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

Question

Determine if the systems in Exercises 15 and 16 are consistent. Do not completely solve the systems.
$$
\begin{aligned} x_{1} &+3 x_{3} \quad=2 \ x_{2} &-3 x_{4}=3 \-2 x_{2}+3 x_{3}+2 x_{4} &=1 \ 3 x_{1} &+7 x_{4}=-5 \end{aligned}
$$

Samantha Lincroft · Accepted 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

Alina Krajewski · 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

this problem. We&#x27;re going to be solving a system of linear equations, and we&#x27;re going to do that through the form of writing and as a matrix and performing their operations. And so the first thing that we need to do is write this problem as Matrix. So we&#x27;re going to do that by creating a matrix and inside this matrix, we&#x27;re going to put all of the coefficients of the original problem, and we&#x27;re going to multiply that by all of the variables from the original problem. And that&#x27;s ex one next to an ex 30. And finally, we&#x27;re going to use that to equal all of the answers from the original problem. And now we&#x27;re going to manipulate this matrix using the row operations in order to manipulate the left side into the identity matrix and manipulate the right side into the final answer is that we&#x27;re looking for. So the first thing that we&#x27;re going to do is take this one up here and use it to kill the two underneath it, and we&#x27;ll do that by replacing row to Withrow to minus two row one, and when we do that, we will have this new matrix and I will be abbreviating the x one x two x three as just X to make things a little easier to follow. And so now that we have 100 in this first column, we&#x27;re going to get rid of the two in the second column and will do that. Just how we did the last one by replacing Row two with row to minus twice pro three. And when we do that, we will have this new equation and you can see that we&#x27;re doing the same operation on this right hand side. So in this case, that meant replacing negative nine with negative nine plus four, which gives us negative five. So moving over here, we&#x27;ll just continue the same process. Except that we&#x27;re in a little bit of a tricky spot because we don&#x27;t have a one anymore. So these ones were really, really helpful in getting rid of the twos. But we don&#x27;t have anyone&#x27;s in the third column, And so, in order to get a one, we&#x27;re going to use our second row operation Skilling to scale the second row by 1/5 and that will give us a one in that column and once again we scale the answer as well. So we have a negative one where we used to have a negative five. And so now we finally have a one in every column. And so the only thing left is we have to use the one in the third column to kill the three and the five. So let&#x27;s start with the three. So we&#x27;ll do that by doing Row One equals, uh, grow one minus three times wrote to. And when we do that, we will now have a zero in that first row and you&#x27;ll notice that we&#x27;ve gotten the first row now to be the first row of the identity matrix. So we&#x27;re gonna do the same thing and use the one to kill five. And we&#x27;ll do that by doing Row three equals ro three minus five times wrote to. So finally, we are almost at the identity matrix with the second the last one, but not quite. And so, in order to get the identity matrix, we just have to swap rose to and rose three. And that will do that with our final row operation into change. So when we do that, we have gray and remember that when we&#x27;ve been writing X here is really excellent. Xbox three. So we&#x27;ll write that out this time. And when we write out our anti matrix, we can see that we&#x27;ve solved by system of equations because if we perform matrix multiplication on the Lexx will have ah x one plus zero x two plus ear extra e equals five and then zero x one plus x two. Plus Iraq&#x27;s three is three and, uh, zero x one plus zero x two plus X three is negative one. And that means that excellent equals five X two equals three, and Exley equals negative one. So finally, we can check our solutions to that by just starting. So we start with seeing that Yes, indeed. We did solve this and will now show that this works in our original set of equations so that for the first equation that we had, that would mean that we want five minus three times negative one or plus three to be eight. Indeed it is. We want two times five might less two times three plus nine times negative one to be seven, which it is, and three plus five times negative one should be negative to which indeed it is. So this all checks out, which means that we have successfully solved I system of equations and you don&#x27;t done so using matrix, uh, operations to manipulate our original matrix into the identity matrix and our right hand side into our solutions. Thank you.

