Transcript for:
Elementary Matrices and Row Operations

section 1.5 talks about elementary matrices and a method for finding the inverse of a now this is something that we've been doing for a bit but we wanted to put them in order Elementary R operations these are the elementary operations of Matrix a you could multiply a row by nonzero constant that's Elementary you could interchange two rows if you wish you could add a constant Time 1 row to another kind of like saying take Row one * 3 plus row two put the row two so that's stuff that we know I just wanted to point that out now we have couple of definitions we want to talk about definition one says matrices A and B are set to be row equivalent if either and hence each other if either can be obtained from the other by a sequence of Elementary row operations and the elementary row operations are right there also a matrix e and this one we're going to find a matrix e is called an elementary Matrix if it can be obtained from an identity matrix by for performing a single Elementary row operation just one so you take the identity say we're talking about 2 by two you could so we say e is called an elementary Matrix if can you obtained from this ident by applying on it just one grow operation exactly one looking in the book at examples one and two it says determine whether the given Matrix is Elementary so we're thinking okay can we just do one operation and again if you notice this does look like the identity so assume this is the identity Matrix what what do you think you could do to this to get that well wouldn't you say it is Elementary uh yeah it this and multiply root here we could write it this way radical 3 * Row 2 and row two there you go so a single operation that's it just multiply by a single by a number that's one how about B we say B is also an elementary Matrix so to get from this to what they have all what I have to do is interchange Row one with Row three and since that's the only operation that we do pretty much that's Elementary three if I take one z z 0 1 0 and 0 0 1 if I take 9 row three added to row two put the result in row two isn't that a single operation multiplying by n and just added to the first row so this is elementary and if I get to the last one well here we have a problem I would have here I would have to switch Row one with row two and no row two with Row three but also I would have to multiply a row one by a negative put it back in row one that's more than one operation so this is not Elementary it has to be a single operation what would be a single operation doing just one of those three single Elementary operation I'm sorry so how about this problem right there it's asking find a row operation and the corresponding Elementary Matrix that will restore the given Elementary Matrix to the ID so you want this to become the identity how would you do that well for this to become for this to to uh become the identity we would take uh three row one added to row two put the result in row two that will do the trick here so let me see if we take one Z 01 so if we take one Z 0 one and if you multiply Row one by three and add it to row two that will become 3 one there so when they say find a row operations and the corresponding Elementary Matrix there's the corresponding Elementary Matrix and I'll tell you what we're going to do with that there's a theorem of how to use those Elementary matrices to do something else for now let's just find it here I am given one Z z0 one01 so what would you do to that what would you do to this to get that well here we would take uh Row three divided by three put the result in row three so we we take the first we leave alone the second we leave alone and the third we divide by three and that's really how this is going to work so every time you do one of these do it on the identity Matrix well here we have the identity to be now normally I try not to have you work on big matrices but I wanted to get used to the identity Matrix and hopefully you mess that part up now what do we do here we want this to become that well I could see if I could take Row one and switch it with Row three isn't that the only no I'm sorry that's row four not three right so what would that do to my Matrix and we call that e that will make this 0 z0 one the second alone the third y alone and one Z and if you're thinking again why are rewriting those I'm going to show you there's a theorem that says you could get the same result by multiplying e times something else who and there it is and on the last one let's see how that works if I take the identity oh let me just leave some space if I take the identity Matrix I notice the only thing in there that's A- 1 over 7 If I multiply this by ative 1 over7 and add it to the first row take negative 1 over 7 row three and add it to row one so let's see how this works uh that means I had one zero okay so this is z z0 one this is 0 0 1 0o this is 0 1 0 0 and if you multiply that by uh uh let me see multiply that by 1 over s my mistake so that'll be e right there so that's pretty much all you have to do pretty much whatever you do the operation you do it on the identity Matrix and keep that as e for now and let's see how that plays right there now they're