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Problem &Write Your D System of linnr oquations with 2 eqquations ndl variables unknowns. Each of the (quations IUSt contain Ith Variabilcs Iu other #ors whike ...

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Problem &Write Your D System of linnr oquations with 2 eqquations ndl variables unknowns. Each of the (quations IUSt contain Ith Variabilcs Iu other #ors whike somthing likeis Actually sstem of lincnt CquAtiOns with ukowus it will get You credit . An icxeptable exutuple iFuYou just cpy this one Vo will get crexlit Similarly; if rou copy system from the Handouts or Homework ron will also get credit ,Write tho system fTom DT ["S A nuEmente MatrixUw tlie ( 'ues - Jordlan lituiut ion p

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Answers

(a) Use Cramer's rule to show that the solution to the system (8.7.22) can be written in the form $u_{k}^{\prime}=\frac{F(x) W_{k}(x)}{W\left[y_{1}, y_{2}, \ldots, y_{n}\right](x)}, \quad k=1,2, \ldots, n$ where $W_{k}(x)$ denotes the determinant that is obtained when the $k$ th column of $W\left[y_{1}, y_{2}, \ldots, y_{n}\right](x)$ is replaced by $\left[\begin{array}{c}0 \\ 0 \\ \vdots \\ 1\end{array}\right]$ (b) Use your result from (a) to show that a particular solution to Equation ( 8.7 .23 ) is $$ y_{p}(x)=\int_{x_{0}}^{x} K(x, t) F(t) d t $$ where $K(x, t)$ is given by $K(x, t)=$ $\frac{y_{1}(x) W_{1}(t)+y_{2}(x) W_{2}(t)+\cdots+y_{n}(x) W_{n}(t)}{W\left[y_{1}, y_{2}, \ldots, y_{n}\right](t)}$ and $x_{0}$ is an arbitrary point in the interval of interest.

