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Q3a) Applying Gaussian Cumarion with ad without partial pivoting and digit fleating puint aritlunetic, solve e Follewing -ystem COIpATC' the Tesule Shcnw all t...

Question

Q3a) Applying Gaussian Cumarion with ad without partial pivoting and digit fleating puint aritlunetic, solve e Follewing -ystem COIpATC' the Tesule Shcnw all the' intermecliate conptation- (2+2)1036T1 0.212272 7381 0.2081T1 0. 124772 0.9327b) Let Anxux SVSteML with quiqu solution. Compare thee nmher of divisions, multiplications AIc ackditions used solving the system uing OSI elimination Gauss Jordan methods Give the exact Hber: ad not the orters

Q3a) Applying Gaussian Cumarion with ad without partial pivoting and digit fleating puint aritlunetic, solve e Follewing -ystem COIpATC' the Tesule Shcnw all the' intermecliate conptation- (2+2) 1036T1 0.212272 7381 0.2081T1 0. 124772 0.9327 b) Let Anxux SVSteML with quiqu solution. Compare thee nmher of divisions, multiplications AIc ackditions used solving the system uing OSI elimination Gauss Jordan methods Give the exact Hber: ad not the orters



Answers

Calculate the eigenvalues and the corresponding eigenvectors of the given matrix. All matrices have integer eigenvalues

{{0,-2,1}, {-7,-1,3}, {-11,2,4}}

Here we are going to solve each of the following system of linear equations. Using the coast Jordan elimination method. We have four systems. We start with number one, which is uh huh. Three X plus Y equals one and negative seven eggs minus two Y equals negative one. Okay, and for this system we write the augmented metrics. Mhm. In this case you will be 31 that is the coefficients of the variables in order X. Y. So in the first equation here, provision of eggs, three condition of wire is one. And then we put the right hand side one. And it's the first roll of the atlantic metrics. Then we do the same for second equation. So we write the coefficient of X is negative, seven coefficient of Y is negative two and right hand side is negative one. And if we put this start line here, we can see that here we have the coefficient metrics, that is the metrics of the coefficients of the variable X and Y. And here we have the right hand side victor which is one everyone but the dogmatic metrics. Yes, one metric, that is all the coefficients and ran side have written in the same metrics. Well, now we try to put once on the diagonal of t coefficient matrix and zeros elsewhere, that is. We try to convert or transform these proficient metrics here into the identity metrics. And for that we can only use elementary operations. That is linear combinations of the of the rows of the matrix. And with that we are going to transform these coefficient matrix into an identity matrix in this case two by two. So the first operation, elementary operation we gotta do is it's used to transform this negative seven here into a zero and that will be that the road to is going to be its actual value. Blust 7/3 row one with this operation. First of all, the first equation, the first role remains the same and the second role is going to be transformed is in this operation here. So the first term is negative seven plus 7/3 times three which is seven. And that is negative seven plus 70 And with shows this proficient in order to that to happen, that is 20 P appear here in this position. So now we do the same operation for the rest of the terms. So we get for example here negative two plus 7/3 times this correspondent entering in first row is one, so we get negative two plus 7/3 and that is negative six with 7/3, that is one third. Get rid of this is one third, so you could hear one third and the same thing for the next entry. So we get negative one plus 7/3 times one. That is 7/3 minus minus one is four thirds we have here the next form of the dogmatic metrics after a plot having applied this elementary bridge. Now we want to nullified this one here. And for that we use as pivot. The second role pivot is row, we use to transform in zeros the interests of the metric. So using the second role as a pivot, we're going to notify this one and it's important that we use the second role as pivot because having this zero here, we won't change anymore to value three we get here and that's very important for the process. So here we are going to transform roll on into its actual value minus three times wrote to this operation here is not going to change the three we have over here because that operation will be three by the three times zero is three again. So it's very important that for the process to succeed and we choose degree. Uh huh. He would metric. So have doing this operation we get three minus three times series three as we said before and we have one minus three times one. Third is one minus one is zero and that's what we wanted. And the third one is one minus three times war too. That is four thirds. So it's one minus four is 93. The first four modified the other road remains the same. 11 3rd, four thirds. And we have already not to fight these entries here. This one and this one and now we want to transform these diagonal entries in order to put value one and that will be multiplication of the road by convenient factor in the case of the first role, we multiply by one third so no one will be one third of its actual value. We're good 10 and negative one second role remains the same. Tend to put a one here. We got to multiply the third world by three. So