taking this in steps and that's fine because if you put it all together it's kind of confusing now let's see where this is going to go observe if I was looking at the identity Matrix do you see what we did we must have multiplied Row one by A6 isn't that what it is if you multiply a row one by A6 you'll get that notice what happened if you take EA if you multiply those two this is 2 by two this is two by 2x4 the outcome is 2x4 I have all the space let me make those bigger so so so what do we get let's see to generate the first row we use the first row multiply accordingly and I'm not going to keep on doing that but for now so again 6 + 0 is a 6 12 + 0 is a 12 -30 + 0 is -30 and 6 + 0 is a 6 for the bottom row you notice that when you multiply those 0 times the first is going to always be a zero and just going to add the bottom you end up the bottom so 0 + 3 0 - 6 0 - 6 and 0 - 6 and do you see that all what we did to this was we took Row one and multiply it by a -6 put it back in row one and notice what happened when you multiply the two matrices if I was to multiply this top Pro by a -6 wouldn't that be the results and as you see a single operation or a matrix operation perform the same outcome and of course multiplying Bro One by a six is a lot easier than multiplying those two if I looked at this one look how messy that looks now if I looked here and I'm trying to figure out what did you do to the identity Matrix to get a 4 there you must have taken -4 Row one and you added that to row two put the result in row two that's what you must have did well if that's the case then guess what If I multiply a row one by A4 and add it to row two they're saying if you multiply those if you take EA e is 3x3 3x five so the outcome is 3x five so to generate the first row I use the first row to generate the second row I use the second row and to generate the third row I use the third row and accordingly multiply in order by this to get the first and let's see since that's one z z you're just going to get zero two on the first you see that and on the second it's going to be negative one and those are going to be Z zero so pretty much you're going to get the first row only on the second row do you see that you're going to get -4 * 2 + 1 * 1 and the bottom is not going to count because that's zero so that's -8 + 1 which is -7 and for this it's going to be -4 * -1 4 - 3 is a 1 and then 0 - 1 is a negative 1 and 16 + 5 is 21 and 16 for the last 16 plus 3 is 19 and for the third row when you multiply by 0 0 by 0 01 multiply the first by 0o that's a z the second by 0o that's zero that's going to happen in all of those the only thing that's going to remain would be that bottom row where's that R oh here it is I cluttered this I just want to H come on here clean it up there it is all right and again if you really think about it multiplying those two matrices is the equivalent of me multiplying so this is what I did to the elementary well do that for this and you get the same result multiply this by a -4 and add it to the second row it should give you that row as well8 + 1 4 - 3 uh uh 0 minus one uh 4 16 + 5 and 16 + 3 so just making sure that you see that and again we need this for theoretical purposes but for now let's look at this do you notice here what you did all what you did to the identity Matrix you took row two multiply by five put the result in row two now they want to show that look EA gives the same results if you take an elementary Matrix and multiply it by a it has the same outcome as you applying that operation on a if I was to multiply a Row 2 by five and put a book back in there wouldn't that be a 10 and 25 and they're saying look multiply 3x3 3x two the outcome is a 3X two and to generate the first row I use the first row to generate the second row oops it didn't register I use the second and to generate the third I use the third multiply accordingly well here 1 0 0 multiplied by those 1 * 1 is 1+ 0 + 0 and 4 + 0 + 0 when I get to the second 0 * 1 is 0 5 * 2 is 10 + 0 and when I get to this 0 * 4 is 0 5 * 5 is 25 when I get to the last 0 * 1 0 * 2 that's a three and that's a six and do you see it's as if multiplying those two matrices has is the same operation and me just multiplying that by a five and this is something we're going to come back to later on and use whenever the time comes so it's much easier if you could find such Arrow operations to just multiply the single matrix by that operation I apply that single Matrix apply the operation of the single Matrix instead of multiplying two matri now here they say look here they're saying hey tell me what e is so that when I multiply B I get that Matrix well that's the same thing as looking at those say look the first is the same the last is the same what did they do from this to this then they multiply Row Two by -3 so if I take the identity Matrix multiply Row Two by a -3 wouldn't that be it so this homework not a whole bunch of work at least not in the first problems but very important for what's coming we want to make sure we get these Elementary mates out of the way if you multiply e and there's D to get B and here we took e time B to get D well again look at this that's the same that's the same what did they do