hey folks my name is nathan johnston welcome to lecture 25 of introductory linear algebra this video is going to be our last lecture about invertibility of matrices so we're just going to introduce two final results about matrix inverses just to round out our toolbox concerning the subject okay so to motivate our first theorem that we're going to go through in today's lecture i'm going to start off by going all the way back up to the definition of the inverse of a matrix that we saw originally a few lectures ago okay so remember the inverse of a matrix it is a matrix a inverse with a property that if you multiply it by a on either side you get the identity matrix okay so we always had to be careful that we checked whether or not it was an inverse on the left and on the right so when we went through this example for example you know we multiplied both ways and we had to check that we got the identity matrix both ways no matter what order you multiply the matrix matrices in okay well now we're going to introduce a theorem that says well no actually that's overkill you don't have to multiply it both ways as long as it looks like an inverse from one side or the other then it actually is the inverse okay so here's the theorem suppose you got some square matrix okay well if you ever find any matrix b such that a b equals the identity or b a equals the identity then actually that matrix b must be the inverse the matrix a must be invertible and b is the inverse of a in other words you don't have to check multiplication on both sides you only have to check one or the other if either of them is the identity then the other one must be the identity automatically for free so another way of thinking about this is even though matrix multiplication in general is not commutative this is sort of one special case where it is commutative okay if you get a b equals the identity you know right away that b a must also be ident the identity so in this special case commutativity does work all right so let's go through a proof of this okay and we're going to prove it assuming that b times a equals the identity okay you can prove it assuming that a times b equals the idea of d2 i mean nothing drastically changes but we're just going to pick one of the starting points and go through that okay so suppose that b times a equals the identity our goal is to show that b is the inverse of a okay well consider this equation here ax equals zero that's a system of linear equations right and we saw a couple theorems ago that there's this nice connection between invertibility and systems of linear equations so if we can prove something about this linear system here in particular if we can show that it has a unique solution then we'll know that a must be invertible so that's our goal here let's show that this linear system must have a unique solution okay well the way that i'm going to do that is i'm going to take this linear system ax equals 0. i'm going to multiply it on the left by b and see what happens okay well one thing that happens is bax that's if i just introduce parentheses here that's b times ax and ax equals zero right so i can just sub that in so it's b times zero and anything times the zero vector is just a zero vector okay so bax must equal zero but on the other hand we can also group parentheses in bax a little bit differently if we group parentheses differently then we can group the parentheses around ba and then what we get is well b a times x well what is ba though ba is the identity matrix so i can plug that in okay ba is the identity matrix so i get that's equal to identity times x which of course is just x itself okay now we play our usual game of sort of tracing the equalities through x equals this equals this equals equals zero okay so if ax equals zero and ba equals i then x must equal 0 as well in other words what we've shown here is that this linear system ax equals 0 has a unique solution we've shown that if this happens if ax equals 0 then x must equal 0. so that's the unique solution okay but wait now we go back up to our characterization of invertible matrices so by our big theorem that had like a laundry list of invertibility is equivalent to this and this and this and this and this that theorem told that told us that hey if ax equals 0 has a unique solution then a must be invertible okay so we've shown that now we've shown that hey if ba equals i then a is invertible we haven't quite shown that b is the inverse yet though but fortunately we can do that easily now okay now that we know that a is invertible we can multiply a by its inverse okay so in particular i'm going to take this equation here ba equals i and multiply it by the inverse a inverse now that i know that it exists if i multiply on the right in particular by a inverse i got an a inverse over here and i get an a inverse over here and the a and the a inverse cancel i'm just left with b equals a inverse okay so just take this equation and multiply on the right by a inverse and you're going to find that yeah b equals a inverse and then you're done okay that's all we wanted to show okay we've shown that a is invertible and b really is the inverse just based on that one-sided product this time though and you can prove it the other way as well okay so you can also prove the case that if a b equals the identity matrix then a is invertible and b is the inverse as well all right so all those times that we did checked two-sided multiplication it was overkill don't need to do that for inverses one-sided easy enough all right and then the final theorem for this week says that well if you're working with two by two matrices you can actually get an explicit formula for the inverse of that matrix so if you don't like doing that method