the next major operation is grow three becomes three times rule three. So the first one remains the same and three times throw 3014 And as we can see here we have the identity metrics in the place in the city metrics which corresponds to the coefficients of the variable. So now we can say that this is the solution of the system. That is because if we transform again this into a system that will be one times X plus zero times Y equals negative one. And the other equation will be zero time sex plus one times Y equals four. That is X equals negative one and Y equals four. We don't do that because we know already that is the case. That is if we have the identity metrics, what we have here is the solution of the problem. But we have the solution in order of physicians, that is the first value is X. And the second one is why? So we can say he here that solution is the column vector 91 4. The solution of the first system of equations. We go to the second one mm. Uh huh. And we write the augmented metric system is we write it here X. That's why let's see equals zero. Then we have two, X minus Y plus E. Equal one. And then we have X plus y minus dizzy equal. Yeah. Uh huh. Yeah. And uh now we write the Augmentin metrics and that will be the coefficients of the variables first. For the first equation here coefficient of X. One coefficient of Y. One, Z one ran side cereal. For the second equation we have coefficient of X to commission. Why now they want coefficient coefficient of C. One coefficient right inside one. And the third equation here coefficient of X. One question of Y. One question of dizzy and attitude coefficient. Right hand side too. And this is the automatic metrics. And we want to transform these sub metrics here into the identity mathematics in this case. Three right three. So we have to another five first these two here and for that we do the following elementary operation Road to is road to its actual value minus to Rwanda. So the first thing because we are going to modify here the road to the wrong number two, we write first He rose. That does not change in this step. And they are the first one and the third one. So we write them here as they are at that moment. And now we calculate the second room. So we have zero here as we expected and then negative one minus two times one. That is negative three. Then we have negative, we have one minus two times one is one minus two, is 91 and finally one minus two times serious one. This is the second metric. Now we get general if I this one here, so we're going to modify the third row using the actual value of that role mine. In order to put a syrah here, we only had him subtract this first one of the first row to the third one that is row three minutes for one. So the first two rows are the same in this step. Once, 110 and 0 93 91 1. Now we modify the third row is one minus 101 minus 10 negative two minus 1 83. And too many series too. And as we can see here, we have already known to fight this position. That's good. That's not happening all the time. But in this case it happened so we use it and and now we get to nullify these one here. If that we use as a pivot second role in order not to modify this one, we obtain here. And then the operation we're going to do now is and we find row one, row one will be its actual value, last one, third of row two. And doing that, we get so wrote two and three are the same. We write them right here while Syrian native 3 91 1 and 00 and the 32 And what we get here is one plus one third times zero is one, one plus one third times 93 is one minutes 10 Then we get one minus one blast sorry one third times 91 is one minus one third two thirds and then uh zero plus one third one is one third. Just the next version of the of mathematics here. Yeah. Now we got to help unify this term here. So we do the following. We do row one is Rohan blessed to over nine times grow three so row two and three remained the same. 0 93 91 1 and 00 up nearly 32 here Here we get one minus sorry one plus 2/19 0 is one zero plus two of an attempt 00 and then two thirds. Let's do it here, blast to over nine times negative. One is two thirds minus 2/9 is nine over three is three times to six minus two. Is 4/9 here. Yeah. Sorry I think I think it made a mistake here. So expected zero team. Yeah. And yeah. Okay. My mistake because throw three so we get all right, we get here two thirds. Yeah. Okay. Plus 2/9 times to row three here. So it's 93. Okay, that's it. And that to over three minus six over nine. If you want six of the nine, threes to over three so we get zero as respected here. And Yeah, we don't have to calculate that if we know that we're gonna get zero. I did anyway. And the last one is important is one third plus two of her nine times roll threes to at that position and that is one third plus 4/9. That is nine of the 33 times 23 plus four is seven overnight overnight. That's it. And now that's correct. So we have zero here and now we want to zero over here and for that we do the following, we do wrote to Israel to minus one third row three. So roll one and three are the same. So we get 100 7/9. And the third 00 93 to we calculate the second role and then we have zero. My name is one third times zero is zero 93 minus 1. 13. 