here they got they had a -6 so wouldn't that mean they divided by a-3 except we don't divide well you could divide we take row2 divided by -3 put the result in row two well that's 1 0 0 that's 0 0 1 the middle will be zero if you divide it by - 1/3 it would look like that and again we have a use for those just bear with it and you'll get how this works all right now the next set of matrices again I really want you to get the hang of this really quick and get used to this forward and backwards and that's where we're doing now again tell me what is e so that you take e * B and you set it equal to F and again I copied those for you well look that's the same that's and I don't want you to think over we're always doing something for the second row that's that's not true to the second row yeah so let's see how this works well I went from 6 to 8 how how would I do that I went from six negative -6 to 8 no negative 6 to 8 remember this is Elementary you must have did one thing so either you could switch rows that's allowed or you could either you could interchange rows that's allowed you could multiply a row by an constant or you could multiply a row by a number and add it to the second row that falls under the elementary row operations and that's how you get the elementary Matrix so if I'm looking at this here I can't multiply you know like like for example this is one this is 21 you can't say I divided by 21 because -6 divided by 21 is not eight so here it looks like they multiplied a row by a quantity and added to the second row well let's see what they did uh h I thought they would add Row one but that's not working why is that you multiply so if you take this to get an eight oh boy uh okay B let me make sure b b e b equal f b and f b and f b b is oh okay cut it let me fix that made a mistake hold on okay I fixed the problem I didn't I actually had D instead of B I'm like wait that's not working so in this case they're saying this is Elementary again you can't always find an elementary operation otherwise life would have been easy so if I take uh uh what did I do now let's see I did something to this Matrix so to get an eight right there well to get an eight right there uh I could have multiplied this by two and add it to that would that work on this eight and two yep there you go so I must have did twice row two twice Row three plus row two and row two well what does that mean if I'm looking at 1 z0 0 1 0 0 1 If I multiply Row 3 by two and add it to row two wouldn't that become one two 0 0 one and there it is that's e similarly if I take okay let me write that for you can see it two and two if I want to look at the last row oh you know what let's leave it there that's okay look at the last and figure out what would be the elementary Matrix that you multiply f with to get B again well let's see again they left the first and the last it seems all the work is done right there so how do you get that how do you get an eight well H I could say okay there are many operations remember you're only allowed one operation and that's it now oh what do you do to this If I multiply this by a -2 that would be -6 8 + 1 is -7 and -2 plus okay there it is so if you take Nega to row two Row three add it to row two put the result in row two now one z0 0 1 0er Z 0 one If I multiply by a -2 and add it to that the only thing that will change is that and we already showed in the first example that if you multiply those two matrices it's the same as doing this operation to that Matrix now recall from the previous section how to find an inverse now there was something I didn't talk much about because there was no point then but this is what it says to find an inverse you take one over the determinant now you're dividing by this determinant what happen if the determinant equals zero this will be undefined so notice what they say the Matrix is invertible if and only if this denominator does not equal to zero and if it equal to zero so if it doesn't if and only if if it doesn't equal to zero it is invertible and it's invertible it doesn't it doesn't equal to zero if it equal to zero this is not invert you can't find an inverse recall so they say okay find a inverse if it exist well let's see first things first the product of those is that a zero that's -6 minus 15 no it's not so the inverse of this will be 1 over the determinant 1 over -16 + 15 minus A-5 swap the diagonals and multiply the non diagonals by a negative that's a negative one so the inverse of this Matrix is simply 16 -53 N1 and let's do it again on Part B if I cross those that'll be a -12 minus a -12 so the determinant of a is -12 minus a -12 which is zero so this is not invertible and that means an inverse to that does not exist and therefore I cannot I cannot come up with it now remember this trick only works on a 2X two I just wanted to show what happen happen remember there was another way of finding an inverse if you want to find the inverse of a you make a line right there and you put the identity Matrix on the left do row operations and when you get the identity on the left the enr will be on the right well let me just show you I just want to make sure you see what happens because this only works on a 2 by two when you deal