in-exercises-1-6-solve-the-equation-a-mathbfxmathbfb-by-using-the-lu-factorization-given-for-a-in-ex

Question

In Exercises $1-6,$ solve the equation $A \mathbf{x}=\mathbf{b}$ by using the LU factorization given for $A .$ In Exercises 1 and $2,$ also solve $A \mathbf{x}=\mathbf{b}$ by ordinary row reduction.
$$
A=\left[\begin{array}{rrr}{3} & {-7} & {-2} \ {-3} & {5} & {1} \ {6} & {-4} & {0}\end{array}ight], \mathbf{b}=\left[\begin{array}{r}{-7} \ {5} \ {2}\end{array}ight]
$$
$$
A=\left[\begin{array}{rrr}{1} & {0} & {0} \ {-1} & {1} & {0} \ {2} & {-5} & {1}\end{array}ight]\left[\begin{array}{rrr}{3} & {-7} & {-2} \ {0} & {-2} & {-1} \ {0} & {0} & {-1}\end{array}ight]
$$

Chris Trentman · Accepted Answer

we were asked to solve the equation A X equals be using the l U factories ation give it from the Matrix A and to solve the equation X plus B for X equals B by ordinary rail reduction as well. So the Matrix forgiven a is the Matrix three negative seven. Negative to negative. 3516 Negative for zero and the column. Vector B is negative. 752 and we see that the l U factories ation given for a is a three by three matrix 100 negative 110 to negative 51 times the three by three matrix three negative seven negative 20 negative to negative. 100 Negative one. So, looking at the factory ization, we see that the unit lower triangular matrix l He&#x27;s going to be 100 negative 110 and to negative 51 And we have that The row echelon matrix you is three negative seven negative too. Zero negative to negative one and 00 negative one. And once again we have the column. Vector B is the three by one matrix. Negative. 752 So first we&#x27;re going to solve the equation. El y equals B for why so To do this, we use rail reduction. So we have that l. B. This is three augmented day tricks, which is 100 negative. Seven negative 1105 and to negative 512 And we see that after performing arithmetic in column four. This can be row reduced to the matrix. 100 Negative seven 010 Negative too. And zero negative. 51 16 and we see that it&#x27;s a matrix can be reduced with arithmetic. Onley in column four to the Matrix 100 Negative seven 010 Negative too. And zero zero 16 And we see that this is the Matrix I and then a column vector. And so it follows that. Why is this column vector which is negative seven negative to six and now will solve the equation You exit was why using back substitution. So we have the augmented matrix you why this is going to be the matrix three negative seven negative two negative seven zero negative to negative one negative too. And 00 negative 16 This could be row reduced to the Matrix. Three. Negative seven Negative two Negative. Seven zero Negative to negative. One negative, too. 001 negative six. This could be row reduced 23 Negative 70 Negative. 19 zero Negative 20 Negative eight and zero zero one Negative six. This matrix can be reproduced to three negative 70 Negative 19 zero 104 and 001 Negative six and this could be reproduced to the Matrix. Three zero zero nine 0104 001 negative six and finally, this could be reduced to the Matrix. 1003 0104 001 negative. Six. You notice that this is the augmented matrix I followed by a column vector, which is X and therefore it follows that the column Vector X is three for negative six. This is one way to find solution to the equation. X equals B in other ways to simply ro reduce a. So consider the augmented matrix A B. This is the Matrix three negative seven Negative two negative seven Negative 3515 six Negative 402 And this could be road reduced to the Matrix three Negative seven Negative. Two negative. Seven zero Negative to negative. One. Negative too. Zero 10. 4. 16. This could be reproduced. 23 negative. Seven Negative. Two negative. Seven zero Negative. Too negative. One negative too. 00 negative. One six. And so notice that this is the same as the matrix. You Why And so following the row reduction, you will obtain I X again as above.

