In this presentation, we will make a reversing entry related to accrued interest time to engage with sage 50 cloud accounting Here we are in our get great guitars file We're gonna start off by opening up our reports going to the reports drop down looking at those financial statement reports We want that balance sheet report. So we're gonna double click on it so we can get it We're gonna have period two. That's gonna be February. So we'll start off there gonna say, okay And there is our report now You'll recall in a prior presentation that we had an adjusting entry related to the Interest payable accrued interest in other words if we go to our loan document We looked at our loan dot document over here And we saw that this this loan isn't going to be paid back to the end of five periods or six periods six months that is and after the first month we owe 250 and therefore we had incurred it just like if we incurred rent and we hadn't yet paid it so we had to do an Adjusting entry to record the fact that we had a payable of the interest payable and the interest expense on the other side of things Now we're gonna reverse that now just a quick caveat here to the reversing of this entry Note that we could have had the same kind of thing happen on this loan It just depends on when you know how much of the interest was accrued where we pay it monthly, right? So we paid the loan the loan monthly we might have some accrued interest And you can have the similar process and the reason I point that out I'm gonna go back to this loan is That if you have the reversing entry, it's the same kind of thing with the reversing entry I want to make it as easy as possible on the the difference between the accounting department and the adjusting department So in other words if I put in accrued interest here And then the accounting department had to make a normal type of loan payment and now they got this accrued interest on the books Well, it's already made it complex Let's go back over here and just imagine this situation where I had you know this kind of loan the more standard kind of loan well then you'd have to make the loan payment and then you'd have part of the interest which is on the balance sheet and then part of the Interest that might be in the current period that you'd have to record and then the principal reduction You made an already kind of complex type of transaction for the accounting department more complex and just writing a check more difficult And that's what we're gonna try to avoid so we're gonna say okay Well, let's just reverse all the adjusting entries so that when the loan payments made let's just go back on to this example Then they could just do it according to the amortization table and they don't have to think about you know The accrued interest thing that's gonna be the general idea now just note on this loan It's not the same kind of standard loan because we're not actually gonna pay it off until The end of six months and you're gonna pay the entire thing off So in this case it may not be beneficial to do the reversing entry because it's not gonna be something paid off till the end of The process, but I just want to give an example of Of the reversing entry so that it's gonna be something that the loan balance the accrued interests Could confuse things if the payment was to come up And this it could still confuse things if it wasn't to come up We're just gonna say hey We're gonna do the accrued interest as the end of the period Then we'll reverse it as of the end of the next period and then we'll make an adjustment You know each time as of the end of the period to record the accrued interest So this is what the reversing entry would be like if we go back on over We're gonna say then That we're gonna enter a reversing entry and let's just look at the the actual entry we put in place first if I go to this this 250 And double click on this entry what happened this is the actual entry we put in place We're just gonna simply reverse it exactly as of the first day of the next period that being March 1st So i'm gonna close this back out and then close this back out and then we're gonna go back to our Over here to our information and then i'm gonna go to the tasks Drop down so we can go back to our journal entry in this format and then go down to the journal Of the general journal entry now the general journal entry is going to be as of the first day of the following time period So it's going to be as of March 1st. So i'm going to go to March First and we'll pick up that date and i'm just going to simply reverse exactly what happened similar processes last time I'm not going to reverse the order of the account. I'm going to put 705 5 Which is the interest expense. I'm going to put that on top even though i'm going to credit it Because i'm going to reverse it exactly and that's going to be the 250. I'm going to put the description of a reversing entry And then i'm going to put the other account, which was 2420 wasn't it was 24 something i believe let me just Double check it here. Well, let's go ahead and find it. I'm going to scroll down. It's going to be The payable account. So there it is the interest payable 2410 2410 that's going to be the debit of the 250. All right, let's record this. I'm going to say save And there we have it. We're going to close this back out. I'm going to go back into the balance sheet report going into the balance sheet report and Scrolling down to the liabilities section. It's still there as of the end of february Now let's go to march. So i'm going to go back up to the options up top And we're going to change the time period to March let's go to the range And i'm going to make this march 1st to The 31st first to the 31st Okay, and now it's gone. So you can see it disappeared. Let's go to the other side the income statement This is going to be a little funny looking on the income statement And so i'm going to go back on over and open up the income statement And let's look at it first for February i'm going to remove these tick marks. I don't need the zero balances And there it is now the interest is included