Mirza Abdul Waris Begh · Answer

clay slits sold this set of Lena equations. No, we have three set up the immigration&#x27;s here, If a question 12 and three. And to solve these as we seem we have embrace me how hair through variables X one x to an extreme and we have three minute bulls and we&#x27;re three questions, which means we can solve. For the three unknowns now, a simple way to solve the three set of integrations is to take two of them in piers off, for example, everything. The 1st 2 of the 2nd 2 you saw for we tried to eliminate one off the variables. And so as toe be able to compare the other three questions now, as we can see in the question Number two and in the question Number three, we do not see the bearable X one. So let&#x27;s try and eliminate X one from these two equations so as to be able to compare the first and the question that we can. So what we can do is we can take this equation and we can multiply this equation with a minus negative off. Three. Human divide This in question The negative of three. What we essentially get is we will get any into off three here. And when we add them up, we would be ending up zero or hit right Tum. So we have thanks Focus from ending in two or three x one loves. Then we multiply a negative three and dream with me mind Get off nine. Get you off nine. Thanks. Home X one and then a gain. Sorry. And then me Mum, multiply my and you get all three with five is minus 15. Extreme shouldn&#x27;t be in Quito. Um, sorry. It&#x27;s been driving me the metric bust. Pain has become exhaust tree and finally being get where here is, which is equal the rightness. I don&#x27;t be negative off to a negative three multiply. It will become positive off six for six. So from this equation, we get from second question while deploying. Make it make it to off three. So sorry. We get this question and then we take this aggression as it is on DDE. That&#x27;s right. It tome. It&#x27;s fine. Write it down three times. Thinks one for us seven times. Thanks. Troll with us seven times. Thanks. Three is equal to Thanks. All right, add these three questions off left inside in the right hand side. So these two guys against loud and minus off nine plus off seven becomes minus off toe storm minus or 15 plus of seven becomes minus off aimed. Thanks. Three. And this is equal to six and six. Added up together becomes 12. So this is the question that we get from him. And then we have this equation. First question we think the first ignition and we mounted by the first a question with a negative off to know a positive off so that we can really said we can animal and her removed the 1st 0 so we get you here a minus off. Sorry, I&#x27;m sorry. Two times ex off too. Plus to wanted by before his 88 times makes off. Three mutual is equal to two multiplied by five is minus off 10. What we get over here is when we god first toe up came. So what we see is these two guys cancel out these two Alaskans love what we get is zero is equal to two, which is row Frances. It means that the system of ignitions does not have any solution Because this is never possible. Zero can never be equal to. So ideally, what we mean from here is that thes three creations This one this one and this one. They represent planes in a three dimensional access. For example, if we take Thea excess X one extraordinary and X tree And if you compare them with the normal away off three dimensions X why and z So we have a three dimensional access numbers in space and in three emission space. These three equations this 1st 1 the 2nd 1 in 3rd 1 It essentially represents a plane. And what we see from here that since there is no solution possible here, it means that these two planes do not intersect each other. Since these two planes no intersect each other, we do not have any solution. So the answer is there is no solution. No, Tony, let me Right in. Tom. No. No solution exists. This is young

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.

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

Problem

Determine if the systems in Exercises 15 and 16 a…

Problem 15 Medium Difficulty

Determine if the systems in Exercises 15 and 16 are consistent. Do not completely solve the systems.
$$
\begin{aligned} x_{1} &+3 x_{3} \quad=2 \\ x_{2} &-3 x_{4}=3 \\-2 x_{2}+3 x_{3}+2 x_{4} &=1 \\ 3 x_{1} &+7 x_{4}=-5 \end{aligned}
$$

Answer

Consistent.

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

Linear Algebra and Its Applications

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Alayna H.

Kayleah T.

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Absolute Value - Example 1

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Consistent.

Linear Algebra and Its Applications

Catherine R.

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Michael J.

Absolute Value - Example 1

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

Catherine R.

Alayna H.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications