find-a-different-factorization-of-the-a-in-exercise-29-and-thereby-design-a-different-ladder-network

Name: find a different factorization of the a in exercise 29 and thereby design a different ladder network
Uploaded: 2020-09-17
Duration: 5 min 21 s
Channel: Numerade
Description: Find a different factorization of the A in Exercise 29, and thereby design a different ladder network whose transfer matrix is A.

Question

Find a different factorization of the A in Exercise 29, and thereby design a different ladder network whose transfer matrix is A.

Chris Trentman · Accepted Answer

we were asked to find a different factories ation of the Given matrix from Exercise 29 and then design a different letter network with the transfer matrix of a so recall that from the previous problem transfer matrix A was four thirds negative. 12 negative 1/4 3 recall also from the previous exercise we had that when we constructed a circuit we had resistance are one of 36 owns distance are too of 12 homes and resistance R three of six homes and we saw that are one was a shunt circuit are two was a serious circuit and our three was a shunt circuit And so they had the factory ization 10 negative 1/6 0 one negative 12 01 times 10 negative 1/36 0. So now we can actually use this factory ization to find a new factory ization. So one of the easiest things to Dio is just simply break up this matrix in the middle by factoring into two different matrices. So this could be factored as 10 negative 16 zero one I mean, times matrix one negative 601 one Negative 601 You can just check this on your own. Recall that for unit upper and unit lower triangular matrices. They&#x27;re multiplied by adding their non unit number if there are two by two together times 10 negative 1/36. So this is a new factory ization for a and we see that the network corresponding to this factor ization, has well looking on the right. This is a shunt circuit followed by to Siri&#x27;s circuits, followed by a shunt circuit again with resistance is, uh, r one equals six are to equals. Sorry, Are one is 36 are two is six are three is six and our four is six.

Chris Trentman · Answer

in part, they were asked to compute the transfer matrix. Uh, the network given in the figure for problem 29. So looking at figure, we see that it is a network with we have a shunt circuit, followed by a serious circuit, followed by another shun circuit. So it follows the transfer matrix of the network is the product of matrices for each circuit and rightto left order. So we&#x27;re starting with the shunts circuit. Resistance are three. This is 10 negative. One of our our three one times. And this is going to be the serious circuit with resistance. R two, this is one in the Siri&#x27;s circuit has transfer matrix one negative are too 01 and followed by the transfer matrix for the Sean circuit with resistance R one just 10 negative one for our one one. Then we see that multiplying these matrices together we get if I start with the one here on the left in the middle, we have one negative are too negative. One over r three and are two of our three plus one. Multiply this by 10 negative one over our 11 and we get one plus are two over our one negative are too negative. One over R three minus are to over. Our one are three minus one over R one and finally are two of our our three plus one. And this could be rewritten if we&#x27;d like as r one plus R two over our one negative are too the opposite of here. We have least common denominator is R one R three speak it R one plus r two plus our three over are one are three and then we have our two plus r three over our three. So this is the transfer matrix for the whole network now in part B, whereas to design a ladder network with a transfer matrix of A by finding a suitable matrix factory ization of the matrix A now a is the matrix four thirds negative, 12 negative. 1/4 three. So we want to find resistances R one R two and r three such that a equals this matrix up here. So we see that immediately are two must be 12 homes and plugging that in. We see that four thirds must equal r one plus 12 over our one so one plus 12 over our one or thirds minus one is one third So 1 30 equals 12 over our one. So our one must be equal to 36 homes and we see that are two of our our three plus one has to equal three. So are to ever are three as equal to or are three as two equal are to over to which is 12 over to or six owns. And so the matrix a could be factored since this, it&#x27;s this transfer matrix. They can be written as 10 negative one of her are three. Where are three with six homes? One followed by one negative are too so negative. 12 01 followed by 10 negative one over R one, which is negative. One of her 36 one. So we have given a factor ization for a

