determine-if-the-systems-in-exercises-15-and-16-are-consistent-do-not-completely-solve-the-systems-2

Question

Determine if the systems in Exercises 15 and 16 are consistent. Do not completely solve the systems.
$$
\begin{aligned} x_{1} &-2 x_{4}=-3 \ 2 x_{2}+2 x_{3} &=0 \ x_{3}+3 x_{4} &=1 \-2 x_{1}+3 x_{2}+2 x_{3}+x_{4} &=5 \end{aligned}
$$

Alina Krajewski · Accepted Answer

and this problem were given this system and were asked to determine if it&#x27;s consistent, which by definition, means whether have city shits up as a first step ever in the system as this matrix right here now to the terminal that has a solution set, we&#x27;re not gonna have to fully solve the Matrix. We&#x27;re just gonna have to prove that it could be Britain introduced to root for. What that means is that these numbers over here going on diagonal are gonna correspond to our solution set and all the other numbers, with the exception of last column, of course are gonna be going to zero. So in this matrix, we can see that these three variables already equal to zero and where we&#x27;re gonna try to do is ah, half this negative to this three. And this too, turned two zeros as well. So, for the negative, too, we can add two times were one 24 to get a new road for of zero three to negative three negative one. Now, for the three, we can add negative three over two times wrote to to row for to get a new roof or of 00 negative. One negative three. Negative one. Lastly, for the negative one, we can look at our row three right here and simply add a row three chiro for and we could see that that&#x27;s gonna give us a new role for of simply all zeros. So now we have a new matrix of our original 1st 3 rows and then a new row, four of all zeros. And this right here is enough to prove that the system can be written in reduced reform. Because we have these zeroes over here in a triangular form and these numbers over here on the diagonal that are going to give us our solution set. Um, what having the zeroes in a triangle proves is that these numbers over here can also be turned into zero. And now this is gonna be our solution for X one. This is gonna be your social Rex, too x three and X four. So, again, the system is consistent

Samantha Lincroft · Answer

So for this question were asked to show that this system is or is not consistent and were given a matrix that&#x27;s already partially reduced. So we&#x27;re just going to reduce this a little bit further to get into triangular form so that we can see whether an optic system is consistent. So we&#x27;re gonna do that first by using these to ones in the upper right hand corner, and we&#x27;re gonna use that to get rid of the three and the negative to over on the side here. And when we do that, we&#x27;ll see that we now have three rows that are all good and triangular. Unfortunately, when we do this to the third to the fourth row, we have this negative nine here, which means that we&#x27;re going to need to reduce one more time, and we&#x27;ll use that three. That&#x27;s right above the nine to get rid of it. And once we do that, we now have a system that is in triangular form and since we can see that we have numbers and all the left hand side and triangular form, and we have all non zero constants on the right, then we can tell that this system is consistent

Viswesh Uppalapati · Answer

in this video, I&#x27;m gonna be going over how to do problem number six. Uh, session They&#x27;re going three, which is based on determines in the Cramer&#x27;s rule. And I did everything beforehand because the determinants of three by three take are very time consuming and very long, So I&#x27;ll just go through how I did this problem. Um, I didn&#x27;t go over it once, but the video, then upward properly. So, uh, I&#x27;ll just go over roughly. Um, it&#x27;s pretty simple is just like solving a problem, too. And for except this time instead of two by two, it&#x27;s a three by three. So that&#x27;s more mathematically, a little bit difficult. So to start off were given this set of system of equations and that would make this problem this have a problem a lot easier if you were toe. Visualize it in a X equals B matrix form. Like this, uh, that I underline just now, um, so here the problem is asking us to solve, uh, system using the Cramer&#x27;s rule, which is basically says that if we were a substitute 4 to 2, which is, uh, here is be label be into each column of a and take it. The determinant. The rest of the matrix is a, um, and divided by the initial initial determinant when, without substituting be into it. Um, then we would get every single solution to the system equations. So here to start off, I first took the initial determinant of A without swapping anything. I&#x27;m to take the determinant of the three right to be, um, you&#x27;re basically for the for the first moment you do one times the determinant of this square two by two, right here and then minus three times the determinant of the outer two by two closer, one times the determinant of the 1st 2 columns. Right here. Well, sorry if it&#x27;s a little bit messy, cause I keep going over what I&#x27;ve drawn over previously. So here we got, um, negative to minus negative 18 minus one, which is which ends up being a negative too close. 18 minus one, which is 15 for the initial determinant. And then we, um and then for the first time of the creamers are always swap the first row with 4 to 2 and take it to determine which comes out to be six, and we repeat the same for the other two rows where when we swapped the second row with 4 to 2, Um, which is which is Ah Bee column. We got the determinant to be 12. And when we do so with the third column, we get the determined to be 18 um, and take out the solution to the system of equations. We just divide these determinant values by the initial determinant to get each value of the X column vector. That would be this. That was service a solution to the system equations he wrecks. One is six, divided by 15 extra. Strolled the water by 15 Extras 18 Divided by 15 um, and which gives us these Nissen, while is, respectively.