with a 3X3 a 4x4 now the reason we didn't do that on those two problems because it's a lot quicker to use the formula if I am to do raw operations my first goal will be to make data zero so I want this to be reduced raw Eon form I'll take twice a row two plus Row one put the result in row two again the goal is to make that down that number down there is zero If I multiply this by at two that becomes -6 -4 0 and a two and if I add it to the first row and I'm going to put the result in the first row in the second row I'm sorry because I want that to be a zero I'll get a zero zero a one and a two well what do you notice do you see that that's z z there's no way of getting the identity anymore for this to work I need the left to be one z01 if that happens whatever is on the right hand side those four by 2 by two that will be the inverse anytime you cannot find an inverse an inverse does not exist and we'll talk about what that means as far as Solutions pretty soon that will happen one of the row zero out in the original which means there's no way of getting the inverse and therefore you cannot just find the inverse so every Elementary remember what elementary is every Elementary Matrix is invertible and its inverse is also an elementary Matrix so here's one again let's see what they want use the inversion algorithm to find the inverse of the Matrix if they exist now this process is really tedious and long and the chances of making an error is very high so just work with it work slowly as you're doing this so there is no shortcut to this this is the only method we have so far my goal is to make the left this exactly like this Matrix which is the identity so I start with the same thing what we did before I want to make that a zero and that a zero how do you do that so you take negative Row one plus row two put the result in row two take negative Row 1 plus Row three put the result in row three and what did we get the remaining outcome would be what uh Row one I'm using but I'm not changing now if you read the book they might multiply Row one by a five first you could do that if you wish If I multiply this by a negative that becomes -5 since I'm doing the same that's zero negative one5 plus one5 that's a zero -5 minus 45 that's -5 over five If I multiply this by-1 that becomes two fths if I take 2 over 5 and add it to 1 over 10 well that is 4 over 5 + 1 uh 4 over 10 + 1 over 10 is 5 over 10 5 over 10 is a half and that's the same now if If I multiply this by -1 and add it to this row that will become -1 1 0 If I multiply this by a negative one and add it to this row that'll become one01 I just want to make sure you see it I'm going to do that one so if I multiply r one by negative 1 that's - one5 - one5 two fths -10 0 and if I add that to row two that Zero's out that Zero's out that's 3 over five I'm sorry * 2 * 2 that's four 5 over 10 which is a half that's a negative one that's a one and that's a zero and that's what I wrote there so you could put it on the side write it out if you want that's not an issue actually I might do that for the next step now the next step I want to make that a zero well there's a problem if I use Row one that will change this zero so the only way to really make that look like the identity is to swap the rows take row two and swap it with Row three but remember now it's going to apply to the entire R Row one if I switch those two rows that will be 0 -1 12 1 0 1 and this row will become the last 0 0 121 zero okay now the goal was to make those three zeros I did now the reverse steps two more steps First Step I'm going to make the those two zeros and then I'll make the last one zero right there and I'm done well I don't know why I wrote that so small I'm going to rewrite it a bit bigger give me a sec now again I could multiply this by a negative one2 and add it to that and proceed I have a suggestion this is only me though you don't have to do this if you don't want to I strongly suggest from the beginning before I even started I would have multiplied this by five this by five and this by five so here I would say you know what I would take I'll take an extra step take Row one multiply it by five put it back in row one take row two multiply it by two put it back in row two and take Row three multiplied by two put it back in row three I would rather deal with integers than deal with fractions this is one one -2 5 0 0 this is 0 0 1 -22 0 and this is 0 -2 one uh I forgot that I switched them we switched Row one with row two crap oh boy if we switch them then that will become zero negative one again when I erased it I I just copied the old I copied that pretty much which is a mistake Z we go much better okay so if we do that then that'll become multiply by a two multiply by a two okay now it's much easier to play with again you're don't have to what I prefer that I'm going to take negative Row one and negative Row three add it to row two put the result in row two and I'm going to take twice Row three plus Row one put the result in row one so I'm looking at If I multiply this by a negative 1 and add it to the second 0 + 0 is 0 0 - 2 is -2 -1 + 1 is a zero 2 plus a -2 is a 0 -2 + 0 is a 2 and 0 + 2 is a two If I multiply Row three by a two then that row