Sean Thrasher · Answer

Okay, so we&#x27;re given this matrix A and this is back from exercise 12 on were asked to find the Q R decomposition. So let&#x27;s go back to page 250 nine&#x27;s. Figure Out what the cure factories ation is. So we have a M by N Matrix A. And it must have linearly independent columns. And you should check that the columns are indeed linearly independent. Then I can factor out A as cute times are where Q is an M by N. Matrix, whose columns form on Ortho. Normal basis for the column Space of A and Our is an M by n upper triangular in vertebral matrix, where the entries are positives of positives entries on the diagonal. So the first big step and our little adventure to solve this question is to figure out what Q is. Okay, So we&#x27;re looking for the Matrix, whose columns former North of normal basis for the column. Space of a Well, let me take S. So let me take this to be x one. This to be extra and this is this to be x three, then all I&#x27;m saying by column A is column mazes the span this substance is the subspace, spanned by x one x two x three and I&#x27;m looking for an Ortho normal basis for the subspace column space of A. And so I must use the Gram Schmidt procedure too. Make this basis into a North a normal basis. Okay? And so we go back to exercise 12 where we figured out a nor Fogel basis for the column space of A and it was given by this. This was a perfectly good orthogonal basis, right? The span of the one V two V three was the same as column A on dhe. We won&#x27;t be too V three are Fogell. Okay, so this is my orthogonal basis. But I need I want an ortho normal basis for the column. Space of a Okay, then, you know, and then I&#x27;m gonna put those or or thermal basis as my columns of Q. Okay, Well, to find the north Tunnel north a normal basis, simply take the results that you get from applying ground schmidt and normalize each of the vectors at the very end. So the magnitude of this guy is too Okay. Magnitude of this guy. Well, when you divide it by the magnitude. You get this vector when you divide this vector by its magnitude, you get this so on. So these guys are the unit vectors in the same direction as these guys here. So when I take this and normalize it, I get this. Take this and normalize it to get this. Take this and normalize it. I get that. Okay? Meaning I divide by the magnitude. Okay, so then Q is going to be this guy right here. Okay, because you know Q is an M by N. Matrix, whose columns? Former North normal basis for column Call him space of A And we figured out that this is a perfectly good example. All right, We&#x27;re not done, though. Okay? Now, the harder part is to figure out what our is. Okay? No, um, let me just make a little aside, okay? So a Z Q. Times are right now I want to figure out what our is, so Q. Well, um, it&#x27;s tempting to write down this immediately. Right? But first of all, you need to check that Q inverse exists, but Q in verse indeed exists because the columns Forman, or so normal basis and So you know, it&#x27;s linearly independent in particular, right? The columns on the nearly independent and so cues in vertebral. Okay. And so I can write this down. I can solve for our That&#x27;s fine. Okay, however, figuring out Q and versus a little bit complicated. So there&#x27;s a little bit of a trick, right? Andi, this is mentioning the textbook, but I&#x27;m going over this again. So since cute has columns that are also normal cute is on orthogonal matrix. Okay, He was an orthogonal matrix since as columns that are that former North a normal you know, basis. And so this means that Q times cute transposed is the identity matrix. And so in particular, you know, in particular Q is in vertical. Okay, in particular cues in versatile on dde que in verse que in verse is the same as cute, transposed. And so we&#x27;re going to use this fact right here. It&#x27;s gonna save us a lot of work. Okay, So far, North Diagonal matrix. If you want to figure out its inverse simply takes that simply take the transpose. Okay. You don&#x27;t have to use the formula for finding the inverse of matrix. Um, just a little side note. Okay, so it&#x27;s tempting to think that it&#x27;s tempting to think this This is just a side note. I&#x27;m going on a little bit of around here, but I think it&#x27;s important. Um, let&#x27;s say that I had a matrix stem where the columns of em were orthogonal. Right. So it could be something like 00 0 to 0003 Right. Let&#x27;s say that such a matrix is this matrix on orthogonal matrix. The answer is no. Okay, so you know what is an orthogonal matrix on Orthogonal matrix North Organon Matrix has to have columns that are ortho normal. Okay, so it&#x27;s not good enough for them to just be orthogonal. So that&#x27;s something you should be careful about. Yeah. Okay, so let&#x27;s go back here. So are is Q inverse times, eh? But I have already just mentioned that that&#x27;s the same thing is cute transpose times a day and be careful. The Q has to the cute transpose has to be