Gennady Notowidigdo · Answer

Okay, so we&#x27;ve got this tree by three matrix A with this l u decomposition l u factory ization. And what we going to do is given this matrix after, Given this three dimensional vector B, you&#x27;re going to solve the system a X equal to be or the matrix equation exit B four D make a vector X gets. We&#x27;re trying to find solutions to assist them up in your equation, so to speak. A case are I will a shoe that you have watched armed by video with regards to question two from the same section of this book. So just as a quick review to solve this using l u factory ization, we going to augment the matrix l with the Matrix, I put the vector B and this gives us a solution that&#x27;s called this vector. Why? So once we obtained this y vector, then we augment the Matrix year with defect. Why? And then solving that&#x27;s make it quick. Gives us the vector X, sir. Four bit more insight into how this works. Just go to my veer with regards to problem too. I&#x27;m just gonna go straight in. No. So I went in the matrix lb My old man dimensions held with defective be Use me one, huh? Three halves minus five one sarah 001 and then zero minus five and seven. So doing the role reduction on this Or gas elimination The 1st 1st States, as it is identified, do the second row minus 1/2 off the first row. I get that. So right two years wrote to minus half one and similarly, I&#x27;ll do row three minus 3/2 times one. So that zero there Mona&#x27;s 51 and seven. Okay, sir, if I leave the 1st 2 rows as it is and do so, I&#x27;m going to the rose three eyes out of you. Five road, too. Plus rose three. And that should give me mon ist 25 plus seven, which is minus 18. So have a well, precisely have a diagonal matrix. But the main thing is, you have a lower triangular matrix here, So upper triangular matrix at the very least, then if we read off the final column, that is going to be our wife Victor. Okay, so that&#x27;s step one omitting l with B to get the y vector. Then we all spent you with the Y vector that we obtained, sir. Thief eyes off here. So I have two mon n&#x27;est tu on four and I have zero there. Zero mon ist too modest. One at minus five 00 minus six and minus 18. Okay, sir, I&#x27;m going to divide the bottom road by six minus six to be more precise. And then if I do, eso haven&#x27;t got to do this. So it will be, um, six times the second runner minus rose three. So I end up having zero mon ist wealth and Monta six plus 60 You got minus 30 plus the 18. So that gives you minus 12 over there on. Then if I do, that&#x27;s right too. If Roe one then becomes Ah, so it will be Wait. Oh, my God. It&#x27;s I have I have the bottom row off the new system here off New matrix here. So if I used that which is really just a stick minus 1/6 of row three or if I do a sixer if I added by 1/6 off for three or 46 to be precise and I put road one there. So what happens our case. Our the 1st 2 entries do not change. Fortunately, and maybe if I do sort of all this divided by two. Sorry for the mess. Okay, So if I duel that divider I realized I mention doing, uh, I&#x27;ll just rub it out. So if I rub this part out, if I divide a ll that by two So I get, um, want em one and modest one day. So all this divided by two. So four plus four six. If that is +40 sir. And then you&#x27;ve got zero plus 2/3 of that, which is mine ist wealth. And then when I divide that by two, I get minus six. Okay, that&#x27;s all right. Sorry for bit of that mess. I&#x27;m thought of doing so many things at once, but I sort of want to keep everything to one page case are, too. I&#x27;m just gonna divide it by monness 12th. And then if I just add wrote one, two, um, minus a tour one on minus 12 times. Right, too. So, Maddie, this row here to this road there, Then I should get 100 minus five. Okay, so again, we have the identity matrix, which is a diagonal matrix, or at least an upper or lower trying to the Matrix. Then if we read off the last column, the Vector X, which is minus 51 and three is a solution to the system. Okay, just one more thing. Before I finish off this video, I want you have the solution. It&#x27;s always important if you can. And I&#x27;ll leave this as sort of a personal exercise verified that this is indeed a solution to that. Okay, so you got minus 10 um, minus twos. 12 plus 2 12 0 And I&#x27;ll leave it again for you to show whether the other two components check out. All right, thank you very much.