in The interest expense this 846 now. Let's change the date to march So i'm going to select the cog up top to do so We're going to pick the range. I'm going to change this to march march 1st Through march 31st and then say okay And so now all we have is this negative interest Expense and again that that will look funny because you wouldn't have a negative expense now. Normally it would wash out Once you make the payment. So once I made the payment and again, I think it's easier to see actually with this loan Are you right? Once I made the payment if I was to record all the interests to to the To the expense account then the two would net out and it would be correct at that point in time Would we look at this loan that actual payment's going to happen at the end of of six months? So if I was to make the payment At this point then The the normal process of making the payment would be to credit cash for the the 51 519 and then to Reduce the loan balance, which is going to be on the books for 50 000 And then the difference just go to the interest expense, which would be 1 519 would go to interest expense, but I'd have to worry about that payable Well, if I reverse the payable, I don't have to worry about it And then it'll take that 250 and apply it back to the prior period In in the proper format. So again, it wouldn't really happen in march. However, so if I was if I was to to Do this process in march Then it's not going to happen in march not going to happen for this particular loan until after six months So just remember on this particular loan it may not be necessary to do the reversing entry But just want to give a an example of what that reversing entry would look like So I'm going to go back on on over. So there's the 250 and again, it would look funny until Until the actual payment was made and then it would be proper would net out In the proper at that point in time. So that's going to be it for now. Let's get out of here


Chak chak Ya wan you don't want Ya zeme Ya boi kayo Ya chakia ya polushon Ya chakia ya polushon Kutia wa veru From Majiku From Vwandani Rungalunga Aya network is a project Where artists and Scientists came together And find out ways or solutions And how they can deal with the Polushon around me Me personally have affected me In a way that When my neighbor pollutes It affects me because I have respect A little bit challenges If you have affected the community Economically because People survive With the little money And when they get sick They direct their income To the health issues That leaves them with nothing For education or for food We hope to find solutions And educate the community Always they can keep themselves safe And away from this risk of pollution My way forward is to see people Really change And work toward a clean air And a better life


Hey, everyone. Welcome to this week's video update. Today's Friday, February 24th. Hope everybody had an excellent week of trading. A couple quick announcements. One, just five days left on our pre-launch for the new course, Trading Butterfly Spreads for Income. If you haven't already gotten that, make sure you go check that out. Got a ton of awesome feedback from our students. Everybody's loving it. It's a great strategy to have in your toolbox, so make sure you check that out. After March 1st, it's going to go up to the normal price, so make sure you get that now for just 49 bucks. Navigationtrading.com forward slash butterfly course to check that out. Second announcement, if you were not on our live stream yesterday, Facebook and YouTube have created this live streaming platform that's really cool. We're going to start doing more on that. Starting with, every Monday morning when the market opens, we are going to do some live trading. We're basically going to do a live trade of the week. You're going to see me go over my thoughts on the market on different symbols. You'll watch me place an actual trade that we'll call the trade of the week. Make sure you put that on your calendar. Every Monday morning at 8.30 central time, go to our Facebook page. Make sure you connect with us on Facebook so that you can see that coming. Or just go to our Facebook page at facebook.com forward slash navigation trading or youtube.com forward slash navigation trading. You can watch it live on either those platforms. Really excited about that. I think that's going to provide a lot of cool value and it's all free. So hopefully you enjoy that. Please leave comments, likes or anything in the comments section to help let us know what you think. Also, if you have any topics or specific strategies or things that you want me to cover on those live sessions, let me know. You can email us or just post comments there on facebook and youtube during the event. So look forward to that. All right, let's jump into the trade alert. So remember, President's Day was Monday, so market was closed. First trade was on Tuesday, February 21st and it was an adjusting trade in natural gas. For those of you that were in natural gas, you saw we had it and I think it ended up around 10% down in one day. Huge move in natural gas. Anytime that happens, you're typically going to have to make some adjustments. But just stay calm, stay mechanical, just make the adjustments and keep your position size small, not a big deal. So let's go to the platform and take a look at this one first. So essentially, we just rolled down our calls on our natural gas strangle. So this is one that had already been adjusted once, so now we are actually inverted. So we've got the 3.2 put and the, that's this one here, which is the 2.9 call. So remember, the calls are usually above the puts on a typical strangle, but we've had such a big move down in natty gas. See this is that, this is that day here, huge move. It's since come back and helped out our position. So nothing to do at this point. So we've got this inverted strangle, still got 32 days left. So I mean, that's a ton of time. We could have it bounce back and kind of get out of that trade for a break even, or if we need to adjust again and roll to the next expiration