solve-each-system-in-exercises-1-4-by-using-elementary-row-operations-on-the-equations-or-on-the-a-3

Question

Solve each system in Exercises $1-4$ by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure described in this section.
Find the point $\left(x_{1}, x_{2}\right)$ that lies on the line $x_{1}+5 x_{2}=7$ and on the line $x_{1}-2 x_{2}=-2 .$ See the figure.

Samantha Lincroft · Accepted Answer

So for this problem, we&#x27;re trying to find the intersection of two lines, which means that we&#x27;re trying to find the point where abuse intersect by solving the system of equations using matrices. So the first thing that we&#x27;re gonna do to solve system is right that as a matrix, And so we&#x27;ll have the matrix 157 one minus two, minus two. And next, we&#x27;re going to start solving this. So the first thing that we&#x27;re going to do is go to equals Roh to minus for one, which will give us a zero in the bottom left. And now we need to have a one in the second row in order to cancel out back five. So we&#x27;re going to multiply, wrote to you by 17 it&#x27;ll give us a little fraction there. But that&#x27;s okay, because what we&#x27;re going to use is use that one to cancel out the five. And when we do that, we will have 10 4/7 and 01 9/7. And we can see now that we have the identity matrix in the left hand side, which means that we can show from this that X one is 4/7 and X two is 9/7 and therefore the solution to our intersecting lines is 47 9/7.

Alina Krajewski · Answer

Okay, so this problem are given these two equations and asked to find the point of intersection just basically just asking us to solve the system. We&#x27;re gonna be using elementary grow operations for this. First things first, we&#x27;re going to try to get rid of this three export in the second room. Did you that we see the one x one in the first row and we can figure out that to get rid of it, we can add negative three times were one to row two, and that&#x27;s going to give us a new road two of eight. X two is equal to two, and we&#x27;re still gonna have our original road one right here. Next we can solve for X two by dividing by eight on both sides. And that gives us a new row. Two of X two is equal to one over four, and we&#x27;re still gonna have our original first row right here. So next we&#x27;re going to get rid of this negative five. Excuse yourself for X one, and to do that, we can add five times wrote to t grow one, and that&#x27;s going to give us a new one of X one is equal to nine over four. And our second row, of course, we found to be ex Cheek was one over four. And then our answer in X one comma X two is gonna be 904 comma, one over four.

Samantha Lincroft · Answer

So for this problem, we&#x27;re solving a system of equations, and we&#x27;re going to do this using row operations in Matrix. So we&#x27;re going to start by writing the problem as Matrix. And we&#x27;re going to do that just by copying all of the coefficients into this matrix, which will give us one by seven in the top row and minus two minus seven minus five in the bottom row. So now we can start doing our operations, and the first thing that we&#x27;re going to do is use this one in the top right corner to cancel out the negative to underneath it. And we&#x27;ll do that by doing routine equals ro two plus two Road one. And when we do that, we will now have a zero there and a three and a nine after doing those operations to the other entries of the second row. So now we&#x27;re going to scale this row Ah, by 1/3 in order to get a one in this second column here, and so will he wrote out two times 1/3 which will give us one and three in those positions. So finally, we&#x27;re going to use this bottom one to kill the five on top of it. And when we do that, we&#x27;ll get a zero in the top row. Which means that the left hand side of our matrix is now the identity matrix 1001 And that means that this minus 83 is our answers. So that means that we can use this to show that X one is negative eight and X two is three. So what that means is, basically, if we have these two lines, the point where they intersect would be minus eight and three.

Bobby Barnes · Answer

So they want us to solve this by either using elementary grow operations or an augmented metrics. Um, so what should do both? So you can kind of see how this actually plays out in real time. Like how us applying these to the Matrix is also doing the same thing to this system of equations we have here. So first, let&#x27;s set up the system of equations. So the first thing you need to make sure is that your variables are lined up, so x one x two lined up right on top of each other. That&#x27;s good. And then we have our equal sign and all the constants on the right side, so that&#x27;s good. And now all we&#x27;re going to do is so in this first position here, we&#x27;re going to put a two because that&#x27;s the coefficient of our first variable. Then we put a four because that&#x27;s the coefficient of our second variable. And instead of putting an equal side, we&#x27;re going to put this line here and then on the right. We have negative for, and then we&#x27;ll close it and then we&#x27;ll do the same thing for the second row, so it would be 57 and then 11. Okay, Now, if I was going about trying to do this just by addition or subtraction, the first thing I would want to do is maybe divide this first row by two, because that way we&#x27;ll just have X one. And then once we get X one, we could go ahead and clear out the X one below it. Five x one. Ah, lot easier. So let&#x27;s do that. So I&#x27;m going to multiply this top room by one half. I&#x27;m going toe right that result down here. And so the notation. I don&#x27;t think they write this notation in the book. But the notation I will normally uses one half row one because that&#x27;s the first room. So let&#x27;s go ahead and apply that here. So that&#x27;s going to become X one plus two. X two is equal to negative two, and then the second equation states unchanged five x one plus seven x two is equal to 11. Okay, now let&#x27;s do that same thing over here, so I&#x27;m going to multiply the first row by one half. So the only thing that&#x27;s going to change is the first road, and we multiply each of the components here by one half. So two gives us one or gives us too. And then negative two gives us or negative forgives us. Negative two. And then again, we just right that equation down there exactly the same as we did before. So now if we were to go ahead and add negative five times Row one to Row two, that&#x27;s going to cancel out that negative access. So what, this is going to be? So we&#x27;d multiply this top row by negative five, and then we add these and then that gives us our new road to. So let&#x27;s just go ahead and do that really quickly. So that&#x27;s going to give us zero x one plus s o negative 10 plus seven. So that would be negative. Three x two and then we&#x27;ll have negative 10 plus 11, which is going to be equal toe What? So now this is our new equation, too, because that first one we didn&#x27;t change it. Also, it&#x27;s actually write out what we just did. So x one waas two x 20 tonight, too. So first equation still stays the same. But then this one here become zero x one minus three X two is equal toe one. So the way we would write that in this notation over here is going to be what we can&#x27;t just kind of combine these two get this in the same way. So we write it instead. So we say what we multiply the first row by. So we multiply the first row by negative five row one, and we&#x27;re going to add this to row two. So the notation is kind of weird. Um, but the operation you&#x27;re applying is the first one, and what role you apply to is the second one. Er, least that&#x27;s how I was taught to do. So the first row stays the same. We didn&#x27;t do anything to it would be one too negative too. And now this second room. Remember? We did one times negative five. That gives us 02 times negative. Five negative. 10. Add that to three negative three and then negative, too. Times negative five gives us 10 at that to 11. That gives us 21 book actually over here. Seems like I even did this wrong here because that should have been 10. And when we added we got 21. Yeah. So good thing I was doing this on both sides because I was able to catch my error, right? But yeah. So now you can see how these two matrices kind of correspond to each other. Now, over here, what we would do. Well, I would multiply row Two by negative three. So at this point, let&#x27;s just get rid of X one, because, I mean, it&#x27;s just zero X. We really don&#x27;t need that anymore. So we are that being gone and we&#x27;re going to multiply this road bye. Negative one third. And if we do that so we get x one plus two, x 20 to negative two. First equation doesn&#x27;t change. And then the second one is going to be X two is equal to negative seven now, to show what we did over here. So we&#x27;re going to do line going down, and this is going to be negative one third row, because, remember, we want to somehow cancel out this to make it a one. So then we can go ahead and try toe, add it to the row above it. So let&#x27;s go ahead and do that. So that&#x27;s going to be so one too negative too. And then Or actually, let me do this in, uh, read to make it stand out from the rest of blue. So negative one third row two. So zero times negative. One third is zero negative. Three times negative. One third is one, and then 21 times negative. One third is negative. Seven. All right. And then at this point, if we want to cancel out the X two above this So if we come back over here, what we would do is what we would want to multiply this whole thing by Negative, too. And then add this. So in this case, the X one just drops down, so it would just be X one loss. So zero x two and then negative two times negative seven is 14. Add that to and we get 12. So if we were to rewrite this equation, it&#x27;s going to be so x one plus zero x two is equal to 12. And in that second equation, there should stay unchanged. So it be X two is equal to negative seven. And so over here, if we repeat that same process. The way we write it is going to be so we want to cancel this to appear so we multiply rogue one. I mean wrote to buy negative too. And then we add that to road one. So negative two times zero is zero at that one. Still 11 times negative to negative. Two had that to to zero negative. Seven times negative too. Is 14 at that, too? We get 12 and in the second row stays the same. So 01 negative seven. And then if we come over here, we can erase that. And what this tells us is, well, X one is eager to 12 X two is equal to negative seven. And over here this tells us the same thing. Because if we write this out in that notation, this actually becomes lost zero x two plus zero x one. So this here is the system equivalent of what we have there. And so that just tells us x one is 12. Next two is equal to negative seven. So you really didn&#x27;t need to do both ways, but I just wanted to kind of drilling the fact of what we were actually doing here since doing operations with matrices may be a new thing,

Uma Kumari · Answer

in this problem, we have given the equations, he explains. Four Way Z equals to one hand. Well, we&#x27;re experts Fight by his equals. Two. Let&#x27;s consider equation bond. I&#x27;ll equation, too. We can also ride the systems, as is equals to 34 One bite B is equals 12 on X is equals to It&#x27;s why first we find that a in booze. For that we died augmented my prick, that is is slash I We did 34 or five slash van 001 venues row operations to convert the a slash I in metrics into I slash a by a blind operation. I went in just 21 by three Carlin Regan one over by three or buy slash one by 3001 Now applied operation I don&#x27;t changes to minus four into carbon plus are too. On are changes to minus three. Harvard Big it 14 by three 01 Flash one by 30 Full one on Finally, by applying operation our religion used to minus four by three r two plus Arman read it. 1001 slash minus five four For my history ends the last metrics is in the hole I slash Impose from they&#x27;ll be big Reinforce is equals to minus 54 Well, in my history now we multiply in goes and be talkin x Since exes equals toe in goes in Toby that is X is equals to minus phi for over minus tree in two one who by multiplying these torment reasons we get three and minus two From here we have X Z equals 23 rise equals to my mr Now the lady five Xs equals 23 and bicycles to minus two Satisfies the given equations by substituting the radio fakes and buying both the exclusions The A three into three plus four into minus two is equals to run But a living this we get one which is equals to one So the solution is true now substitute Were you fixing by In the second division we get for into three plus five into my mystery is equals to prove But holding this we get to is equals to which is also true It&#x27;s excess It was two or three on bicycles to minus two Is the solution of the given equation

Christine Anacker · Answer

All right. So for this question, they&#x27;ve given us a system of equations. They filled out the Matrix, and then they walked you through the process. But they&#x27;ve left blanks rest to fill in along the way. So the system we&#x27;re trying to solve is for X minus, Y equals 14 and X plus y equals six so we can see that we&#x27;ve got it set up his X Y answer longer tops. So if you need to remember, goes X, why answer? And then we have equation one an equation to its wave equation. One an equation, too. So this first equation. If it&#x27;s four for X, then I&#x27;m using my first equation listed in the system. So my wife is negative one, and that&#x27;s what I&#x27;d be missing from that blank. When I get down to the next step, it says, are one exchanged with blank. Well, there&#x27;s only one row I can exchange it with, cause it&#x27;s a two row system, so I must have exchanged it with our two, which means this row 116 move to the top and this blank would have to be a one from there. It&#x27;s telling us we&#x27;re going negative four times the first row and adding it to blank. Well, it&#x27;d be pointless to add it to the same rule when you could do accomplish the same thing by multiplying by a number. So we must be adding that to a different row. So you must be going four times the first road negative four times the first run, adding it to the second room. So if I think about this, I would have I&#x27;d have to look back at the previous matrix to get to my answer so I&#x27;d have one times negative for which is negative for Plus four is where that zero is coming from. So then one times negative four is negative for plus a negative one. This answer here would have to be negative. Five. So then moving on to my next matrix step, it tells me that I&#x27;ve gone negative one fist times my second row. So I&#x27;m pretty much taking my whole second ruin, dividing it by a negative five. So negative five. Divided by negative five. This blank would have to be a positive 10 divided by anything. Stay zero and negative. 10. Divided by negative. Five is a positive, too. That is my last. Up here it has a saying that negative are two plus blank again, we wouldn&#x27;t want to add it to the same road. That&#x27;s pointless. So we must be adding that to our first row, which makes sense because we&#x27;d have negative one plus one is zero, which I have this whole last matrix filled up. Now that I have my diagonals as my ones and everything else is zeros, I&#x27;m done with my Matrix row operations and I can write these out. So my first equation, I have a one in the ex spot and zero in the wife. So that beat one acts. Plus, Cyril Wise is for so Ex would have to be four for my second equation. I&#x27;ve got a one in the wide spot in a zero in the ex spot, so I&#x27;d have zero excess plus one y is equal to two. So then why must be equal to two? So the solution for the system of excess for unwise to my solution would be for two

Problem

Solve each system in Exercises $1-4$ by using ele…

Problem 3 Medium Difficulty

Answer

$x_{1}=4 / 7$
$x_{2}=9 / 7$

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

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$x_{1}=4 / 7$$x_{2}=9 / 7$

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Kayleah T.

Caleb E.

Kristen K.

$x_{1}=4 / 7$
$x_{2}=9 / 7$

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

Linear Algebra and Its Applications