Gennady Notowidigdo · Answer

Okay, sir, we have a matrix A which is given as follows. So you got a three by three matrix are we also have? It&#x27;s what we call it l u decomposition. So this is L. A lower triangle of matrix and you an upper triangle, a matrix. So given these three dimensional vector B, we want to solve this matrix equation a X equal to be be solving for X. Um, sir, we will do it in two ways. So the primary way we want to do it, is he going to do it using its l u factory ization. So what that entails is, if we suppose that a x so because A is l times you, um So Because this equation is true, so to suppose that they&#x27;ll bracket sort of between there, Then what we can do is because of the associative ity off the matrix multiplication operation. We can rearrange the brackets to be over there. So what we can do is start by old menting, l and B, and this gives. So let&#x27;s let&#x27;s call this affect The wife&#x27;s obviously a matrix sounds of vectors a vector. So solving this are the L. B opened its system gives us thief Ector. Why? And then from there we all meant Are you with why? And this should give us the solution for thanks. Okay, so it&#x27;s a two step process. Are the other method, which is sort of normal to most of you is gas and elimination, which is which actually takes longer than the l U factory ization method. But you don&#x27;t have to do a two step operation, but rather just a simple one step operation. So you old meant a with B, and you solved that system to get your ex straightaway. Okay, so we&#x27;re going to start by quickly doing the gas elimination method. Then for this question and for future questions, will specifically be looking at solving it using the l U Factory ization case are if we were to augment a with B, this should give us, uh, the no. This baby&#x27;s gonna be three by four matrix four minus 48 three minus five and six, minus 57 and minus eight to minus four and six. So that just double check that moment in this with our A with me. Okay, so if we do the road reduction, sir, If we look at the second road, if we add the first road to the second row Ah, we&#x27;ll get a zero over there. That gives us minus 22 and minus two over there. And if we perform Ah, so this case, it will be the third road minus two times the first row. Maybe I&#x27;ll write out the operations. So the third row is third row minus two times the up in a second de first word. Okay, so that 00 and minus a plus 10 gives us too. Six minus four should give us too. Um, yeah. Then from here, we reduce further. So you got 0011 If we divide everything by modest too, I get zero, uh, one minus one and one. And I still have full three minus five to Okay, I&#x27;ll probably just speed up at this point, So you get 0011 Um, if we add the second and the third row, that&#x27;s what we get on. If we put a zero this or minus five plus five times the third row, we should get a seven there. Okay? And finally, um, 02 So have full there. Three monastery. So if I do first for minus two times the second row, I should end up with that. Okay, But this tells us that X is going to be 21 out the second and third components, respectively. And the first component will be 1/4. So when we divide the top row by four, this will be covered. One. And whatever entry we get here is going to be our X. So this is our solution using gas and elimination. Okay. So as I said, algorithmic Lee, this will take longer a shooting that, you know, if you do it so far enough for now. Former more computational point of view to be more precise. So if you sort of looked at the number of steps that you need to do, this takes longer. Um, the quicker way to do this is using the l u factory ization. So if we start with so l augmented with B, sir, l is one minus one. So and two, 010 001 two, minus four and six. All we need to do is to remove thes two entries. So let me what we use a different color just to be so. All we need to do is to, ah, rare operations. Sir, we have that row one plus road, too. 010 are modest. Um, there we going to do to roll one, so you know, we&#x27;re gonna do right three minus to write one to baking. More simple. You&#x27;re too modest to zero. Mona&#x27;s 01 minus 06 minus four is too, sir. Affect the Why? Well, read off is to minus two and two. So at this point, as long as you have an identity matrix here, or at least wrote a lower or upper triangle of matrix, then this quantity are this column here. The final column gives us the solution color. So the augmentation is just more. It&#x27;s more of a shorthand to describe the system of equations. So once we&#x27;ve got the wide victor, we plump that into the final column off the augmentation with you, sir, he is the matrix year. This one typically takes longer. Especially given that, um, in this case, you have, um, thes three variables to get rid off and possibly some divisions off the lost heroes to clean. I didn&#x27;t know how to clear the common factors. All right, we&#x27;ve already done this before. We&#x27;ve done this, or, uh hell, well, let me probably use a different color to sort of hard like that. This is precisely this thing here. And in doing all that before he figured out how to reduce this. Um, So if I go back to the originals to slight here, this should tell us. Um, so we found that this was a full 0010102001 What so x is equal to quarter two? And what? That pretty much completes this question, sir. I&#x27;m gonna leave it here. Thank you very much.