00 is re sorry. So is we are seeing the importance of selecting correctly the role we have Syrian three unmodified your negative results because of these two serious here and now we got negative one minus one third times row three is 93. So we get 91 plus 10 as respected. No, the last one will be anyone here. Uh Sorry 11 here plus minus one third times uh Two. So it's one minus two thirds is one sec. Mhm. Okay. So we are almost done because now we have zeros of the diagonal of the coefficient metrics and now we want to put once in the diagonal we have already a one here. So we get a good one here. And that is achieved by multiplying this second row by negative one third. So second role is negative one third times second role. And that we get the first road remains the same. The last row also remains the same. And the second role V 010 and 91 3rd times one third is negative 1/9. Yeah. And the last modification will be this position here is you have to be transformed to one. And for that we multiply the third row by negative one third as well. So row three will be negative one third times throw three. And I give us the following of wanted metrics. 10 0, 7/9 010 negative 1/9 and 001 and negative two thirds. And that's it. We have identity metrics here. So we know that this vector here is the solution of the system of equations. So the solution is 7/9 negative 1/9 and negative two thirds. It means X equals 7/9, Y equals negative 1/9 and Z equals negative desserts. Because that's the order in which we put the coefficients in the atlantic metrics. And we recall that we get to put it also here solution it is X is negative one and why? Well he's war So we got to put it over there. So this is the second example. We go to the third one. Yeah. And the third in the system is three, X minus two. Y equals 93 two, X plus why go three and x minus two? White was 95. And so the augmented matrix in this case is the following metrics coefficient of X in the first equation. Yeah. Is three coefficient of Y 91 coefficient right inside 93. The second equation here coefficient of X two coefficient of Y is one rand side of three handfuls. Third equation we have coefficient of X. Is one coefficient of negative of why it's native to and rancid 95. And as we see here, we don't have a square metrics here. That's because we have three questions to variables. And this is artificial metrics and we want to put one in the diagonal, which is also, we're talking about these two entries and zero as elsewhere. So we tried to do that here using elementary operations, we get the first one will be um we may wrote to which we knew if I hear this entry wrote to his throat too minus two thirds 01 doing that. We get the metrics. First four is the same. Three negative 1 93. The third one is also the same one negative 2 95. And the second one becomes two minus two thirds Rwandan zero was respected and one minus two thirds of 91 is uh 7/3. And the last operation is one is three story minus two thirds 93. Is we do that carefully, is five so we have that and we got to nullify this one here for that. We do modification of row three which is his actual value plus 4/7 times are all too given us 3 91 93. I think I made a mistake here. Sorry. Um yes, look at this. In the first equation we have three coefficient of X and condition of why here, it's important to solve this at this stage because if not everything. So we get negative 23 93 to 1 three and 1 82 5 95. It's a great. So here I put the right coefficient is native to same here. Okay, we are on track. So we got three native to three. The third and the second one is zero, seven thirds and five. And we are modifying the third row by discipline elementary operations. So we get zero and in this case we're going to get also zero here and the last one is negative eight of seven. If we do that carefully we get that and we have a very nullified this entry here particularly in this case. And now we can really if I this to his native to and for that we do the following. We do Rohan is Rohan plus 6/7 row two. We get to this road to in order not to modify this three here and so we get 30 and 9/7 during this operation here. The other two rows are the same 07 of three five 00 98/7. Uh huh. And we have already notified. It turns out of the term of the day going up. So what we need now is to put once in Diagon. Also the first one is row one which we multiply by one third. And that give us 10 and 3/7. The other two remains the same. And now we do. Mhm. Uh brought to Mhm. Which will be 3/7 or two. And that give us um first one is the same 10 3/7, 01 15/7 and 00 98 oh eight or seven. So we have already an identity metrics here. But as we can see we have uh an additional role zeros. So we have we have not sexually that 20 metric but an identity matters with zero rows of zeros underneath. And that means that if we write the equation correspond to this, let's roll here. We had the following equation zero X plus zero Y equals negative 8/7. It doesn't matter which one is of X and Y. We put here, we will have a zero on the left hand side of the equality and on the right hand side we always have negative pay over seven. So if this is impossible, that is, there is no solution to these question. That is for this question here. And when we arrived to something like this in the Calle Shorten method, it means we have an inconsistent system of equations. That is, we have an equation where the coefficients of all variables are zero and the right hand side different from zero. So there is no solution for that particular equation. And of course indeed for the system of equations. So we can say therefore the