Chris Trentman · Answer

were given a system and were asked to use Cramer&#x27;s rule to compute the solution of the system system is three x one minus two x two equals three and negative. Four x one plus six x two equals negative. Five So this system is equivalent to the equation. A X equals B where a is the matrix three. Negative to negative 46 and be is Thekla Vector three Negative five. So we have that a one of B is replacing the first column of A with B, so we get three negative to negative 56 and we have a two of B Replace the second column of A With B. We get 33 negative four, negative five and now continue determinants. We have the determinant of A This is 18 minus gate, which is 10 determinant of a one of B is 18 minus 10 which is eight, and the determinant of a two of the, he says. Negative 15 plus 12 which is negative. Three. So, using Cramer&#x27;s rule, we have that X one. It&#x27;s the determinant of a one of B over the determinant of a, which is 8/10 or 4/5 and X two again, by Cramer&#x27;s rule, is the determinant of a two of B over the determinant of a, which is negative 3/10. So it follows that the answer X is the vector 4/5 negative, 3/10.

Linh Vu · Answer

So the first step will be and we take the one we end with a chew becomes the fi Thanks. Just resent you called Joe one. And now the second case, we want todo on youse want you can, uh, Andi under equation involving ex incense. So I was used to doing three now, but we turned, uh ah. Three times number two plus a minus The far times I&#x27;m, uh, three so And we do. So which again? Uh, three times number two, We&#x27;re going too far X plus The miners missed entering route times. The so the jewelry times three. And, uh gee, with him&#x27;s far. And then we do this Objection here. So we do so and then when she ended six My house. Uh, eight. Thanks for minutes to thanks. And why would be out on ah three? Yes, J said minus 20. Ah, event. So B minus 17 cent. We&#x27;re ankle. And you so Ah, 16 times three. We minus the 20 times we minus the Judy TEM Song will be anything you&#x27;re gonna minister. Did you so the funnest juice it ju them here with way would do one more here. So isn&#x27;t Lewit hams. Uh, isn&#x27;t with him. Jew soon becomes five, so additional becomes ah, 10. Thanks. Plus six. Set it. Could you chew? And hey, wake on minus 10. Thanks. A minus seven, Dean times five something times five. Well defined. Is that what you go? Jew said Jew teams fight when 1 60 then we put a line and we do the ending up. So I am to do so. We got zero on then AA minus 79 set. We go to, uh, way end up. So we&#x27;re gonna minus 1 58 says that we go two months to 1 58 Do you have my 79? And when we have dissent, we go being able to find it. Thanks now, so that next Well, doesn&#x27;t immobilized. So doesn&#x27;t implies done that. Thanks. You go to Ah, to minus six set So it could do Minus six times 1 58 Remember? 79. So we should get. Doesn&#x27;t he? Could chew 17 9 times to minors. Thanks. Terms 1 50 date. So we got minus 790 Remember? 79 on Dan the accident that the why were equal to can too. On then we have this undivided tennis. Well, so we need to develop a 10 uh, about here on then in November 10. So everything here. So in development 10 and then when we revive, it isn&#x27;t big 10. God. Hey, Good. Ah, go to minus one. So doesn&#x27;t Thanks on now from this. Is that your father? Why we can find it Using the first question. First question saying that. Ah, under wine, we got your minus fit then ministry ex ministers that divine by madness. Far. So we should get Manus for Dale. Ah. Blessed three minors. Jew times 1 50 18 I&#x27;m in 79. Remember, man, A scar. So we should get, uh, a good Jew AA minus. What trail? Two terms 1 50 Okay, remember 17. 90. My, my mind is far. So which you can&#x27;t trout time. 79 I June terms 1 58 You&#x27;re right. So we got under time away. Could you? Minus 16 November minutes? Phone. So you got what you could do far. So that&#x27;s three. A solution from exquisite