For part a of this exercise. We have the moment matrix one minus 23 on two minus 19 We can start by subtracting twice. The first wrote to the second row on DSO. We obtained zero three on three. Now we can divide the second row by three and we have 011 From here, we can see that the system is consistent. The second road means that extra vehicles one on from the first row. We have extreme Minister Sorry x one minus two, x two Because three which means that X one equals five. So there's only one solution for the system which is X one equals five X two equals one. Now for system be for party. We have documented matrix two minutes 35 hard minus 46 8 on if we add twice. The first wrote the second road We have seen zero zero on 18 on from here we can see that the second row is a bathroom. So this system is inconsistent. Now notice that in part C, we have on imagining a system so we can reduce the coefficient matrix. So we have the metrics. One, one 233 minus two If we tracked twice the Frist wrote the second road we have 01 on. Then if we should craft three times first wrote the second group, we get to remain its life. No, Uh, notice that if we add five times a second round the last year we have 00 on. This means that well, the system is consistent on from the second row, we have X triple zero. I'm from the first row. We have X one plus x two equals cereal, which means that X two is also uh since 6 to 0. X one is also zero. So the solution for the system is unique. X one equals six to it consumers. Now we're part the We have the augmented matrix three to minus 14 One man stood to one 11 to 1 14 Onda. We can interchange the 1st and 2nd rooms. We have one minutes 2 to 1 three to minus 4 11 to 1 14. Now we can subtract three times. The first wrote to the second row to obtain zero eight minus seven on one on. We can have tracked 11 times the first road for the last road to obtain Cirio 24 minus 21 on day three. Now notice that the last row is a multiple of the second room. In fact, this three times a second. Rosa, if we subtract within the second road to the last row, we obtained a road form of C Rose on. From here, we can be by the second row by 1/8. So we have one minister to one 1.7818 in a row of syrups. Um, all the form of this matrix means that the system is consistent on. In fact, it has infinite solutions. So the third, uh, column, which for response to the variable X three is gonna be a parameter, which we are going to the note by S on the second room means that X two equals 18 plus 7/8 off S on the first room means that X one minus Well, if we put everything on the other side is equal to one minus two s plus to extend. Now, if we write instead of next to the expression we found before we have two times 18 class 7/8 affairs on When we rearrange this, simplify this expression We obtained five words minus on fourth of s So the solutions for the system are x one equals 5/4 5 4th minus 1/4 off S X two equals 18 plus 7/8 of S on extreme calls s where the parameter s can be. Any riel number, foot part E. We have the May drinks. 2311 1113 34 to 4 on. We can interchange the first on the second row. So we have this matrix here and now to the second row. We can subtract ties. The first row to obtain 01 minutes one on minus five to the last wrote we subtract three times the restaurant. No, we see that the second last rooms are the same. So the last room we can subtract it. Second room. We have a room of sea routes. Um, the shape of this matrix means that the system is consistent on there are infinite solutions again. The last variable is a free variable From here. From this second room, we have exterior close five plus 63 on from the first row. We have X one equals three minus X three minus X two. And when we use the expression we found for X two, we have the following minus two minus two x three. So the solutions for the system are X one equals two minus two s X two equals five plus s on X three equals less where it's gonna be Any riel number. Okay for the system. In part f, we have the metrics one minus 12 four 23 minutes. 11 7347 If we subtract twice the first wrote to the second road we have obtained 05 I miss five on board in the seventh. And if we have tracked seven times the friends wrote the last room we have seen serial 10 minus stand Andi minus 21. Now we can subtract twice a second road to the death row on DSO In the last row, we're going to have 000 aunt minus seven. So now notice that this is a bad robe. So the system is inconsistent In party, we have the following augmented matrix 11110 23 minus one minus 12 32115 36 minus one minus one four. Um, we are going to use the first road to, uh, we're going to serve twice the first road to the second room. We have seen this. We should crack three times difference. Wrote the third row to obtain the full of on. We also subtract three times for throughout the last road. So we get this matrix. Now, if we add the second row to the third row, Yeah, we're gonna have sea route. She room minus five, minus five on seven. On. If we subtract three times a second, read the last room we have seen 005 five on my honesty. Now we can add the third growth to the forth road on. So we're going to have 00005 on again. This last row is the back rows of the system isn't consistent for a part age. We have the matrix one and 23211 minus 584 on if we should track toys. The first room to the second row, we obtained zero five on minus five and if we had five times first wrote to the second road we obtained zero minus 2. 19. Now we can divide the second row by five. Next, we're going to add twice a second road to the third growth. So we have 00 at 17. Notice This last room is also a bathroom. So the system is also inconsistent. Now the system in part, I has augmented matrix minus one to minus 12 minus 2 to 14 3225 minus 385 17 If we multiply the breast road times minus one, we obtain one minus 212 Sorry, minus two. And then all the other rooms has stayed the same on Now we can add twice a free through the second room to obtain zero minus two 30 We can subtract with them. First wrote to the third row. So we have 08 minus one. Aunt 11. We can actually times the first wrote the last road to obtain zero to a on 11 again. Now, if we divide the last group by two a new interchange it with the second room. We're gonna have 014 The 11 halfs 08 minus one and 11. On what Waas before the second road serial minus 230 Now, using the second row, we can subtract eight times a second row to the third growth. So we get 00 my necessary three. Andi a minus 33 On board, we can add twice the second road to the fourth row tow obtain 00 11 and 11. Now notice that the third row on the fourth row are equivalent, so we can just keep one of them on divide to obtain the road off Serial serial 11 at a row of cigarettes at the end. Now, from the third row, we have that X three equals one. From the second room we have X two equals 11 halfs minus four times 63 In this case, X two will be three halfs onda from the first that we have X one equals minus two minus 63 plus two x two Now with the bodies were found that up this ISS minister minus one plus three, which is Syria. So the system is consistent on there is on Lee one solution which ISS X one equals zero x two equals three halfs. A next three equals one. Okay. Now for the system in part Gay, We have to make creeks. 12311 minus one, minus one for minus 16 minus two minus 47. Minus 11 Andi, if we add the first row to the second growth, we have been 01 107 And if we are twice the first road to the third growth, we get zero. She wrote 113 Now, if we subtract the third road to the second road, we've been 01 c zero SMI nous one on Ford on the last row stays the same. Well, from here, we can see that the system is consistent on it has infinite solutions. So, in fact, variable X four is gonna be a parameter s from the last road. We have X three equals three minus 64 From the second room, we have X two equals four, 64 I'm from the first row. We have X one equals one minus six words plus two extreme minus two x two. Now, if we use expressions we found before you have the following. Okay, Andi, once we simplify, we obtain to minus six x four. So the solutions for this system are X one equals two minus six s X two equals four plus s X three equals three minus s on X four equals s. Where s gonna be any riel number? Okay, in part. Okay, we have the following matrix. 1331 three to I understood. Want to hate one minus 5015 on in with the track twice a free throw to the second row. We obtained spro minus eight, minus 102 And if we crack the first road to the last road, we obtained serial minus eight minus 102 Now, notice that, uh, we can subtract the second road to the last road to obtain a row of see roofs on. Then we can divide the second road by 18 So we have these matrix. Uh, this means that the system is consistent on. There are infinite solutions. Now, in the second room, you can see that X two equals minus 1/4 minus 1/8 off x three on board. From the first row, we have X one. It was three minus 64 minus 63 minus three times X two. Now, with the expression we have for X two, we can simplify days. So we have 15 forwards, minus X four Mm minus five. Eights affects three now. So this system has infinite solutions. There are two parameters which are gonna be S and T. So x one is going to be 15 4th minus s minus 58 off T X two is going to be minus 1/4 minus 18 of tea. X three is gonna be t on X four is the parameter ISS. So here t on s can be any riel numbers In part l we have the matrix one minus 311 to one minus 12 14 Understood. One five minus 8 to 5 on Invasive Trapped twice The first road to the second row we get She will seven minus 30 If we subtract The first wrote of third robe We have 07 minus 30 on If we subtract five the first through five times the first wrote to the last row we obtained Syria of seven minus three Syria. Now we can see here that the third, 2nd, 3rd and 4th groups are equivalent. So we have two rows of zeros on the second row. We can divide by 1/7 on these matrix means that system is consistent. Uh, for the second road, we have X two equals 3/7 time 63 from the first world we have X one equals one minus 63 plus three weeks to on. We can simplify days to obtain one plus 2/7 off X three. So the solutions for the system our X one equals one place to seventh off s on X two equals 3/7 off s and extreme equals s Where s is a parameter. It can be any riel number.


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