given system has no solution and that can be approved easily, although around by trained to solve this system by traditional methods. For example, if we if we do first equation man is the third equation, we get two X equal to x equal one then. And that value means using for example the first equation, uh that negative two Y is equal to negative six. That is why is three. And that is inconsistent with the fact that the Mexico want to the second equation, why must be white one also. So we can prove easily doing some algebra with these three equations that we have no solution because there is a contradiction between the values that can be uh get off from these equations. And in general if we have more conditions that these equations and variables in general, we will have no solution. That is because we are imposing more conditions than the number of free variables we have. It is possible even though that there is a solution or even infinite solutions and that's because one of the equations can be the same as the other. That is there can be one or more equivalent to gration. In that case there we cannot have one solution or infinitely many solutions. So it depends on the equations we have. But in general a system with more equations and variables in general tends to have the situation of not having any solutions. So we're going to the false system. And here we have an example of what we have said before that is we have this system eggs minus two Y equals negative two equal to. Sorry, seven X minus 14 Y. He was 14 and three eggs when a C. Y equal Thank you. And here the kinetic metrics. Okay, he's one negative 22 vision of X. We've seen as white brand side for the same for the other equation 7 80 14, 14 and three Native 66 So we got to notify the seven here and for that we do elementary operation wrote to becomes grow to minus 71 and that gave us one negative two to the third one is the same three negative 66 And here we get zero all over zero. So this is an example. We have serious all the way they say grow and now we notify these three here. And for that we do row three he's row three minus three year old one. And that give us one negative 22000 And we get again zeros in all positions of the third world. And the gas or in method can proceed anymore here because we cannot nullified these two here using elementary operations. So cats got shorten methods stuck here and proceed any further. And so we can say that here, we have two equations of the form. In fact the same equation zero, X plus zero, Y. Zero. And it doesn't matter which values affects. And what we put here, the question is true because on the left side we will have zero there, I have we have zero. And that's an example of a consistent systems that is X and Y can have any value. As we look at the other aggressions to see if we can find X and Y. So hear any X and Y. Our solutions. So we we get only the first equation. Mhm. We are left with one equation right, which is X. Here we are looking at this coefficient x minus two, Y equals two. And if we look carefully at the given system, we can see that the second equation, for example, can be divided both sides by seven and we get x minus two Y. It was seven, sorry x minus two, Y equals two. And if we divide the third one by three we get x minus to y equals two. So looking at the given system, we can verify Italy that the three equations are the same in fact, so we have only one equation, which is this. And is that the method the application of the method says to us that is there's only one important equation because the other two are the same. As is there is there are um satisfy as long as the first one is satisfied. So from here we can say that for example X is to y plus two, that is X is two plus two times one plus one. And this equation give us infinitely many solutions by putting any values why we get a valuable effects. So this system has infinitely many solutions of the form X. And sorry because I sold for X, we write that this way we got to ride two Y plus one Y. For any real number. Why you can solve for Y. It's the same thing. So here we should write X. And the question resulting from solving this equation here for and why? But so for X. And so the question is this formula here which is the value of X and why and why? It can be any value. So putting any real number here, we get the corresponding X coordinate that solves this question. So we have infinitely many solutions and we had uh we have been given three questions with two variables. So it's the same situation as the third example. So this shows that we can have infinitely many solutions and that's because there are equivalent equations inside the system. That is, some equations are really the same. In that case is like we had less equations effect in this case we have only one effective equation. The other two are the same as the first one, and we can also have for example, one of these two equations, we have to uh equivalent equation and the other one is different. So we could have even only one solution to the system in that situation, so we can have all the cases going down. But the most likely to happen is what we saw in example three. That is three conditions, there is three questions and two variables and general more equations and variables tend to give us a non consistent system that is a system with no solutions, but we can have certainly infinitely many solutions as we saw in this example here, Number Four.