on the left. Right. Okay, so when I multiply this out, I&#x27;m gonna get a three by three matrix. Okay? And so let&#x27;s figure out what the entries are So this is gonna take a little bit of time. So a 11 almost by the sounds, you get 1/2 plus 1/2 plus 1/2 plus 1/2 which gives you two a one to this guy right here. So that&#x27;s this role with this column. Okay, so that&#x27;s three halves plus three halves plus five House Plus five House, which gives you eight A 13 So it&#x27;s this role times this column. So it&#x27;s five halves minus 1/2 plus two halves plus eight lives, which gives you some seven a 21 So that&#x27;s going to be this guy with this guy. So that gives you a minus. Route two by four minus were to buy floor plus word to buy floor plus word two by four, which is zero. Okay, um a 22 that&#x27;s AA minus 3 to 4, minus three. Ritual before and then plus five to before. Um, sorry. Wait a minute. Plus two route to over two plus five route to over four plus five or two before. And so when you simplify this, you&#x27;re going to get to route to right, because this is route, too. This is five halves. Sorry. Um, yeah. This is five hops route, too. And this is three minus three halves. Rude, too. Okay, Yeah. And then this simplifies Thio route to So it&#x27;s rude. Two plus two. Get to root too. Okay. Um Okay. Now I&#x27;m starting to run out of space. So this is we just said that that was too, too. And then this was zero 22 Okay, on. Now, what is a 23? That&#x27;s, um, minus five. Route to over four. Plus retune for plus three. We&#x27;re two for two. Plus two. Route to over four plus eight. Route to over for I&#x27;m just writing this out so that if you have any, if you made any mistakes, you can see where you went wrong. Now, that is going to be equal to three root too. Okay, so three were to again Let me get rid of all of this. So it&#x27;s three root too. Now a 31 That&#x27;s ah, half minus 1/2. That zero minus 1/2 plus 1/2. About zero. A 32 That&#x27;s three halves minus three halves. That zero minus five halves. Plus five House That zero again. A 33 That&#x27;s five halves plus 1/2 minus 1/2 plus eight halves. So that gives you Is that writes. Sorry. Um, so it&#x27;s this one, right? So it&#x27;s minus two. House. Be careful. And so that gives you 16 by this story. That gives you 66 12 by two, which is just six. Okay. All right. So now we figured out what our is are is given by this. Okay, Now question. Well, let&#x27;s check. All right. So first of all, our has to be n by n right? A was m so a was m by n yeah, and m was what? Five and end waas three. Right? So a r is indeed a three by three matrix. We&#x27;re good with that. Are is upper triangular. That&#x27;s good. Are has positive entries on diagonal. Correct. Is our inverts a ble? Yes, it is, right. It&#x27;s a convertible because the determinant of our is the product of the entries on the diagonal for an upper triangular matrix. And that&#x27;s certainly non zero. So you good with that or another way to see it is that clearly the columns are linearly independent. So just a question for you can could our have any zeros on the diagonal? The answer is no. Because if I had any zeros on the diagnosed, then the determinant would be zero. And so I would not be invincible. Okay, finally, um, where are we? So finally, we should check. You know, the question asks you to check your answer and so multiply queue with our and check that you get a back. So let&#x27;s quickly do that. So that&#x27;s one on this side. That&#x27;s four minus one and notice that I&#x27;ve written the roots on the bottom square. It&#x27;s on the bottom because it&#x27;s easier to deal that so four minus one that gives you three seven halves. So seven house minus three house by to have and then plus six halves. That&#x27;s three. So that&#x27;s seven halves minus three hubs, which is two plus three, which is five. It&#x27;s looking good. All right, so here, that&#x27;s minus one. Here s O here is minus one here. That&#x27;s minus four plus ones. That&#x27;s minus three here. That&#x27;s minus seven halves plus three. So, plus three halves minus seven halves plus three halves, which is, too, which is minus two plus three. That gives you one. Uh, this one, that zero Here, here. That&#x27;s going to be two right here. That&#x27;s going to be three here. That&#x27;s going to be one here. That&#x27;s going to be for plus one. That&#x27;s five looking great. Um, and then Sorry. Where were we? Were this? We&#x27;re doing this, bro with this column now. So that&#x27;s seven halves, seven halves plus, um, three halves minus three again. So that gives us, uh, sorry it&#x27;s Yeah. So that&#x27;s five minus three, which is to here, That&#x27;s one again during this road with this column. Now that&#x27;s four plus one. That&#x27;s gives us five. That&#x27;s seven halves plus three halves plus three. Is that right? Yeah. So that&#x27;s five plus three, which gives us eight. And indeed, that&#x27;s the same thing&#x27;s a onda. We are done. Hey,