becomes 0 0 2 -4 4 and a zero and if I add it to row one I'll get a one one 0 1 four and a zero so as you see 3x3 will take four steps pretty much five because at the end you're going to have to reduce each one and make it a one last step is to make that a zero and to do that I would say take I meant I meant this and you should use that R so the last step I would say take row two okay so take Row one multiply it by two add it to row two put the result in row one this is set I'm not going to use it this one I have to modify in a minute the only that's going to changes Row one If I multiply Row one by two I'll get 2 2 0 2 8 0 add it to row two I'll get two 0 0 2 6 and two now hopefully I didn't make a mistake like I said the chances of making a mistake is very high and if I do now the last step take Row one divided by two put it back in row one and take row two divided by neg or multiply by negative one2 to be politically correct and the outcome would be one Z 0 and if I divide that by two that's 13 1 0 1 0 divide by -2 0 1 1 and the last one is okay and guess what that's the inverse provided we made no mistakes again the chances of making a mistake is pretty high you should work very slowly but that's pretty much how you run these now I do have the answer okay perfect I got the right answer here's another one now again before I start before you do any modifications you have to write and this is the only way we could find the inverse up till now right it's not a 2 by two so we can't find a shortcut well if it was you I would say look before I start take Row one time 5 put it back in row one take row two * 5 put it back in row two and take Row three * 5 put it back in Row three and what will happen I'll get a 1 1 -25 0 I'll get a two -3 -3 050 0 and a 1 -4 1 0 0 five and now I'll proceed just like we did I want to make those two zero first so the way that would happen I will take negative twice Row one + row two in row two and I'll take negative Row 1 + Row three in row3 so Row one I am using but I'm not changing If I multiply Row one by -2 that's a -2 -24 -10 0 0 add that to row two 2 -3 -3 05 0 I'll get a 0 -5 1 -10 5 and a zero okay and if I multiply Row one by a negative I'm looking at -11 -2 -5 0 add that to the third row I'll get a z51 uh -55 let me make sure of the numbers crap I see the mistake let me fix it the mistake was uh that's a 10 you want to multiply this by a 10 because again I have the answer to those and it wasn't working like it is right now so let me fix that really quick that shouldn't take long and let's see if I multiply by a 10 that will be a 4 -6 -3 0 10 Z multiply this by a 10 that's a 2 8 uh uh 1 z0 10 okay now that'll change few things uh if actually now if I take -4 Row one plus row two and row two and if I take -2 Row one plus Row three and Row three I don't know why I'm writing so small I have all the space on the planet take -4 Row one plus row two in row two and take -2 Row 1 plus row Row three and Row three well let's see multiply this by a-4 -4 + 4 that's a0 neg that's -10 that's 8 8 - 3 is 5 -20 uh 10 and zero and If I multiply by a -2 -2 + 2 is0 -2 that's -10 -2 that's four that's 5 10 a zero and a 10 oh and obviously if I take negative row two plus Row three and Row three what would happen if I add those two wouldn't that become 0 0 0 with a minus of course that become POS 10 - 10 5 + 5 it doesn't matter what that is once this zero is out we say this is uh not possible this is not invertible again a lot of work to get there okay and here I'm not going to work this out all the way this is easy to play but this one but I just wanted to point out that what you could do you could before you start again make sure you put those as zeros so you could say you know what take Row one divided by radical 2 put it back in row one and take row two divided by radical two put it back in row two and what would happen you're looking at now 1 3 0 but this will be 1 over radical 2 0 and this will be a -4 1 0 0 1 over radical to0 and this Remains the Same the work here is not intense First Step multiply Row one by by four added to row two that will become a zero and then all what you have to do is you know what let's do it might as well so if I take -4 Row one plus row two and row two 1 3 0 1 over radical 2 0 0 multiply by a 4 that's -4 -12 and that is a 0 -4 over radical 2 0 0 if I add those two that's a zero that's a -1 that's a zero that's -4 over radical 2 1 over radal 2 0 and this Remains the same and now if I could make that a zero I'm set how do you do that if you multiply Row 2 by 3 Row 1 by 11 and put the result in draw one and that should do the trick and if I look at 3 * row two 3 * row two will be uh 0 three 33 Z -12 over radal 2 3 over radal 2 and a zero and If I multiply 11 * Row 1 that's 11 that's 33 that's zero that's 11 over radical 2 that's 0 and a zero and if you add those you get 11 you get zero get zero 11 radical over radical 2 - 12 over radical 2 is -1 over radical 2 3 over radical 2 and a zero the numbers are not pleasant they're kind of annoying but you get the idea now all what you have to do take Row one divided by 11 put it back and Row one take row two /1 put it back and row two and basically you're done so you end up