from the previous class where you take the two by two matrix put on the left augment with an identity and row reduce you can use this method instead to get an explicit formula for the inverse of a two by two matrix if it exists okay for three by three and larger turns out there's an explicit formula for the inverse of a three by three matrix and four by four and so on for no matter what size you like but they're actually absolutely horrendous to use no one uses them in practice okay so if you're working with a three by three matrix or larger just do the row reduction method from previous class okay augment with identity and row reduce all right well what is the explicit formula for the inverse of a two by two matrix okay so suppose that your matrix just given names to its entries a b c and d then that matrix it turns out it's invertible if and only if a times d minus b times c it does not equal zero so this funny little quantity a times d minus b minus sorry a times d minus b times c that has to be non-zero okay and if it is invertible in partic in other words if this quantity here is non-zero then we have an explicit formula for the inverse a inverse equals this junk here and you can sort of see now why oh we need a d minus bc to be non-zero so i could divide by it here okay and the inverse you just divide by that funny little quantity there and you swap the diagonal entries right originally the diagonal entries were a and d now they're dna and you stick a minus sign on the off diagonal entries so b and c became minus b and minus c all right so let's see where this comes from let's show that this actually is the inverse of the two by two matrix a that we started with okay and remember the way that you can show matrices are inverses of each other is you just multiply them together right if i give you a proposed inverse you can check whether or not it actually is the inverse just by doing the multiplication and seeing if you get the identity matrix after you multiplying it after you multiply them together okay so that's all i'm going to do i'm going to do a inverse times a equals this junk here so here's a inverse here's a let's do our matrix multiplication multiply them together okay you get a big ugly mess right you get d times a minus b times c in the top left entry and similarly d times b minus b times d in the top right entry and so on for all the other entries but the miraculous thing that happens is hey look in the top left entry here that's exactly a d minus bc so the thing that i'm dividing by cancels out with it i get a 1 in that top left corner here i've got db minus db oh 0 in the top right corner they cancel with each other here's ac minus ac 0 in the bottom left and here's a d minus bc in the bottom right again the thing that i'm dividing by so in the top left and bottom right i do get one and in the bottom left and top right i do get zero i get the identity matrix okay and by the previous theorem it's enough to check that one-sided product if i wanted to i could check a times a inverse as well but you don't need to okay you don't have to check the multiplication the other way anymore all right so that proved sort of half of the theorem that showed that here hey if ad minus bc does not equal zero then the matrix is invertible and this is the inverse okay i also have to show that if ad minus bc equals zero then the matrix is not invertible okay and be a little bit careful here it's tempting just to say that oh well this formula doesn't work because you can't divide by zero but that's not a valid proof because it doesn't show that there's not some other matrix that's the inverse in this case okay we have to show that if a d minus bc equals zero then there's no inverse whatsoever okay not just this formula but no inverse anywhere all right so that's what we've got to show now to round out this proof and finish it off okay so if ad minus bc equals zero i'll just rearrange that a little bit that means that a d equals b c and i'm going to split into a couple cases and show that in each of these cases now the matrix isn't invertible all right so case one what happens if either a equals zero or b equals zero okay well let's imagine a equals zero then on the left hand side i get a times d well that's zero equals bc so at least one of b or c must be zero as well so a equals zero and either b or c is zero so i go back up to my matrix what's that tell me a is zero and either b or c is zero so either i get a zero row at the top or i get a zero column at the left okay so either i get a zero row or zero column and something similar happens if b equals zero okay if b equals zero then this right hand side is zero so then either a or d is zero so b is zero and a or d is zero okay let's go back and look at our matrix b is zero and either a or d are zero so again either you get a zero row or you get a zero column okay so in case one here if a or b equals zero then you know that your original matrix a has a zero row or column okay now i'll come back to that later okay just keep that in the back of your minds for now case two if a and b are both non-zero okay so the other possibility if they're both non-zero then i can divide by them so i'm going to divide this equation a d equals bc by a and also by b i'm going to divide it by a b and when i do that on the left here i get rid of the a and i have a b on the bottom so i get d over b and on the right the b goes away and i get a c and i get an a on the bottom okay so i get c over a on the right okay and this appears a little useless at first glance but what this equation says is that the bottom row is a multiple of the top row okay and that matrix a the bottom row is a multiple of the top row okay because these ratios are the same