And this problem we're calculating the Eigen vectors and the Eigen values for a given matrix. And we have our matrix A equal to the following. Uh huh, wow minus 71 minus 13 and minus 11, 2. And for okay so our strategy here is going to be well set up are characteristic equation where we take our matrix A. Subtract the identity matrix, uh find the determined determinant of that and saw that setting it equal to zero. And so that will give us a cubic equation. Uh So we just need to solve that cubic equation to find our three Eigen values. And then for each again value we check with our matrix A two and solve a set of three questions to find the I in vectors. Okay, so let's set up our characteristic equation. So are a minus lambda identity matrix. Going to look like the following. We're gonna have a minus lambda minus 21 minus seven minus one minus lambda three minus 11 two and minus four. Mine is lambda. Okay. And we want to set up determinant of our a minus, I am the I equal to and south or equal to zero and so are determinant. So first we'll do cross a first row first column. So we'll get ah minus lambda chimes to have a minus one minus and uh whoa for mine, islam and minus six, wow. Then next so we go move on to second column, first row. So crossing out the corresponding ah second column and first row. Oh and so that's gonna give us a plus two. We're gonna get a minus seven. Mhm. Four minus land. Ah Plus thirties three and then going to the third column and crossing out first row third column and we'll get plus a minus 14 minus one minus land. Uh It's a minus one minus lambda and a minus 11, horrible. Mhm. Okay And we've got adding all this up. All right, we got a minus lambda uh and a minus four minus three lambda. Yeah, plus lambda squared minus six. Uh Plus two. The minus 28 plus seven seven. Land plus 33 wow. Plus A When I was 14 minus 11 plus 11 lambda, wow, mm. All right. And if we sum all that up, we get a ah minus lambda, cute. Plus three. Land is squared plus 13 and ah minus 15 and we want to set that equal to zero. Ah So we have this cubic equation to solve and we could solve it ah numerically or graphically or whatever. If we need to solve it by hand, it's actually not too bad to solve. Ah We could see that ah If we're given that the Eigen values are all integers, so we're gonna have three roots and it's gonna be one, it has to be a one, a three and a five to get our landed to the zero term of 15. So we have to have something like ah one, A three and a five. Uh So where are possible routes are plus minus one plus minus three and plus minus five. And so we can play around with a little bit and we'll find that our ah lambda solutions for lambda are one minus three and five. So those are our Eigen values. Okay, so now we'll need to uh for each Eigen value will need to go through and find the corresponding Eigen vectors. And so that is according to are a so each a ah times are vector equals two uh lambda vector. Right? So ah well so we'll go through here for our first one, lambda one equals to one. And so we want to solve that. So taking are a uh so a times V. One equals two lambda one times V. One. So that's gonna be our our Eigen vector V. One. Uh And the corresponding with its corresponding Eigen value lambda one and so are a. Right? So we've got our a vector ah zero minus 21 minus seven minus 13 minus 11 2. For. And it's criminal applying that times are vector. And so I'll give those an X. And a. Y. And Z. And that equals to. Ah So our value here is just a. One, so X. Y. Z. Okay, so that's going to set us up with a system of three equations to solve. And so we get a minus two. Uh Of course the equals X. We get a minus seven, X minus Y. Plus three Z equals two Y. And a minus 11 X plus two. Y plus four, Z equals two Z. Okay. Ah We'll start here uh with the we've got a substitution for X from the first equation X minus two Y. Posey. So plug that into the second equation and we'll get a minus seven times a minus two. Ah Plus Z minus Y plus three, Z equals two. Why? And I'm gonna skip a few intermediate rearranging of equations here and go to that, giving us Z equals 23 Y. And then we can we can just plug that back into our first equation here and now we'll go X equals two minus two. I plus Z. When Z equals +23 Y. It's two Y plus three wise. So we get X equal to Y. Okay, so now we've got our solution so we've got uh X equals X. Yeah X equals X. Why equals X and z