with uh one Z 0 if you multiply by 1 over 11 you get -1 over 11 radal 2 3 over 11 radal 2 and a zero divide by A1 that will be 4 over 11 Ral 2 And1 over 11 variable two hopefully I didn't make a mistake but you get the idea of how that works and this is the inverse of a now we'll do the same here one z0 z0 1 0 z0 Z again I I'm not really a big fan of doing a 4x4 but some of those are really easy to do I want to make sure that you believe that as well I want to make all of these zero well step one take negative Row one add it to row two and row two take negative Row one add it to row three in row three and take negative Row one add it to row four in row four what that does so 1 z0 0o 1 0o z0 if you multiply Row one by a negative it's the same one add it to the second row 0 3 0 0 -1 1 0 0 if you add it to this row that's 0350 01 0 1 0 and if you add it to the last that will be 0357 -1 0 0 1 now second step make those two zero take negative row two plus Row three and Row three and take negative row two plus row four and row four and what that does multiply this by a -3 well what's going to happen by negative 1 I mean that's going to be Zer -3 + 3 is zero the five remains now -1 + 1 is 0 -1 + 0 is1 1 0 and if you take negative row two and add it to row four that's 0 0 0 7 those will give you a zero that will give you a one a zero and a one and as you see all what I have to do now is take Row one oh no need row two divided by three put it back in row three row four oh boy row two divided by three put it back in row two Row three divided by 5 put it back in row three and row four divided by seven put it back and go forth so what will happen and this will be my invers provided we didn't make any mistakes on any of those that's pretty much it all right now we're almost done and this is the only section we're going to do today I'd like to look at number 20 you know I would have done it says find the inverse for each row by4 where K1 through K sub four are not zero all of them all of them are non zero because if one of them is zero you can no longer do it well aren't you g to First Take Row one and switch that with row four and take row two and switch it with Row three if that's the first step then I'm going to get K4 K sub4 0000 0 0000001 and if I put the first one last it looks like this and if I switch those two then I'm looking at zero K3 is 0 0 0 0 1 0 and I'm looking at 0 0 K sub two 0 0 1 0 0 all what I have to do right now is take Row one multiply it by 1 over4 put it in row one take row two multiply it by 1 over K sub3 put it in row two take Row three div multiply it by K sub 2 put it back in row three and take row sub four multiply it by 1 over K1 or divide it by K1 put it back in RS of four now I know that these are going to become all ones so the identity Matrix will automatically be 0 0 0 one over case of 4 0 0 1 over K sub 3 0 0 1 over K sub 2 0 0 and 1 over K sub 1 0 0 0 all right and the last example we have would be the following now the goal is to make again the identity you should write the identity first and then you begin oops messed up so if I want to make that a zero I would take Row one multiply or divide it by K and add it to row two put the result in row two that's my first step one uh K 0000 0 one Z z0 so if I multiply Row one by 1 over K that becomes one oh negative negative 1 over K that becomes negative 1 0 0 0ga 1 over k0 0 Z if I add it to one k0 0 0 1 0 0 what would happen I'll get a zero I'll get a k 0 0 -1 over K one0 0 and now as you guess we're going to do the same now I'm going to go to the second row say hey let's make that a zero how do you do that well obviously you take negative row sub two / by K but not that c plus row sub three put the result in row sub three so row sub one is k0 0 0 one z0 now of course you could divide row One by K right now and put it back in row one that'll save you time on the long one so you make that basically one and that's one over K now uh row two if I take row sub row two multiply by1 over K it becomes 0 - 1 0 0 1 / k^ 2 1 over K negative 0 0 and if I add that to row three I'm not lining those up properly so I have 0 1 0 0 1 over k^ 2 1 over K 0 0 and if I add to that 0 1 k z z0 one0 I'll get 0 0 k0 so if I divided that that's one Z 0 0 1 over k0 z0 now uh I lost track all right 0 k0 01 over K 1 0 0 so if I multiply Row Two by negative 1 over K and add it to row three put there Row three this is the product by 1 over K of this negative Z negative okay good and if I add it to this I'll get zero 0 k0 1 / k^ 2 And1 over K 1 and a zero okay and also as I proceed a bit longer than I anticipated but it's okay okay now I would move to that top say okay now let's look at what happens uh if I take 1 over K times oh that's not working okay if I take 1 over K * row 3+ row four negative put the result in row four I'll notice that I'm looking at one z0 Z and I could you know I'll do that at the end instead of keep on adding to the list so if I multiply this by 1 over K 0 0 1 01 over K cubed 1 over k 2 1 over k and a zero and if I add that 0 0 1 K 0 Z 0 1 I'll get z0 0 K - 1 over K cubed 1 over k 2 1 over k and a one Cub s 1 And1 1/ K cubed 1 over k 2 1 over K and one now take Row Two ided by K put it back in row two