cycle, collect more credit, give ourselves more time to be right. We can do that as well. So that's the strangle. We'll come back to the other nat gas positions here in just a second. So the next one was an adjusting trade in our nat gas iron condor. So we had an iron condor on, big move down, and so essentially what we did was we just closed out the call side. We do that, you know, when it moves down past our break even, typically we're going to close out the untested side, which in this case was the call side, we're going to close that side out for max profit. And there's the bell, markets closed. So let's take a look at that. So that's this one here. So we took off the call side. Now price is down here. So we're just waiting for a little bit of a pullback. If we get some more up here, we'll get out of that trade for a break even or potentially a winner still. Obviously if it goes down, you know, we have with our risk is, risk is to the downside. So if natural gas keeps moving now, we're going to have some risk there. So to help offset that loss, I got some questions on this.  And that's why you need to be comfortable with the amount of risk when you put the trade on it at order entry. Okay, you've got to stay small enough so that you're not stressing out about this when you put on a defined risk trade. You need to have the mindset that if I take a total loss on this, it's going to be okay in regards to my overall account size. And if it's not okay, that means you're trading too big.  So, you know, you may, you know, just do one contract or if it's feel it's too big, too much risk, you know, don't don't do it at all. You don't have to do every single trade. There's always going to be opportunity every single day the market opens. But in this case, you know, natural gas is down here looking for a move up. But to help offset that, simultaneously after we took off the call side, we added another iron condor. So you can see this one is still very centered. And so we'll just continue to wait. So we, you know, we could, we can end up being a winner on all these if we get a little bit of a move up in natural gas. Or if we're a loser on the other one, we'll win in this one maybe and help offset that. So we're collecting more credit, giving ourselves more time to be right. And that's the name of the game. Just just stay mechanical and keep making the necessary adjustments and trades. Let's see. Next pay next trade was the one I just mentioned. We sold that new iron condor in that gas. EWW, we had a an adjusting trade. So I sent this adjusting trade out in two separate alerts. And so essentially we rolled EWW has been moving up. So we rolled our puts up to the 45 strike and we were in March. So we moved, we moved them out to the April only 23 days to left, left to expiration at that point. There's under 60 days in the next cycle. So instead of doing it, you know, keeping these in the March cycle and then, you know, waiting a few days and then rolling them out to the next one, it just, you know, save on commission made sense to just roll not only up, but out to the next cycle. And then the second part of that adjustment, go ahead and get to that before I go to the platform. Second part was we went ahead and rolled the calls also to April. That way we have our puts and calls both in April. So now we're holding the April strangle and it's the 45-47. So if we take a look at that, go to EWW. Now we've got this here. Price is well centered. We've got a nice profit potential there to get back, you know, some of the downside that we've incurred in EWW because it, you know, made a pretty massive move up over that period. So looking for a little bit of a pullback contraction in IV and then we'll bank that profit. Or if we need to adjust, obviously we'll do that too. Next trade was an opening trade in the Qs. So take a look at QQQ and look at it. This mark has just been parabolic to the upside, which causes implied volatility to be extremely low. So we put on a calendar trade already up a little bit in that one, not enough profit to take off. So we'll continue to monitor that one. We bought back a strangle, closed out a strangle in TLT for a nice profit made over 45% of max profit there. Current IV dropped down to 22. So in TLT we took off our strangle and then we still have our calendar spread on, which at that point was just the 119 calendar. Price continued to move up so outside of our break even. So first adjustment for calendar spreads is to add another one in the direction of the move. So we added the 123. So now we've got this double calendar, got a nice wide range for price to move around in. So we'll continue to to monitor that one. I had another opening trade sold a strangle in XRT. IV got up to 56 at that point, actually got even higher. So hopefully if you didn't catch this right away, you might have gotten in for more premium at an even better price. We entered in the April cycle with 58 days to expiration. So if we take a look at XRT, still still right in the middle of our range here, nothing to do there. But if you look at XRT, I mean, I got in this when IV percentile was at like 56. And then it spiked spiked a lot higher with this down move that we saw here. And I think that was just in the anticipation of some retail number, retail sales numbers coming out. This is the retail S&P retail ETF. And then as soon as those numbers came out, price moved up a little bit and implied volatility got crushed down from basically 100. So if you got in a little bit later than me, hopefully you got in when IV was a little bit higher, got even a little bit more a better pricing than me. So we'll continue to monitor XRT. And then let's see, we did a closing adjusting trade in DIA. So essentially, we had to make a second adjustment on our DIA calendar. So let's take a look there. So essentially, all we did was rolled our calendar. So real quick here, let me see, remember what the strike was. So yeah, so we closed out our 200 and we rolled it up to the 210 because price has just continued to move up. So we went ahead and rolled that calendar up. So if we go back here, so we had the 200, which would have been a calendar right here.  