equals 23? Y. Okay, so that gives us our the one. So Eigen vector for our first agon value of one? Going to be 11 and three? Okay, so that will be our first solution. So we got lambda. Uh oh uh so we've got our lambda one equals to one and are corresponding Eigen vector of 11 and three. All right, next step, let's check. I mean vector for I can value of lambda too equal to minus three. Okay, so all right on our a at r zero minus 21 minus seven minus one. Three minus 11 minus to four, wow. And are again times are Eigen vector equals or I can value minus three times are again vector? Yeah. Okay. And so we'll set up those three equations, we get minus two plus C equals minus three X minus seven X minus y plus. Crazy. It's minus three Y minus 11 X. Uh Plus sweeps. Um This should be a positive too. Are they negative too? So we've got minus 11 X plus two. Uh uh Plus for Z equals two minus three. Z. Okay for this one I'll just take that first equation. Uh and use our Z as a substitution. Uh So take C equals to two y minus three X. And we'll plug that into our second equation here, which will give us a minus seven X minus Y plus three times to I minus three X. Uh huh. And skipping a few intermediate steps that leaves us with Y equals to two X. And then we can just put that back into our Z expression. So we've got Z equals to two times two X minus three X. Uh So Z equals two X. So we've got for this one, ah We had our wow, X. X equals X, Y equals two X. And Z equals two X. So are RV two. I can vector going to be a one 21 All right. So that so for our again our second Eigen value of minus three, the corresponding Eigen vector V two will be 1 to 1. Oh all right. Now we just got our third Eigen value to do. Okay, so lambda three equals 25 And our matrix a zero minus 21 minus seven minus 13 minus 11 to 4. And I in vector X Y. Z equals I can value five times Eigen vector X, Y. The animal. Mhm. And that gives us the equations a minus two plus Z equals five, X a minus seven, X minus Y plus three, Z equals five Y and a minus 11 X plus two. Y plus for z equals 25 Z. And I'll take that first equation and go Z equals five X plus two. Why? And plug that into our second equation, which will give us a minus seven, X minus Y plus 35 X plus two, Y equals 25 Y. And that gives us an X equal to zero. And then we can plug that into our first equation here, we get Z equals to two Y. And so we've got A. X equals zero, Why equals Y. And the equals to two Y. And so our V. Three, I am our third Eigen vector is going to be a zero 12 and that's going with our the corresponding Eigen vector for our, I can value lambda three equals 25 mm. So we've got our three Eigen uh three Eigen values, so we've got our lambda one equals to one corresponding Eigen vector of 11 and three. We've got our second agon value lambda two equals minus three, corresponding Eigen vector 1 to 1 and our third agon value. Lambda three equals 25 and corresponding Eigen vector 012


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4 answers
The graph of g consists of two Ooctags-butosave#C straight lines and semicircle_ Use it to evaluate each integral,Y = gx)g(x) dx(b)K 96) dx9u ) &
The graph of g consists of two Ooctags-butosave#C straight lines and semicircle_ Use it to evaluate each integral, Y = gx) g(x) dx (b) K 96) dx 9u ) &...
5 answers
If you throw a ball vertically up with velocity 19.6 m/s; what is the maximum height it will reach (assume it was thrown from height 0 m)
If you throw a ball vertically up with velocity 19.6 m/s; what is the maximum height it will reach (assume it was thrown from height 0 m)...
5 answers
Y= f(x)=sin.xGraph vi10}y=fr)-4r-12-2 -1 24 5 6 7 8 9 1055 (18)-10 Graph v.13 . Use Graphs v and vi on this page to find the value of each definite integral.fsin x dxJustify the answer using complete sentences
y= f(x)=sin.x Graph vi 10} y=fr)-4r-12 -2 - 1 2 4 5 6 7 8 9 10 55 (18) -10 Graph v. 13 . Use Graphs v and vi on this page to find the value of each definite integral. f sin x dx Justify the answer using complete sentences...
5 answers
Arrange these elements according to electron affinity:Most energy released by gaining an electronMost energy absorbed by gaining an electronAnswer BankClBHe
Arrange these elements according to electron affinity: Most energy released by gaining an electron Most energy absorbed by gaining an electron Answer Bank Cl B He...

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