in-exercises-5-and-6-write-a-system-of-equations-that-is-equivalent-to-the-given-vector-equation-x-2

Question

In Exercises 5 and $6,$ write a system of equations that is equivalent to the given vector equation.
$x_{1}\left[\begin{array}{r}{-2} \ {3}\end{array}ight]+x_{2}\left[\begin{array}{l}{8} \ {5}\end{array}ight]+x_{3}\left[\begin{array}{r}{1} \ {-6}\end{array}ight]=\left[\begin{array}{l}{0} \ {0}\end{array}ight]$

Runpeng Li · Accepted Answer

Okay, so for problem six work even the vector vector addition, that is X one times Vector Negative two and three plus X two times eight and five, which is equal to sorry. We still have X three so plus x three times 161 negative six, which is equal to zero. So if we write down this as it&#x27;ll set of Lena you questions, then this will give us connected to x one plus eight x two. Thus, X three is zero and another question is three x one thus five x two minus 663 his zero. So that is how early.

Sean Thrasher · Answer

Okay, so we got this vector equation right here and were asked to write it in on equivalent system of equations which have done here. So this is the answer How I get to that while the left inside this is six x one minus x 15 X one, Right for this vector right here. I factored this X one inside, and then this is similarly this Okay. And then the left hand side you, When you add these two vectors together, you get this. Okay? And this is equal to this on the right hand side and for two vectors to be equal, they have to be equal component by component. So this right here has to equal this. I&#x27;ve done that there. This has to equal this. That&#x27;s that. This has to equal this, and that&#x27;s that. And we&#x27;re done.

Runpeng Li · Answer

All right, So for this problem, we&#x27;re given three questions. That is X one last three X two minus five x three zero Second equation this ex Juan class for X two minus eight x three Equal cereal and third equation Issa connective three. Ex Juan minus seven. Next to plus nine x three, which is zero Sorry. Three x three already down clearly and, uh, zero. All right, so we ready down, inform off me treats. That is 13 negative. Five 14 naked eight and elected three. Active seven and nine. Now we&#x27;re gonna use the gushing elimination. So we first use the second row minus the first roll. So the first entry will be zero. Second entry will be one, and then give eight miners naked. Five, that is, uh, five minus a day. So it&#x27;s negative. Three and negative. Five. All right. So for the third rule, we can at the first roll times three. So that&#x27;s Ah, neck. A three pack of three plus three times one So that it zero and back of the neck of seven plus three times three. That is nine. It&#x27;s nine minus seven. It&#x27;s too. And nine minus three times Negative fight, that is 15. So that&#x27;s nine minus 15. So it will be next. Six. Okay, so right now we have 13 negative five. So I can see that if we usedto third role minus two times the second road, then we will cancel all the all the coma patients here. So that will be 0000 All of yours. On the third role? Yes. We used a two times one two times. One is two. So two time to minus two is zero and two time. State of threes. Neck of three and three minus Selective. Three is zero. So that&#x27;s 01 like three. All right, Right now, remember, we need Thio transform me too. Transformed this matrix into reduce station and poor. So that is one. We used a first rule minus three times the second row. So that is one, um, zero and next. Five miners neck of three times three so that it&#x27;s a negative on I and, uh, minus neck. Then I would be positive. Nine. So there&#x27;s nine minus five. That is four 013 93 000 Right. All right, So now we have X one plus four x three. It&#x27;s zero and next to minus three. Excrete is zero. That means X three is a free venerable because X two is dependent on next three and X one is dependent on X three. Two. All right, so we can write down the set up solutions in such a form, that is x one ISS negative four x three and x two is three x tree and x three. We take out the scaler, which is X three. Now we have 43 one. So that is X three times Vector be where be iss connected or 31 and we&#x27;re done.

Michael Jacobsen · Answer

in this exercise, we have a linear combination that&#x27;s given on the left hand side of equation set equal to a vector. So all together, we call this a vector equation. This vector equation could be written as a system of equations. It can be placed into augmented matrix, but what we want instead is a matrix equation. So let&#x27;s see how we can build a matrix equation starting from the vector equation. Well, the first thing we&#x27;re going to do is go to each of the columns we see on the left hand side and former Matrix out of them, so it&#x27;s quick enough we just have to copy four negative 17 negative four for column one negative 53 negative 51 for column, too, and seven negative. 802 for the Third column. Next, we&#x27;re going to use the scale urz heer and form a Vector. This is a vector that contains all of our unknowns, which are x one x two x three, and finally were ready to say that this is equal to Don&#x27;t forget equality since we need an equation and it&#x27;s equal to the right hand side, which we copy in as six Negative 80 negative seven. Then this is our matrix equation. But if we were to label this as a a, this as X, the unknown vector and this is B, then it&#x27;s off the form eight times X equals B. So this is our matrix equation, and you&#x27;ll often see it presented in this way as well.

Sean Thrasher · Answer

Okay, so we&#x27;re asked to take this system of equations and write as a vector equation. So this right here is going to be Well, okay, so this, in terms of actors is just x two plus five, x three and then for this one is for ex woman plus six x two minus sechs three minus x +13 x two minus takes three is equal to zero vectors. +000 Right. So here to hear it just translated as a vector, all right? And then I just split it up, right? So I can write this as so Paul. The X one terms into one vector all the x two terms, uh, into one vector. So this should be books, too. Six x 23 X two. Right. And then minus. So it was a plus zero minus x three minus a text. Three equals to 000 And then when you factor this X one from here x to from here. Next three. From here, you should get this is the final answer. But next to 163 Um, sorry. This should be five next three. Here. That is plus X three into five, minus one. Minus eight 000 That&#x27;s the final answer.

Bobby Barnes · Answer

Yeah, So they want us to rewrite this from this vector form into the matrix form. So remember the former wanting to rewrite this into is that a X is equal to be. Let me write that ever. So what we want is a X is equal to be, since we have, like, the vector over here. So remember X is going to be made up of Z one z to C three and C four. So over here X is going to be so we take the first one. So z one and then it&#x27;s Z two, c three and C four. And I guess technically, I should be putting the little vector hat on this. And then our matrix A is just going to be these kind of just squished together into one, as if, like the Z&#x27;s want their it all. So a is going to be so or negative to been negative for five and then negative 54 and then 30 and then be is just what we have on the right side here. So if we go ahead and rewrite this now, we&#x27;re going tohave four Negative four negative. Five three Negative two by 40 and then we multiply by our vector. So Z one z two z three c four and then this is going to be able to for 13. And so then this here is how we rewrite that into the matrix form.

Problem

Use the accompanying figure to write each vector …

Problem 6 Easy Difficulty

Answer

$- 2 x _ { 1 } + 8 x _ { 2 } + x _ { 3 } = 0$
$3 x _ { 1 } + 5 x _ { 2 } - 6 x 3 = 0$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Catherine R.

Anna Marie V.

Kayleah T.

Maria G.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$- 2 x _ { 1 } + 8 x _ { 2 } + x _ { 3 } = 0$$3 x _ { 1 } + 5 x _ { 2 } - 6 x 3 = 0$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Kayleah T.

Maria G.

$- 2 x _ { 1 } + 8 x _ { 2 } + x _ { 3 } = 0$
$3 x _ { 1 } + 5 x _ { 2 } - 6 x 3 = 0$

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

Linear Algebra and Its Applications