Chris Trentman · Answer

were given in Matrix A and L U factory ization for this matrix and we were asked to solve the equation. X equals b using this factory ization for this matrix. So the matrix a is three by three matrix to negative one to negative 60 negative to eight. Negative. 15 The column Vector B is 104 and we have that. The l u factory ization is a equals 33 by three matrix 100 negative 3104 negative 11 times the three by three matrix to negative 1 to 0 negative. 34001 So, just by looking at the l u factory ization, we see that the unit lower triangular matrix l is 100 negative 210 three negative 11 in the upper for the row echelon matrix you is to negative four to zero negative 36 00 one. And the column vector B, as we were given, is 606 So we&#x27;ll solve this first using Alya factory ization First I want to solve the equation. L y equals B. So we have that the augmented matrix L B this is the Matrix 1006 negative 2100 and negative or three negative. 116 in this matrix can be reduced to 1006 01 zero 12 zero Negative 11 negative 12 in this matrix can be reproduced to the Matrix. 1006 010 12 00 10 Which this is dog Minton Matrix. I followed by the column Vector y So it follows that. Why is the vector 6 12 zero next? Want to solve the equation? You x equals y with back substitution. So consider the augmented matrix you Why, this is the Matrix to negative 4 to 6 zero Negative. 36 12 0010 This could be reduced to one negative, too. One three warm Before I do that. Actually, let&#x27;s do back substitution Instead, it&#x27;s really only going to do arithmetic in column three. So this is going to be 200 negative for negative three zero and then 001 6 12 0. Let me see this matrix can be reduced using column operations to the Matrix. 200 Negative 10 010 negative four and 0010 and dividing the top row by two. We simply obtain 100 negative. 5010 Negative four 0010 So the identity matrix followed by a column vector, which is X So you obtain the column. Vector X is negative. Five Negative four zero.

Chris Trentman · Answer

We were given a matrix and Blu factory ization for this matrix. Hey, and we were asked to solve the equation. A X equals B using the value factories ation given for the Matrix. A. So the matrix that were given a is a four by four matrix with entries one negative to negative. Four negative three to negative seven Negative seven Negative six Negative. 1264 and the negative four. Negative. 198 and B is the column. Vector 1703 and we have that Blu factories. Ation given for a is before by four matrix 10002100 Negative. 1010 Negative. 43 negative 51 times the four by four matrix one negative to negative. Four negative 30 Negative. 310 00 to 1 0001 So, first of all, from the l U factory ization, even for a we see that the unit lower triangular matrix l is the matrix. 100 2100 Negative. 1010 Negative. 43 Negative 51 And we see that the row echelon matrix you is the Matrix one negative to negative. Four Negative three zero Negative 310 00 to 1 0001 And he Colin Victor B is 101703 So first I want to solve the equation. El y equals B for Why so consider the augmented matrix L B. This is the Matrix with entries. 1000 one 31006 Negative 101010 and negative 34 Negative 213 This could be reduced to the Matrix 10001 01003 00101 and 04 Negative 216 and this matrix can be reduced. The Matrix one 00 zero one 01003 0010 one and 00 Negative to one negative six This could be reduced to the Matrix. 1000 one 01003 00101 and 0001 Negative four And this is the identity matrix augmented matrix containing the identity matrix and the column. Vector y. So why is the Vector 13 one negative. Four. Using this information, you now want to solve the equation You x equals Y for X using back substitution. So, first, consider that the augmented matrix you. Why This is the Matrix, which has entries one negative two negative to negative. 31 zero Negative. 3603 00241 0001 Negative four. This can be reduced to the Matrix. They simply foreign calculations in the 4th and 5th column. One negative to negative. 20 Negative. 11 zero Negative 3603 002 0 17 In 0001 Negative four This matrix can be reduced to the Matrix one. Negative, too. Performing operations Sorry in the third and fifth columns, so we get one negative too negative, too. Zero negative 11. First time action is going to divide wrote three by two and zero Negative. Three 603 001 0 17 Over to and then 000 one. Negative. Four Okay, and now performing calculations In the 3rd and 5th column, we obtain one negative. 2006 zero Negative 300 Negative 48 0010 17 halves. 0001 negative four and we see that dividing the second row by negative three. You obtained the Matrix one negative, 20 Negative six or just positive. Six. Sorry. 0100 16 0010 17 halves and 0001 negative four and performing calculations in columns two and five. We obtain the Matrix. 100 0 38 0100 16 0010 17 halves and 0001 negative. Four We see that we now have and augmented matrix with the identity matrix and left, and they call him Victor on the right. In this column Victor, in fact, his ex. So it follows the X is the victor 38 16. 17 halves, negative four.