take Row three / by K put it back in row three and take row four divided by K put it back in row four that'll make all of these being one across the diagonals which is the identity and the inverse will simply be Row one is 1 over K 0 0 0 If I multiply by 1 over K I'll get - 1 over k^ 2 1 over K 0 0 multiply by 1 over K I'll get 1 over K cubed - 1 over k^ 2 1 over k and a zero and the last one I'll get one 1 over K to 4th 1/ K cubed 1/ k^ 2 and 1/ K and that would be my identity Matrix I know it doesn't look pretty again we not normally don't do a 4x4 I normally don't do any of those but these are the only ones I could find with the K now summing everything up last aspect we have to do if the elementary Matrix e results from performing a certain row operation kind of like what we did one operation on each and if a is an N byn Matrix so it's not a square Matrix then the product EA like we did is the Matrix do result when this same row operation is performed I already mentioned that and showed that instead of multiplying you could just do to that to the operation and there it is Express The Matrix and its inverse as a product of el ment matrices all right this one we have to be patient for so let's look at that and work at it slowly goal is to make this the identity Matrix the first thing I do I would take normally let's say I want to make that a zero so I'll take five times Row one plus row two and row two so here we will say the first Elementary operations we did so again if I had one z0 one if I was to multiply Row one by five and add it to row two wouldn't that be oops wouldn't that be it right there so that's the first operation that we did and what happens that will make this now one 02 then what then what would I do after that I want to make that a one wouldn't you say take Row 2 / by five oh my God take row two divided by two put it in row two well the same deal look at this the second operation will be what the second operation on the identity Matrix if I take row two / two and put it back in wouldn't that be what it looks like you know what I should have left the first one let me write it down so the first operation was take the first operation was take negative five Row one plus row two and row two and that will yield this and there's E2 and that should be it because if you do the actual division you will get a one Z1 well basically we're done but this is what we concluded this is the work now if I take those and I now I could say this Matrix a I could take this Matrix a we applied E1 first to it then we applied E2 in that order if I do that I'll get the identity so we applied E1 first and then E2 I get the identity but I could also say hey I love this part I could say but if I multiply by E2 E1 inverse this will be the identity so basically the left will be the a now do you remember how this works what in those switch order uh right there if you take a the inverse of product of two mates A and B that's becomes B inverse a if you recall and I could say that a which is this Matrix right the initial Matrix is actually E1 inverse E2 inverse because when you multiply by the ENT identity any Matrix you get the Matrix back and what are those now those are very special because they are Elementary operations so the determinant of this is 1 minus 0 which is 1 so basically you only swap the diagonals and multiply the non diagonals by a negative and here one over the determinant well the determinant is 1/2 one over determinant is two you're going to multiply by two swap those so the one2 will be up there and the one will be here times two that's one two and those are zero and did I do this right E1 E2 not only that I could say the inverse well what's the inverse if I take the inverse of this that'll be the inverse of those so those will be E2 E1 again if I take AAL e sub 1 in inverse d sub 2 inverse if I take the inverse of those a inverse will be remember those will swap and the inverse will undo each other and E2 is simply 1 Z 0 12 and 1 151 and if you're not sure you could multiply those and figure out whether this works or not so here if you multiply you'll get 1 + 0 that's one 0 + 0 that's Z -5 + 0 and z and 0 + two and that's really what we want out of it we're going to come back to this and do it again this is an equivalent so this is a theorem that we always get we're going to use that later on we're not going to use it for a proof but we're going to use the conclusion of it and it's very crucial every every section they're going to compile and add to this eventually we're going to get an equivalent statements of about 12 of those which says if you have an M by m Matrix which is a square Matrix then the statements are equivalent all of these are the same either they're all true or they're all false one of the two if you say a is invertible then you're saying ax equals z has a trivial solution guaranteed n the reduced raw Echelon of form of a is the identity Matrix and a is expressed as the product of Elementary matrices that was the very last problem that I did your homework is not as long as the problems I did it turned out all the even problems are long and complex all the odd are more straightforward so there's the homework and that concludes this section