Chris Trentman · Answer

were given a system and were asked to use Cramer&#x27;s rule to compute. The solution of this system system is negative or sorry x one plus x two equals x three Negative three x one plus two x three equals zero and x two minus two x three equals two. This system is equivalent to the equation. A X equals B where a is the three by three matrix 110 negative 302 01 Negative, too, and B is the column Vector 302 So to use Cramer&#x27;s rule first, let&#x27;s find a one of B. This is the Matrix A with the column one replaced by B So we have 30 to here and then 10021 negative, too. A two of B. This is the Matrix A with column to replaced by B So one negative. 30 302 0 to 2. And finally a three of B. This is Matrix A with column three replaced by Vector B. So this is 11 negative 3001 and then 302 And now computing determinants, we have the determinant of a Can you sound in this case, I&#x27;m going to find it by expanding across the top row. So we have the determinant of 0 to 1 negative too. Minus the determinant of negative 3 to 0. Negative too. This is equal to well, we have negative. Too minus. And then we have six minus zero. So minus six, which is negative. Eight. The determinant of a one of B. We&#x27;ll find this by expanding across the second row. So really, we have Let&#x27;s see positive. Negative, positive, negative. So negative. Two times the determinant of 3121 This is negative. Two times three minus two, which is one so simply negative too. And tournament of a two of B. Find this. I&#x27;ll expand across the top row again. So we have determined of 0222 minus three times the determinant of negative 3 to 02 So we get negative four minus three times. This is negative. Six negative four plus 18 which is simply 14 and the determined of a three of B We confined bikes being across the second row. So we get the opposite of negative three, which is simply aerated health times. The termine int of the Matrix 1312 So we get three times two minus three, which is three times negative one or negative three. So now we have all that. We need to calculate the solution using Cramer&#x27;s rule. So it follows that Cramer&#x27;s Rule X one is the determinant uh, a one of B over the determinants of a call that we calculated the determinant of a one of b was negative. Two. So this is negative. Two over negative eight or 1/4 x two is the determinant of a to B over determinant of a call we calculated the determinant of a two of B was 14. This is 14 over negative eight, just the same as negative 7/4. Finally, x three is the determinant of a three of the over the determinant today, and we calculated that the determinant of a three b is negative. Three. So this is negative three over negative eight, which is simply 3/8. And so it follows that the answer by Cramer&#x27;s rule to this system is the vector. 1/4 negative 7/4 3/8

Problem

Do the three lines $x_{1}-4 x_{2}=1,2 x_{1}-x_{2}…

Problem 16 Hard Difficulty

Determine if the systems in Exercises 15 and 16 are consistent. Do not completely solve the systems.
$$
\begin{aligned} x_{1} &-2 x_{4}=-3 \\ 2 x_{2}+2 x_{3} &=0 \\ x_{3}+3 x_{4} &=1 \\-2 x_{1}+3 x_{2}+2 x_{3}+x_{4} &=5 \end{aligned}
$$

Answer

Consistent.

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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Consistent.

Linear Algebra and Its Applications

Grace H.

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Caleb E.

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Absolute Value - Example 1

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

Grace H.

Catherine R.

Caleb E.

Maria G.

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

Linear Algebra and Its Applications