So we had this double calendar on. Price has just continued to have crazy move to the upside, obviously, just like all the other indices. So essentially, we just take off our 200 and we rolled that up to the 210. So now we hold the 205, 210 calendar, got a nice range to get back, took a little bit of a loss on the one we got out of here at the 200. And so now we'll try to make that back with our adjustments. We've still got in March, still got 21 days left. So good amount of time to get some of that back. Next trade was in opening trade. We did a sold and iron condor in FXC. Ivy percentile got right at that, right to that 50 level, went into the April cycle on this one. So that that trade is still fairly centered, nothing to do here except for weight. So we'll continue to monitor FXC. As you can see, you know, Ivy percentile did get up to 50, but it already went down to 37. So we need a little bit of a time decay in there to go by and hopefully take that one off for a nice profit, assuming it stays in a nice range for us. Next trade was in corn. So we had a closing adjusting trade in corn. So the price of corn moved down past our break-even point. So we took off the call side. We closed out the call side for max profit, still holding on to the put side. And then we added another iron condor, a centered iron condor there in corn. And we did that in the next cycle. So gave us more duration, collected more credit. So let's take a look at what that looks like. So and remember, you can't use the Ivy indicator on the grains. So you've just got to kind of monitor. There's some decent premium because we've had this move up in corn, some decent premium in the options there. So it's just kind of a manual process. But if we take a look at corn, so we've got our put side. So it's still in the range there. You know, just need a little bit more of a move up and then we'll take off that whole iron condor position for a profit. Then simultaneously put on another iron condor in the May cycle, collect more credit. Got a max profit on this one of 468. So hopefully corn will kind of stay in that range and give us some more profit in that one. Let's see. Well, last trade was today and that was a, we added an adjusting trade in TLT. So we had a 119 calendar and we added the 123. So let's go to TLT and take a look at that. So we had the, we had the 119. I'm going to reset these so I can uncheck my boxes. So we had the 119. Price moved up today past our break even. So we just added another calendar here. And so now we've got a much wider range and we'll wait to get some of that profit back. So take a look at TLT. Also had a little bit of a spike in IV today. Not enough to sell any premium, but you know implied volatility is so low and so many things that we're really just kind of putting on some more, you know, calendars and start potentially doing some directional trades. I saw a couple of things that I liked today, but I'll wait till next week and potentially look at some, you know, taking some contrarian moves on some, some things that have taken massive price extremes like XLV, for example, you know, this is the healthcare and it's just had a huge run up, you know, just looking for, you know, potentially a pullback. So maybe buy a put debit spread there. We'll look at some of those other things next week too. So let's see if there's any other positions that I hadn't mentioned. Wheat, still, still got an iron condor in that. We've got a nice profit, but we'll wait for a little bit more before we take that off. Apple has continued its, let's see, we've just got this on. So it's continued, it's moved higher. So we're down on this trade here, but still, you know, the main reason I put this on was to add some short delta to our portfolio. So it's doing its purpose. Obviously, we'd love to have it down, a down move in a lot of things would be beneficial, not only from our positions, but also implied volatility increasing. You know, again, just huge, huge move after earnings here. Hopefully we'll get a little pullback. We still have a ton of time on this stuff, 21 days. So don't, don't get scared out here. You know, I mean, there's a chance we could take a full loss on this, but again, stay small and we've still got time left. So don't let, let the probabilities play out. Don't, don't jump the gun. Don't get scared. Just keep, keep the trades on. EWW already mentioned that one and that makes, I mentioned all of them. So that's it for this week. Have a great weekend, everybody. If you have any other questions, let me know. Look forward to talking to you next week.


We are considering Gauss elimination method for solving system of linear equations. Today what we are going to show is that this Gauss elimination method is equivalent to writing l u decomposition of matrix A. So, the A is the coefficient matrix, l is going to be unit lower triangular matrix that means the diagonal entries they are going to be equal to 1 and u is upper triangular matrix. As I said these two methods like Gauss elimination method and l u decomposition they are going to be equivalent. So, in terms of number of computations we do not save anything, but when we consider backward error analysis of Gauss elimination method there this l u decomposition becomes convenient. And for a positive definite matrix we want to show that matrix A has a Cholesky decomposition that means it can be written as G into G transpose where G is going to be lower triangular matrix and N G transpose will be upper triangular matrix. So, for that we are going to need l u decomposition. So, what I am going to do is I will first illustrate it for 3 by 3 matrices and then consider general case n by n matrix. So, let us recall what we have done last time. So, we have a system of equations n equations in n unknowns which we write as n by n in the compact form as A x is equal to b, A is the coefficient matrix, B is right hand side vector which is given to us and x is unknown vector x 1, x 2, x n. We had assumed that matrix A is such that if you