could also write our inverse like this. We could just write it as 1 over ad minus bc times the matrix d minus b minus c and a. And just like that we have come up with a formula for the inverse of a 2 by 2 matrix. You give me any real numbers here and I'm going to give you its inverse. That straightforward. Now one thing you might be saying, hey, but not all 2 by 2 matrices are invertible. How can this be the case for all of them. And I'll give you a question, when will this thing right here not be defined? When is this thing not defined? Every operation I did, I can do with any real numbers, and this applies to any real numbers. But when is this thing not defined? Well it's not defined when I divide by 0. And when would I divide by 0? Everything else you can multiply and subtract and add a zero to anything, but you just can't divide by 0. We've never defined what it means when you divide something by 0. So it's not defined if ad minus bc is equal to 0. So this is an interesting thing. I can always find the inverse of a 2 by 2 matrix as long as ad minus bc is not equal to 0. We came up with all of these fancy things for invertability, you've got to put it into reduced row echelon form, and before that we talked being on 2 and 1 to 1. For at least a 2 by 2 matrix we've really simplified things. As long as ad minus bc does not equal 0, we can use this formula and then we know that a-- and it goes both ways-- a is invertible. And not only is it invertible, but we can just apply this formula to it. So immediately something interesting might-- you might say hey, this is an interesting number. We should come up with some name for it. And lucky for us, we have come up with a name for it. This is called the determinant. Let me write it in pink. Determinant. So the determinant of a, and it's also written like this with these little straight lines around a, and you could also write it like this, a, b, c, d. But most people kind of think this is redundant to have brackets and these lines. So then they just write it like this, this is equal to just, they just write the lines, a, b, c, d. I want to make this very clear. If you have the brackets you're dealing with a matrix. If you have just these straight lines you're talking about the determinant of the matrix. But this is defined for the 2 by 2 case to be equal to ad minus bc. This is a definition of the determinant. So we can re-write, if we have some matrix here, we have some matrix a which is equal to a, b, c, d. We can now write its inverse, a inverse is equal to 1 over this thing, which we've defined as the determinant of a times-- and let's just see a good way of kind of memorizing this. We're swapping these two guys, right, the a and the d get swapped. So you get a d and an a. And then these two guys stay the same, they just become negative. So minus b and minus c. So that's the general formula for the determinant of a 2 by 2 matrix. Let's try to do a couple. Let's try to find the determinant of the matrix 1, 2, 3, 4. Easy enough. So the determinant of-- let's say this is the matrix B. So the determinant of B, or we could write it like that, that's equal to the determinant of B. That is just equal to-- that's this thing right here-- 1 times 4 minus 3 times 2, which is equal to 4 minus 6, which is equal to minus 2. So the determinant is minus 2, so this is invertible. Not only is it invertible, but it's very easy to find its inverse now. We can apply this formula. The inverse of B in this case-- let me do it in this color-- B inverse is equal to 1 over the determinant, so it's 1 over minus 2 times the matrix where we swap-- well, this is the determinant of B. I want to be careful. B is the same thing, but with brackets. 1, 2, 3, 4. So B inverse is going to be 1 over the determinant of B, which is equal to minus 2. So 1 over minus 2. We swap these two guys, so they get a 4 and a 1, and then these two guys become negative-- minus 2 and then minus 3. And then if we were to multiply this out it would be equal to minus 1/2 times 4 is minus 2. Minus 1/2 times minus 2 is 1. Minus 1/2 times minus 3 is 3/2, minus 1/2 times 1 is minus 1/2. So that there is the inverse of B. Now let's say we have another matrix. Let's say we have the matrix C, and C is equal to 1, 2, 3, 6. What is the determinant of C? It is equal to-- we could write this way-- 1, 2, 3, 6. And it is equal to 1 times 6 minus 3 times 2, which is equal to 6 minus 6, which is equal to 0. And there you see it's equal to 0, so you cannot find-- so this is not invertible. So we can't find its inverse because if we would try to apply this formula right here you'd have a 1 over 0. But we know this formula just comes out-- that attempt to put it into reduced row echelon form, and in that last step we just had to essentially divide everything by these terms. So these terms would be 0 in this matrix C that I just constructed for you. And the reason why I knew-- I just pulled this out of my brain-- I knew this wasn't going to be invertible because I constructed a situation where I have columns that are linear combinations of each other. I have 1, 3-- you multiply that by 2 you get 2 and 6. So I knew that these aren't linearly independent columns. So you know that its rank wasn't going to be equal to, so I knew it wasn't going to be invertible, but we see that here by just computing its determinant.