Gennady Notowidigdo · Answer

All right. So just put in the Michael bit. Okay? So this question will be very similar to questions true and for which I have already done so Matrix and this case, a four by four matrix with L u decomposition as fullest gotten l matrix and the U Matrix. And given the vector B, we&#x27;re going to solve a X equal to be are to do this. Um, a word. Keep it too long. So I don&#x27;t have to shoulder steps. We&#x27;re going to use sort of a computer package. So I use Maple. You&#x27;re free to use our Matt lab or mathematical or anything off Sort of that ilk, so to speak. So what was? What I&#x27;m gonna do is, as we have done in the previous failures is the two step perch. The first of all, we augment l with B, sir. That would be the the Matrix. L from here said this one there old meant that with the vector B. So you have a four by five matrix. And when you do gas in elimination on that, so what turns out to happen is the following. So this going too quickly. All right, So what I get is so far duty, gas elimination Just trying to figure out against that 1111 and then just old zeroes over there left with 11 monster and one So this may inspect the hill. We called this vector. Why? Okay to this solves theme system l times Why equal to be? Then we sort of said okay with the Matrix, I was with the vector. Y um we know that&#x27;s not a X, but u x the vector u X has to equal why, right? Or when we went When we put this all together, we get the system a X equal to be knowing that a Ezel times you Okay, so are you created clear demarcation that So now we augment you with why and when we sw when we saw that. So if I do reduce racial for so the solution I get Oh, that&#x27;s got the walking pin on the wrong. And so I get 1111 and mold zeros there Free monster one and one. So the effect the X that we want turned out to be free minus 21 and one. Okay, sir. The answer three mon ist too wanted. Wants it transfers of that. Okay. Again, I&#x27;ll leave it for you to check that. That this matrix a Times X equals the matrix B. Okay, thank you very much.

Problem

In Exercises $1-6,$ solve the equation $A \mathbf…

Problem 1 Easy Difficulty

Answer

$y=\left[\begin{array}{c}{-7} \\ {-2} \\ {6}\end{array}\right]$
$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$
$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Maria G.

Absolute Value - Example 1

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$y=\left[\begin{array}{c}{-7} \\ {-2} \\ {6}\end{array}\right]$$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Anna Marie V.

Maria G.

$y=\left[\begin{array}{c}{-7} \\ {-2} \\ {6}\end{array}\right]$
$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$
$\mathbf{x}=\left[\begin{array}{c}{3} \\ {4} \\ {-6}\end{array}\right]$

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

Linear Algebra and Its Applications