look at principle leading sub matrix which is formed by first k rows and first k columns. So, determinant of this A k is not equal to 0 for k is equal to 1 to up to n this condition is stronger than invertibility. The invertibility it means that determinant of A is equal to not equal to 0. Now, we want to write A is equal to not only determinant A to be non 0, but also determinant of A k not equal to 0. Now, this Gauss elimination method we will write, we will show that it is equivalent to writing A as l into u. So, the Gauss elimination method will be equivalent to saying A is equal to l into u, l is unit lower triangular, u is upper triangular matrix. This is what we want to show and then that will be the LU decomposition. So, we will always assume that LU decomposition means l is unit lower triangular and u is upper triangular matrix. What we are going to show is yesterday we saw that using elementary row transformations we had reduced A to an upper triangular matrix u. So, that is going to be our u upper triangular matrix and l unit lower triangular matrix will consist of the multipliers. In order to introduce zeros or to reduce matrix A to an upper triangular form our first step was multiply the first row by A 2 1 by A 1 1 subtract it from second row. So, this M 2 1 is equal to A 2 1 by A 1 1 is going to be multiplier. So, lower triangular matrix we are going to show that it will be such that it will have one on the diagonal and the remaining entries will be multipliers in the Gauss elimination method. So, l is lower triangular along the diagonal 1 and here you have got M 2 1 M 3 1 M n 1 these were needed to introduce zeros in the first column below the diagonal and so on. So, we are going to first consider 3 by 3 matrix. Now, here are some notations E j will denote canonical vector which has got only one at j th place all the other entries they are 0. When we talk of vector x it is always going to be a column vector. So, this will be equal to x 1 E 1 plus x 2 E 2 plus x n E n i j th entry of n by n matrix we denote by A i j if you look at A multiplied by E j E j is this canonical vector then what we get is j th column of A you can verify this. If you look at E i transpose A then you will get i th row of A and if you consider E i transpose A E j A is n by n matrix E j is n by 1 vector A is n by n matrix E i transpose will be 1 by n vector. So, the result is going to be scalar and that is our A i j. So, now, let us look at 3 by 3 matrix. So, here is a 3 by 3 matrix and in the Gauss elimination what we are doing is the first step is going to be look at A 2 1 by A 1 1 that is our M 2 1 and then R 2 minus M 2 1 R 1. Then you look at A 3 1 by A 3 1 1 that is M 3 1 and then R 3 minus M 3 1 R 1 this is the first step of Gauss elimination method. The way we have chosen M 2 1 and M 3 1 the entries here they are going to be 0. So, you are converting matrix A to this matrix. Now, this entry A 2 2 that will be modified to A 2 2 superscript 1 which will be A 2 2 minus M 2 1 and then the corresponding entry A 1 2. A 2 3 will be modified to A 2 3 minus M 2 1 and then M 3 that is the result of R 2 minus M 2 1 R 1.  Now, these operations we could have performed by multiplying our matrix A by this matrix. So, the matrix has 1 along the diagonal here you have entries to be 0 and here it is minus M 2 1 minus M 3 1. So, when I do pre multiply matrix A by this matrix first row into first column will give us A 1 1 then first row into second column A 1 2 and first row into third column A 1 3. So, no change in the first row then minus M 2 1 times A 1 1 plus A 2 1. Now, what was M 2 1? It is A 2 1 by A 1 1.   So, thus the first step of Gauss elimination method is pre multiplying our matrix A by this matrix. So, now consider matrix E 1 if I define M 1 to be this vector 0 M 2 1 M 3 1. When I look at M 1 E 1 transpose E 1 is our canonical vector E 1 transpose is 1 0 0 M 1 is this vector 0 M 2 1 minus M 1 M 3 1. When you multiply you are going to get matrix of this form this is our E 1 this is M 1 E 1 transpose. So, E 1 will be identity matrix minus M 1 E 1 transpose. If you look at E 1 transpose M 1 that means you are interchanging the order. So, it is going to be 1 0 0 0 M 2 1 M 3 1. So, that is going to be 0 using this you can check that I minus M 1 E 1 transpose into I plus M 1 E 1 transpose is equal to identity which means inverse of this matrix E 1 is going to be identity plus M 1 E 1 transpose. So, this is the first step of Gauss elimination method. Now, we are going to look at the second step. So, in the second step we are looking at only 3 by 3 matrix. So, we will have to make only 1 entry to be 0. We are going to look at the third row minus M 3 2 times R 2. This is after the first step of Gauss elimination method we have got this. Now, you define M 3 2 to be A 3 2 1 divided by A 2 2 1 and subtract from the third row the second row multiplied by M 3 2. You have zeros here. So, these will not be affected this entry will become 0 and this entry will get modified. So, A 3 3 2 is equal to A 3 3 1 minus M 3 2 A 2 3 1 and you can verify that this can be achieved by pre multiplying our matrix A 1 by this matrix. You have got 1 0 0. So, the first row will remain unchanged. Second row here is 0 1 0. So, the second row also remains unchanged. When you look at the third row the third row into first column will give you 0. Third row into second column because of the choice of M 3 2 this becomes 0 and A 3 3 1 gets modified to A 3 3 2. As before let us look at matrix E 2 and define M 2 to be vector 0 0 M 3 2. M 2 E 2 transpose that is going to be a matrix which has 0 everywhere except M 3 2 here and hence E 2 will be identity matrix minus M 2 E 2 transpose. Consider I minus M 2 E 2 transpose into I plus M 2 E 2 transpose this is going to be equal to identity because E 2 transpose M 2 is going to be 0 you will have identity then minus M 2 E 2 transpose and plus M 2 E 2 transpose. So, that gets cancelled and then minus M 2 E 2 transpose M 2 E 2 transpose E 2 transpose M 2 will be 0. So, you are left with identity. So, E 2 inverse is identity plus M 2 E 2 