d over b is the same as c over a if we go back up to that matrix a and see what does that mean d over b is the same as c over a so the ratio of the bottom entries to the top entries is the same in both of the columns right this ratio in the right column is the same as this ratio in the left column okay so the bottom row is a multiple of the top row all right so one of two things happens what we showed is that either there's a zero row or zero column or the bottom row is a multiple of the top row one of those two things has to happen and the point is in either of those two cases the reduced row echelon form of your matrix is going to have a zero row right if your original matrix has a zero row then boom right away you've got a zero row in a reduced row echelon form as well if your original matrix has a zero column then well i mean because your matrix is square you're going to get a zero row as well when you start row reducing it here like imagine your left row was or sorry left column was entirely zeros then when you row reduce it you can use some multiple of b to get d to be zero as well and then you've got all three of these entries being zero okay so you can get a zero row in your row reduce okay similarly if your bottom row is a multiple of the top row then when you row reduce your row operation is gonna cancel out that entire bottom row and it'll just turn it all into zeros okay so in either of these cases you reduce row echelon form it's going to have a zero row and therefore your matrix is not invertible because your reduced rotational form is not the identity matrix okay if it's got zero row it's not the identity matrix so a is not invertible all right and then that's it okay that's that's the proof is now done we've shown both sides of it we've shown that hey if a d minus bc is not zero it's invertible and we've shown that if a d minus b c or sorry if a d minus b c is not zero that is invertible and if a d minus b c equals zero then it's not invertible all right so let's just do a couple quick examples just to sort of illustrate using this theorem okay so let's find the inverse or so that doesn't exist for a bunch of two by two matrices so here's one of them three minus two one 4. if we want to use that theorem to determine invertibility first thing we do is we compute a d minus b c so it's going to be 12 carefully double negative 12 plus 2 right 3 times 4 minus minus 2 times 1. okay and of course that's 14 which does not equal 0 so that matrix is invertible and we can construct it straight away you just do 1 divided by that 14 that we just computed times swap the order of the diagonal entries so 3 4 becomes 4 3 and stick a minus sign on the octagonal entries so 1 minus 2 became -1 2. so that's your inverse and again if you want to double check that this really is the inverse just multiply this by this and you'll see yeah you get the identity matrix or if you wanted to find this inverse the other way via the method from the previous class you could do that too if you take this matrix a and augment with an identity matrix and then row reduce you're going to find that yeah row reduces down to the identity on the left and exactly this junk on the right so you can get the inverse either way you have two different methods for two by two matrices now all right let's do another example so the matrix two minus three minus four six okay again first thing compute that quantity a d minus bc so two times six minus minus three times minus four so careful this is the rare triple negatives triple negative is negative you're going to get 12 minus 3 times 4 so 12 minus 12 which is 0. so the conclusion in this case is no that matrix is an invertible because that quantity a d minus bc equals 0. okay so if you're to try to compute the inverse that other way right augment with an identity matrix and then row reduce you're going to find that you get a zero row down here when you start row reducing you're not going to get the identity matrix on the left all right one final example one place where you know this method this sort of explicit formula for two by two matrices really shines in comparison to the gaussian elimination method uh is you know when your matrix has really ugly entries so something like this pi and 3.5 and minus 2 6. certainly we could find the inverse or determine invertibility by augmenting with an identity matrix and then row reducing but your row operations are going to be ugly right you're going to be doing multiples of pi and multiples with decimals in them and stuff like that okay so using this explicit formula is actually rather advantageous in this case because it's easy uh easy to check invertibility no matter how ugly your entries are okay so start off by checking ad minus bc again we're gonna get 6 pi careful of your double negative six pi plus seven right a d is six times pi bc is minus seven and then you subtract it becomes plus seven okay so that is not zero so the inverse exists and the inverse is just well you do 1 divided by that a d minus bc that we computed so 1 divided by 6 pi plus 7 and then swap the position of the diagonal entries and stick a minus sign in front of the off diagonal entries and you're done that's all there is to it okay but yes please do keep in mind that you can always always always compute the the inverse of a two by two matrix using whichever of these two methods you like okay so you can use this explicit formula or you can row reduce a augmented with the identity matrix down to identity matrix augmented with the inverse okay that this method down here still works no matter what the size of your matrix is 2x2 or otherwise righty so that'll do it for a discussion of invertible matrices so i will see you next class for week seven