Chris Trentman · Answer

in this problem, you&#x27;re introduced to reduced l U factory ization. So given a as in practice problem, whereas to find a five by three matrix B and a three by four matrix C such that a is equal to B. C. And then to generalize this idea to the case where A and M by N Matrix a is factory Izabal and has l u factory ization l u and you has only three non zero robes. So first of all, from a practice problem, we have the A on the practice problem Matrix A is the four by four matrix to negative four negative 23 six naked nine negative 58 to negative seven negative 39 four negative to negative 21 or negative one. In fact, this is actually four by five matrix in the last room is negative. 6334 So to solve this problem first you want to find l u factory ization for a and then we want to roll reduce eight echelon form. Okay, so and roll reduced by eliminating entries under the first column. In doing so, I obtained the matrix with first row to negative four negative 23 031 negative. One zero Negative. Three negative. One six 062 Negative. Seven zero Negative. Nine Negative 3. 13. And in this problem, I can eliminate entries under the three in column two using only raw replaced in operations. In doing so, I obtained the Matrix to negative four. Negative. 23 Uh huh. Excuse me. Uh huh. Zero 31 negative one. Part 0005 000 Negative five became 000 10. And finally, I&#x27;ll ro reduce a using only replacement applications on the roads below road three. And so I obtained the Matrix to negative four. Negative. 23 0311 Negative one. I mean, 0005 0000 and 0000 So we found a matrix. You This is a you from the l u factory ization. And then we&#x27;re going to find El using the algorithm from the book. So the first pivot column This was 2624 negative six in dividing Pivot column by the pivot Which is to we get that The first column l is 1312 negative three. And next The second pivot column was three negative 36 negative nine and dividing by the Pivot, which is three. We obtain zero one negative one to give three In her third pivot column is five Negative 5 10 divided by the pivot, which is five. We obtain one negative 12 00 one negative 12 and will use the last two columns from the identity matrix to fill in l 00010 and 00 001 So we have our value factory ization for a now notice that the bottom two rows of you containing the zeros now for using the Row column method defined L. U. The entries in the final two columns of L won&#x27;t be used because these entries will be multiplied by zeros from the bottom two rows of you. So we&#x27;ll pick be to be the first three columns of l and see will be the top three rows of you. So in other words, we have that be will be matrix. 100 310 one negative 11 to to negative one and negative three negative 32 And likewise, see, this is going to be to negative four Negative. 23 031 negative one and 000 five So we obtained the desired matrices B and C, and we have that b c is equal to l times you by the comment from above, which we know is equal to a So it also has the desired property. And of course, we can generalize when a is m times end. So suppose is an M by N matrix. I am going to suppose that a has an L U factories ation that you has only 390 rose. Well, we know that these three none zero Rose They&#x27;re going to be on the bottom. So we&#x27;ll take be to be the first three columns of l and will take c Yeah, to be the first three rose of you, then we have that&#x27;s and