transpose. So, thus our Gauss elimination is equivalent to doing E 2 E 1 A and then you get matrix A 2 which is equal to our upper triangular matrix U. We have seen that E 1 and E 2 these are invertible matrices. So, we have got E 2 E 1 into A is equal to matrix U which will mean that A is going to be equal to E 1 inverse E 2 inverse U and this E 1 inverse E 2 inverse we will show that that matrix is going to be a lower triangular matrix with one along the diagonal. So, that is going to give us a L U decomposition of our matrix A. It is a bit technical, but the idea is simple. It is just that whatever operations we are doing in the case of Gauss elimination method they can be performed by pre multiplying our matrix A by an appropriate matrix which is say E 1 then E 2 and so on. So, let us look at E 1 inverse E 2 inverse that is identity plus M 1 E 1 transpose and identity plus M 2 E 2 transpose multiply. So, that will be identity plus M 1 E 1 transpose plus M 2 E 2 transpose and E 1 transpose is row vector 1 0 0 M 2 was 0 0 M 3 2. So, that is 0 and hence this term is not there. Now, identity matrix that means you have got these ones and remaining entries 0 M 1 E 1 transpose we had seen that M 1 E 1 transpose is the matrix 0 everywhere except M 1 E 1 and M 2 1 and M 3 1 and hence you have M 1 E 1 transpose will be contributing M 2 1 M 3 1. M 2 E 2 transpose was matrix with 0 everywhere except entry M 3 2 and thus our E 1 inverse E 2 inverse is going to be a lower triangular matrix with diagonal entries to be equal to 1. So, this was for 3 by 3 matrix. Now, I am going to quickly do for n by n matrix, but not going into all the details the idea is similar. So, look at n by n matrix and look at the first step.  This is our matrix a we want to introduce 0's here. So, you consider a 2 1 by a 1 1 multiply the first row and subtract from the second row. So, as we had seen before this first step method can be performed by pre multiplying our matrix a by this matrix call this matrix to be E 1. So, we have E 1 a is equal to A 1 that is the first step of Gauss elimination method this matrix E 1 as before define vector M 1 to be 0 M 2 1 M n 1. So, we had done it for the 3 by 3 matrix. So, in that case our matrix our vector M 1 was 0 M 2 1 M 3 1. So, now the only difference is instead of a 3 by 1 vector you have got n by 1 vector. So, 0 M 2 1 M 3 1 M n 1 then our matrix E 1 is nothing, but identity of the matrix E 1 is identity minus M 1 E 1 transpose exactly same as before only difference is instead of 3 by 1 vector you have got n by 1 vector. So, this matrix E 1 which is identity minus M 1 E 1 transpose it is going to be an invertible matrix and its inverse will be given by E 1 minus M 1 E 1 transpose same proof as before. So, E 1 is identity minus M 1 E 1 transpose and E 1 inverse is going to be equal to identity plus M 1 E 1 transpose. Next in the second step of Gauss elimination method you have a i j 2 to be equal to a 1 minus M i 2 a 2 j 1 i j varying from 3 up to n. So, here now you define your vector M 2 which has the multipliers. So, the multipliers are M 3 2 M 4 2 M n 2 and then E 2 is going to be your matrix which has 1 along the diagonal minus M 3 2 minus M n 2. If you look at E 2 multiplied by A 1 that is going to give you A 2. So, you have started with A you pre multiplied by E 1 and you got a modified matrix A superscript 1. Now, you multiply by E 2 and then you get A 2. This E 2 is also invertible matrix E 2 is identity minus M 2 E 2 transpose its inverse is identity plus M 2 E 2 transpose and in general your E k is going to be identity minus M k E k transpose E k inverse will be identity plus M k E k transpose and then when you look at E n minus 1 into E n minus 1 into E n minus 2 up to E 1 multiplied by A then you are going to get your upper triangular matrix U exactly the same matrix which we had obtained in the Gauss elimination method. In the Gauss elimination method our system A x is equal to b we had converted into an upper triangular system U x is equal to y and if it is a upper triangular system then we can do back substitution. So, we first determine x n then x n minus 1 and so on. So, this U 1 can obtain by pre multiplying A by invertible matrices E n minus 1 E n minus 2 up to E 1 then that gives you A to be equal to E 1 inverse E 2 inverse up to E n minus 1 inverse. Now, if you remember in yesterday's lecture we had said that our aim is to reduce the system A x is equal to b to a system U x is equal to y and that should be a equivalent system that means the original system and the new system they should have the same solution. So, now here is a proof of this that when you do Gauss elimination method then the new system is equivalent to the earlier system. So, we have got matrix A U pre multiply by matrix E. So, E is going to be matrix obtained by multiplying E n minus 1 E n minus 2 up to E 1 each of E j is invertible matrix. So, E is going to be an invertible matrix. So, we have A x is equal to b which is same as E A x is equal to E b where E is invertible and then E into A will be our U. So, you get U x is equal to y. So, if x is a solution of A x is equal to b it is going to be solution of U x is equal to y and the converse is also true. If x is a solution of U x is equal to y then it will be a solution of A x is equal to b. Now, look at E 1 E inverse it will be E 1 inverse E 2 inverse E n minus 1 inverse E 1 inverse is identity plus M 1 E 1 transpose E 2 inverse is identity plus M 2 E 2 transpose and E n minus 1 inverse is this. You multiply and when you simplify what you are going to get is E inverse to be this lower triangular matrix with one along the diagonal and the entries here to be the multiplier. We have E into A is equal to U. So, A is equal to E inverse U and now E inverse is equal to L. So, you get A is equal to L into U. Now, you have to note that the Gauss elimination method it can be performed provided at no stage our pivot is becoming 0 and that means we have proved that at A can be equal to L into U if at no stage the pivot becomes 0. So, let me look at the system A x is equal to b. This A we have written as L into U. So, you have got L into U x is equal to b. Now, this I will split into two systems U x is equal to y and L y is equal to b. So, this is given as B is given vector. So, look at L y is equal to b. L is matrix lower triangular 1 1 1 then you had here M 2 1 M n 1 and so on. These are all entries to be 0 y 1 y 2 up to y n is equal to b 1 b 2 up to b n. So, when we look at the first equation it is y 1 is equal to b 1 vector b is given to us. So, you determine y 1. So, you determine y 2. In the next equation you will have y 1 and y 2 y 1 is already determined. So, you determine y 2. So, this is going to be forward substitution. So, you determine y 1 y 2 and y n. Once you have done that then you look at U x is equal to y. Now, in U x is equal to y the right hand side we have determined and there you are going to do back substitution. So, we have either you consider Gauss elimination method or you look at A is equal to L into U and solve two systems of equations. Once forward substitution, once backward substitution and both of these they will need the number of computations to be of the order of n square. Whereas, finding L and U that will be of the order of n cube by 3. So, what we have done is we have proved that the Gauss elimination method is equivalent to or both of them are the same. So, you know one can show that if you have performed Gauss elimination method. So, you have obtained U, you have got multipliers construct matrix L and then you have got A is equal to L into U. So, now maybe what one can do is try to write this A is equal to L into U directly. So, before we do that trying to write directly, let us show that such a decomposition is unique. If you say that A should be equal to L into U where L is lower triangular U is upper triangular such a decomposition is not unique. But if you say that L should be unit lower triangular that means all the diagonal entries should be equal to 1 that is what makes the decomposition to be unique. Now, the proof of uniqueness is straight forward. There what we are going to use is if you have got two upper triangular matrices and if you take their product that is going to be again an upper triangular matrix. Similarly, if you have got two lower triangular matrices if you take their product then it is going to be again a lower triangular and if these two matrices are unit lower triangular their product is also unit lower triangular. This we will do as a tutorial problem the verification that product of upper triangular matrices is upper triangular. We also have a result that if your matrix is lower triangular and it is invertible then its inverse is also lower triangular. So, using these two results we are going to show that a U decomposition of a matrix A is unique where L is unit lower triangular and U is going to be upper triangular. Our assumption is determinant of A k not equal to 0 where A k is principal leading sub matrix of order k which is formed by first k rows and first k columns of our matrix A. So, under these assumptions our matrix A is going to be invertible matrix. Now, determinant of A will be determinant of L into determinant of U. We have proved existence of L U decomposition. Start with a matrix A with the property that determinant of A k is not equal to 0 perform gauss elimination method. The final matrix upper triangular matrix that is U construct L using multipliers and that gives you L. So, we definitely know that a matrix A can be written as L into U. Now, we want to show that such a decomposition is unique. Determinant of upper triangular matrix or lower triangular matrix is product of the diagonal entry. Now, L has all the diagonal entries to be 1. So, determinant of L is going to be 1. Determinant of U that is same as determinant of A because what we have done is we have used elementary row transformations and elementary row transformation of only one type which is multiplying the row by a non zero constant and subtracting from other row. So, such an operation does not change value of the determinant. So, this determinant of A is going to be determinant of U. So, determinant of U will be not equal to 0 because determinant of A is not equal to 0. So, our U is going to be a invertible matrix and also L will be invertible matrix. So, let us start with 2 decompositions.   Determinant of U 1 is determinant A which is not equal to 0. So, U 1 is invertible determinant of L 2 is equal to 1. So, L 2 is invertible.  U 1 is upper triangular and hence its inverse will be upper triangular. Product of two upper triangular matrices is upper triangular. L 1 is lower triangular, L 2 inverse is lower triangular. So, their product is also going to be lower triangular. So, you have on one hand a lower triangular matrix on another hand an upper triangular matrix. So, these two are equal provided both of them those are diagonal matrices.  In addition L 2 inverse L 1 is going to be unique lower triangular. So, that means all the diagonal entries they are equal to 1. So, your L 2 inverse L 1 then U 1 inverse U 2 both of them they will be diagonal matrices with diagonal entries to be equal to 1 that means identity matrix.  So, that is uniqueness of L U decomposition and this uniqueness we are going to need when we want to show that a positive definite matrix can be written as G G transpose that is the Cholesky decomposition. So, now let us see like we know that a can be written as L into U and such a decomposition is unique. So, let me see whether I can try to determine the elements of L and U directly that means not going through the Gauss elimination method and then multipliers and then constructing L and all. What I can try to do is write a as L into U where L is unit lower triangular matrix U is upper triangular matrix and try to determine the entries of matrix L and matrix U. So, matrix A is given to me the entries of L and U these are not known. So, you will multiply the corresponding entries and then try to determine. So, again I am going to quickly do this. So, we are writing A as a unit lower triangular matrix. So, you have got one along the diagonal and then L 2 1 L 3 1 L n 1 that will be the first column and so on. All the entries above the diagonal they are going to be 0. U will be upper triangular matrix.  Some of the entries above the diagonal in U also can be 0, but what we know is definitely below the diagonal they are all 0. Now, you look at first row into first column. So, that is going to be U 1 1. So, that U 1 1 will be equal to A 1 1 like that then first row into second column, first row into third column and so on that will give you A 1 j to be equal to U 1 j that means you have determined the first row of U. Our L the first row is only has only one and remaining entries 0. So, when I consider first row of L multiplied by various columns of U what comes into picture are only the entries of the first row of U and then you get A 1 j is equal to U 1 j. So, you have determined first row of U. Now, you consider the L 2 1 U 1 1 that is going to be equal to A 2 1 then third row first column, fourth row first column and so on. When you do that you are going to have L I 1 U 1 1 is equal to A I 1 and L I 1 is equal to A I 1 divided by U 1 1. So, that means you have determined first column of L. So, we write A as L into U L unit lower triangular U upper triangular and then the first row of U is determined first column of L is determined. Now, we will determine the second row of U second column of L third row of U third column of L and so on. In this order we can determine all the entries of L into U. Now, what you have to notice is that when you do this thing all the diagonal entries of U they have to be not equal to 0 because in the first one we notice that we are dividing by U 1 1 the matrix A is given to us and we are trying to determine the entries of L and U. Now, in the second step the first column of L is known first row of U is known. So, you consider second row of L and jth column of U that is going to give us A 2 j. So, this A 2 j is going to be equal to L 2 1 U 1 j plus U 2 j j going from 2 up to n. A 2 j's are given to us L 2 1 we have already determined U 1 j elements of the first row are determined. So, that will determine the second row of U then Ith row of L into second column of U that will determine the second column of L. In the second row of L these being all zeroes we have already determined second column and second row of L. So, you have so far determined first row of U first column of U second row of U second column of U and similarly for the matrix L. So, one continues this and one determines L and U. You can do the second row of U then number computation of number of operations. We have already computed the operations in the Gauss elimination method and we saw that they were of the order of n cube by 3. Now, we are doing it differently, but you will see that here also the total number of operations they are going to be n cube by 3. So, we do not gain as such in the number of operations whether you do Gauss elimination method or whether you determine L and U directly you are going to be the number of operations they are going to be the same, but as I said this writing A as L into U that is useful in doing the backward error analysis for Gauss elimination method and for considering the Cholesky decomposition of the matrix A. We have got L U decomposition. Now, we have got another decomposition which is known as L D V decomposition. So, what are this? L is unit lower triangular, V is unit upper triangular and D is going to be diagonal matrix.  And then V is equal to D 1 inverse U. So, this is my definition when you try to look at the entries of V.  So, you get V I I to be equal to 1. So, that means if you define V to be equal to D inverse U where D is diagonal entries diagonal matrix consisting of D inverse I I. So, we have diagonal entries of U then our matrix V it becomes an upper triangular matrix with diagonal entries to be equal to 1. So, we have A is equal to L into U V is equal to D inverse U that means U is going to be equal to D into V. So, we have A is equal to L D V. So, starting with L U decomposition we have proved existence of L D V decomposition where L is unit lower triangular V is unit upper triangular and D is diagonal matrix. Now, soon it will become clear to you why we are going to this that we had L U decomposition that why not be satisfied with it, but go now to this new decomposition L D V. So, let us show that this L D V decomposition is also going to be unique. We have proved the existence and uniqueness of L U decomposition from the L U decomposition we deduced L D V decomposition, but there can be some other way of finding L D V. So, we want to show the existence. So, we have A is equal to L D 1 V 1 is equal to L 2 D 2 V 2 where L 1 L 2 these are unit lower triangular D 1 D 2 are diagonal matrices and V 1 V 2 are upper triangular and unit. The diagonal entries to be equal to 1. So, now, let me look at this way. So, by uniqueness of L U decomposition what we get is L 1 is equal to L 2 and D 1 V 1 is equal to D 2 V 2. This will imply that D 2 inverse D 1 is equal to V 1 inverse V 2. Now, is V 1 invertible yes, because V 1 is going to be unit upper triangular. So, determinant of V 1 is going to be equal to 1. So, we have got D 2 inverse D 1 that means a diagonal matrix V 1 inverse V 2 that means it is going to be a unit upper triangular matrix. So, both of them they have to be diagonal matrices and since the diagonal entries of V 1 inverse V 2 they are going to be all equal to 1. All both of these they have to be equal to identity matrix. So, that gives you D 1 is equal to D 2 and V 1 is equal to V 2. Today's lecture we have shown that the Gauss elimination method can be written can be expressed as a L U decomposition of the matrix. Then we proved uniqueness of L U decomposition and then we also saw how to determine L and U by starting with the formula A is equal to L into U and then taking the matrix multiplication and identifying various entries. Then we have talked about L D V decomposition we proved is uniqueness. So, in tomorrow's lecture we will use this for showing that a positive definite matrix has got a Cholesky decomposition. So, thank you.