solve-the-systems-in-exercises-11-14-beginaligned-x_24-x_3-5-x_13-x_25-x_3-2-3-x_17-x_27-x_3-6-endal

Question

Solve the systems in Exercises $11-14$
$$
\begin{aligned} x_{2}+4 x_{3} &=-5 \ x_{1}+3 x_{2}+5 x_{3} &=-2 \ 3 x_{1}+7 x_{2}+7 x_{3} &=6 \end{aligned}
$$

Mirza Abdul Waris Begh · Accepted 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

Alina Krajewski · Answer

Okay, So this problem we&#x27;re asked. She solved this system right here. You simplify the problem a little bit. I&#x27;ve turned the system into an automated matrix right over here. So now I&#x27;m working on the Augmentin matrix. I&#x27;m gonna trash you. Turn this three and true zero. To do that, I&#x27;m going to your place of row two. Bye. Negative. Three times were one plus wrote to you, and that&#x27;s gonna give me a new road, too. As zero two negative. Five four. Now, I&#x27;m gonna try to turn this negative four into a zero by replacing Grow three with four times. Wrote one plus road three and that&#x27;s gonna give me a new row. Three as zero negative. Six 15. Negative. Nine on. We still have our original first row right here. Um, now, since our last row has this, you already let&#x27;s try to govern. Is this 15 right here To do that, we can look at a row two and replace, wrote three by three times, wrote to plus row three. That&#x27;s gonna give us a new road three off zero 003 So we can see that even though we didn&#x27;t get rid of this 15 like we wanted to. We also got rid of this negative six right here since three times wrote to also gives six that limits this terrible right here. So we can see in this last row that this system will require zero to equal three if there was a solution, which is impossible. So what we see here is that this system is inconsistent and what that means is that there is no solution and solution said is empty.

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.

Alina Krajewski · Answer

Okay, so this problem were given this system and were asked to solve it, we could do us using elementary row operations. So is the first step. Let&#x27;s look at X one. We can see that if we add row, Want to row to weaken, Get a new road to with this X one right here. Eliminated. So again, that&#x27;s row one Plus wrote to And that gives us a new row two of negative X two plus five X three equals seven and we still have our original first row and second row right here. Now we can look at, uh, ex too, and we can see that if we add two times Row three Ciro, too. We can get rid off this negative x two right here, and that gives us a new second row of seven. X three is equal to seven now. Using this, we can solve for X three by dividing by seven on both sides. And that gives us X three is equal to one. Now that we&#x27;ve self Rex three, we can plug that into Road three right here and gets X two is equal to negative one since negative one plus run is equal to zero, and now that would have X two. We can plug that into our first equation right here, and that&#x27;s improvised to X. One plus three is equal to five, which gives us an answer of X one is equal to two. So the answer to the system would be too common. Negative one comma one.

Kerry Thornton-Genova · Answer

to solve this equation, we first need to find a common denominator. Since we have three and five. Let&#x27;s make this 15. In order to make this common denominator, we need to multiply X over three by 5/5. This is the equivalent of multiplying by one, but we&#x27;ll make our denominator 15 2 50 x. We need be multiplied by 3/3 and negative 11 5th who need also be multiplied by 3/3. When we do this, we get five x over 15 plus six. X over 15 equals negative 33 15. Now that we have the same denominator on this side, we can add together enumerators. So we have five X plus six x, which is 11 x over 15 equals negative. 33 over 15. We can then multiply both sides by 15. This will eliminate our fraction. We now get 11 times X equals negative 33 In order to get X alone, who must divide both sides by 11. When we do this, we get eggs equals negative three. So the solution to this problem is X equals negative three

Ankit Gupta · Answer

you are in this question. We have given inequality or a question that the question is, who is it? Questo Not off five minus seven eggs, then plus one. So we need to find out the solution for this equation. So from here we can write. It is quite minus seven X off more. Isaac was 21 only to minus minus one. Now again, this value should be because two plus one or minus one so we can write it as five minus seven. X is equals two plus one or ID minus seven. X is equal to minus one. No again. Seven X is equals toe full y minus. One will be four. Access equals toe four divided by seven Or from here, seven x is it questo six. Exit equals to 1600 by seven. So these two will lose off X. Are this solution for miss? A question solution is for direct investment in six developed by seven. This is the answer off this question

Problem

Solve the systems in Exercises $11-14$ $$ \begi…

Problem 11 Easy Difficulty

Solve the systems in Exercises $11-14$
$$
\begin{aligned} x_{2}+4 x_{3} &=-5 \\ x_{1}+3 x_{2}+5 x_{3} &=-2 \\ 3 x_{1}+7 x_{2}+7 x_{3} &=6 \end{aligned}
$$

Answer

The matrix is inconsistent

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

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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The matrix is inconsistent

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Maria G.

Michael J.

Absolute Value - Example 1

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

Alayna H.

Kayleah T.

Maria G.

Michael J.

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

Linear Algebra and Its Applications