Darsh Patel · Answer

So it&#x27;s but the question we have been asked that you some from off the technology group, remind the authorization off given metrics and really five with the help of Ellicott. Fornication. So what about you Wouldn&#x27;t make tricks? Hey, it was 2. 27 15 23 minus 19 minus 16 minus tote Dean, 30 do 9. 51. So to find a solution for this before going, Ted, I just want to write down the answer. So uncertain. Leader, What about the triangle off Your matrix Will be the 100 five by nine. 30 23 23 by seven After for the 01 minus 86 by one for seven and 001 And in where you will be the 27 00 minus 19 when I was 14. 9 by nine Tzeitel Tohti do minus 17 9 by phone, 2735 by one for seven. No doubt it seems quite valuable, but it is not checked off again. This answer. So the first thing that we&#x27;re gonna do is find a port angle off given metrics and to find that possible, we just know down, given my tricks as a taste that is 27 being granted three minus 19 minus 16 minus 13 do 9. 51. So pretend for step that we take is our do minus nine to our world. And that will take place in Gouda, our group. So the doctor creation, we can&#x27;t go over new metrics and a new metrics will be the 27 0 23 minus 19 minus 49 by nine. And my 30 Turkle would do 79 by nine and 51. So step. In addition, we have to make the serial digital started to go like that. We can go into the artery minus 23 by 27 in two are one and tested separate into the road. Three. So for that, we can get down human weeks. And that will be 27 00 minus 19 minus 14. 9 by nine. It is six. But Grant defense 30 Do minus sound. A nine by nine and 641 by 27. So we&#x27;re just at the end off the question and the harsh thinking. We have to take a look in our answer that we&#x27;re gonna make this into zero and connected became simply. Take our three miners minus 86 by 147 in multiplying through the second group. And they still take second row with Tired Group and they were finally answered at our particular for even met makes. That will be the 27 00 my next 19 minus 49 by nine Vito 30 to minus 39 by nine and do 734 by 147 So that is our final upper triangular. And because of this metal, you can also find Lord trying in desk it. We just have to make zero off elements off upper triangular, and by there we can find the Lord angular. And it will be the 15 by nine 23 by 27 01 minus 86 by one for 7001 that so by multiplying everything was too anybody video of this equation, we can clarify Your answer is correct or not.

Darsh Patel · Answer

into the new in question. We have been asked toe some form or not, you get remind the elephant organization off given Matrix so awful. I just already know down given metrics. After that, I just already know down the answer because it will be quite tough to get that answer, So just don&#x27;t know down to the disease over l off. Our answer is lower triangular and in the multiplication, we&#x27;d still be over you. So a predictor. Here we go. So the question that we go through it just don&#x27;t at least entities and no metrics. Easter for 15 21 81 13 17 37 93 nine minus 17 63 minus 47 2050 39. So I predict a vehicle gained the upper kind off your medic&#x27;s there to make tease, tease, tease the taste and taste component zero make take a veritable we&#x27;ll make 300 So make their testing into your genetics entered in our food. Why not three by 34? Multiplying are one that will take place in our group. So after we, our new metrics will be seems like he&#x27;s And so the hill done in our photo. All the maduro&#x27;s off, the metrics will remain same. So you don&#x27;t help. You need to write this again. So we just simple right? Go to 34. My next story for one by 34 game and minus two for 3.70. The next step zero is below. That was 21 to make a difference. Zero became right. That, uh, are three mine, sir. Driven by 34 through the r one interpretative President Buddha artery So expert while the adults rose remains same except the third. So the I told one when we change into zero predictor 985 by 30 four after that 1743 by 34 77 to 17 whereas all the other growth will be made Same. So we&#x27;re just going to repeat it again. So we just make this like this. So after their next thing that will go in this metric is our foreign born by 34 into the are one and go decorative within our food. So I saved with our for for except our for or will be remained same. And I just know I don&#x27;t need to tastes in the fourth row will turn to 109 by 34 minus 3137 34 and 28 by 17. So potato, by falling this matter became gained over upper triangular for your medics and the final answer Bill, Beat it. You really stoop, as I really know, donate a port. But here against Lodeiro 13 minus 1113400 1934 Won by 34 minus 9341 grew by 00 on the It&#x27;s quite hard to just to make it one page. So the fourth column will be the do for three by 17. 3013 by 37 4 to 55 four 987 You waited by 46 709 Together we were you and l expert that I mentioned about end by their age. Their equations. We can clarify. Take over. Answer. It&#x27;s right or

Problem

[M] The solution to the steady-state heat flow pr…

Problem 30 Easy Difficulty

Find a different factorization of the A in Exercise 29, and thereby design a different ladder network whose transfer matrix is A.

Answer

$\left[\begin{array}{cc}1 & 0 \\ -1 / 6 & 1\end{array}\right]\left[\begin{array}{cc}1 & -6 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -6 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 / 36 & 0\end{array}\right]$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Catherine R.

Kayleah T.

Caleb E.

Maria G.

Absolute Value - Example 1

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$\left[\begin{array}{cc}1 & 0 \\ -1 / 6 & 1\end{array}\right]\left[\begin{array}{cc}1 & -6 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -6 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 / 36 & 0\end{array}\right]$

Linear Algebra and Its Applications

Catherine R.

Kayleah T.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Kayleah T.

Caleb E.

Maria G.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications