in-exercises-5-and-6-follow-the-method-of-examples-1-and-2-to-write-the-solution-set-of-the-given-ho

Question

In Exercises 5 and $6,$ follow the method of Examples 1 and 2 to write the solution set of the given homogeneous system in parametric vector form. 
$$
\begin{aligned} x_{1}+3 x_{2}+x_{3} &=0 \-4 x_{1}-9 x_{2}+2 x_{3} &=0 \-3 x_{2}-6 x_{3} &=0 \end{aligned}
$$

Priyanka Sadarangani · Accepted Answer

Okay. Be able to find the solutions have given homogeneous system in parameter vector form. So given situations are x one plus three x two plus X three equals to zero minus four x one minus nine. Next two plus two extra equals to zero minus three X two minus six x three close to zero. Okay, so we have, right. Same in metrics. Phones, metrics one. We can write it as 1310 minus four minus nine 200 minus three minus 60 Okay, now we have to solve this equation. Or that we will do R two equals two R two plus four other one. Okay, for after doing that, we get 1310 This will be same in the art to that will get 03600 minus three minus 60 It will be same now we will further sold it. We will do. Our one equals two r one minus are too. And for our three, we will do our three plus our too. Then what we get? We get one Zito minus 50 Zito titty 600000 Okay, now we will do our two equals toe are two by three. Okay, so what? We get 10 minus 50 01200000 Okay, so what? We get in vinyl desecration. So, for this occasion, we can write that excellent minus five. X three equals to zero. Okay, this will be component off x one. And this will be for exterior. Okay, For the same in the second row, we can write that x two plus two x three close to zero on from 3rd 00 Quest to zero. Okay. On the variable X rays free. You know, X rays free. So in the way, we will get the value of X one in the term off X three like this and XTO in the term off extra. Okay, so what? What we get in the final regard. Keep the legs equals two. We know x one x two and x three and this will be five x three minus two x three and extra. Okay. Actually, it will be minus tree. Okay on. And now they will take X three comin out off this and what we get. Five minus two and one. And this will be over. Final answer. Thank you.

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 video, we have a matrix equation given to be a X equals zero. So it&#x27;s a homogeneous system of equations. And we&#x27;re also told that a ISRO equivalent to the matrix we see displayed here well, using that information, the idea that we would like to pursue is what is the entire solution set to this system of equations? Well, to do that, we&#x27;re going to have to roll, reduce. Let&#x27;s start back on this matrix A that we see displayed. And I&#x27;m going to make my first role operation to divide Row one by three that obtains one naked of three and two and below. We have negative 13 and negative, too. We noticed. Now that Row two is suspicious. It&#x27;s just row one multiplied by negative one. And when that happens, we can immediately say that this is row equivalent to one negative 32 with a zero down below. Or in other words, if the rose are a non constant multiple of each other, one of them is eliminated. So now that we have a in its role reduced echelon form, we&#x27;re ready to go back and consider the solution set to this system of equations indicated here. We have to be careful, though, because our Matrix say has not been augmented. So when we go to Row One, no one is thick. Waysh in x one minus three x two plus two x three equals zero. Row two has no new information, so we&#x27;ll just record that x two is free and x three is free. These air free variables since the Onley pivot column is column one corresponding to variable x one, making both x two and X three free variables. Now we&#x27;re ready to describe our solution set to this system by a step by step process you can write to. The X is equal to x one x two x three So this is our set up and we&#x27;ll aren&#x27;t going to do the following first go to the equation for X one and solve it for X one. We would be forced to add three x two double sides, so that would go here minus two x three. So now if you read this factor equation, we&#x27;re saying X one equals three x two minus two x three And that&#x27;s no different from this equation displayed here. Now the trick is what to do with the free variables X two and x three. Well, there&#x27;s no restriction on next to since it&#x27;s free, so we can say X two equals x two. But the vector equation we have here we would just write in an X two in this place. Likewise, X three is free means there&#x27;s no restriction so we can write the trivial equation. X three equals x three and that means X three gets placed here in that way. Now, with this set up, we&#x27;re ready to begin describing the Parametric Vector form. Now. One trick in the vector form is the free variables are going to be equal to the scale. Er&#x27;s in this linear combination. Let me start out by writing x two and a plus x three here. Now we just need to determine what the vectors that these scale er&#x27;s X two x three would be multiplying well for the first vector. It&#x27;s associated with the X two variable, and we just pull the coefficients, which are 31 and 00 because there is no value here or if you prefer, you could have called it zero times x two just for the sake of organization. Now for the X three variable, we go to the x three column here and take negative too. Zero and one here X two and x three are parameters or free variables. And although those things are just different ways of saying the X two x three are in our so any real number with no restrictions. So this is the parametric vector form for this system of equations.

Runpeng Li · Answer

All right. Um so for these problems were given the Matrix one connected to negative nine and five, and the second row is 01 to negative six. Now we need to transform this, transforming its matrix into the reduced rations for So that is, we had had the first rule by twice up to the second roll. So that&#x27;s one neck of two plus two. So that is zero neck neck of nine plus four. That&#x27;s negative. Five and five and five plus negative. Six times, too. So these five minus 12 it is, uh, neck of seven. And the second rule would still be you. 012 negative six. Okay, now we&#x27;re plugging our variables. So that is X one minus five x three. Find a seven next four, which is so and next to plus two x three minus six x four because zero. All right. So observed that there are two for your valuables here because X one is determined by five x three plus seven x four and x two is also determined by X three ANC exports. That&#x27;s a to X three, first six x four. Okay, so we first right down solutions That will be X one substitute by if this expression. So that will be five x three cause seven Next four second x two will be substituted by this expression. So that is to x two x three. Sorry we missed the X three here and plus six x four and we still have X three and export. Now we want to separate this expression. So we first excuse me. So we first have five x three and two x three and x three and zero plus seven. Next four and six x four zero Next four. Now I noticed that it&#x27;s just a separated These separate the inspectors by the sum of two different vectors. So that&#x27;s Ah five x three plus seven. Next four that it&#x27;s a that it&#x27;s the first entry and two x three plus six export at this that it&#x27;s a second entry and extreme plus zero that is 3rd 1 and zero plus exported. That is 4th 1 So the next we take out the x three. So we have the vector 52 one zero Next three for us 76 zoo one and times exports. So this it is exactly the parametric a defective form off the solutions off exit actually cost zero. Now we&#x27;re

Runpeng Li · Answer

in this problem. We have three equations, and we need to find out solutions and write it down in Para mention vector borne so that your questions are x one plus three x two minus five x three It was four and x one plus four x two minus eight x three You caused seven and the last one is negative three x one minus seven x two plus nine Next three, which is negative. Six. All right, well, first we were set down the trees that it&#x27;s 13 negative five and one for 98 and negative. Seven neck of three. Negative seven on DDE nine. Remember this? Ah, nothing consistent. Sorry, This is not homogeneous. We know equation. So we have to set down the council terms. That is for, um seven. Negative six. Now we need to perform the cultural eliminations first. Uh, we first used the second roll minus the first road to cancel one here, So that gives 01 elected three and three. And we are so we can also perform that we use our three plus three times are one we used to throw three plus three times to rule the personal So that is a neck and three plus three. That&#x27;s zero. And we have to because that it&#x27;s Ah, negative. Seven plus three times three, which is nice. So that&#x27;s as to and so on, so forth. We have next six on six and the first rule keep the same. That is one really negative five. And for just for your information, I all right? I will write down the road crew operations here. So that&#x27;s road too. Minus real one and row three plus three times ago. One. All right, all right. In the next a cz we observe here. The third row is twice on the second row, so the third row can be cancelled here. So that is 13 connected five and or and 01 negative. 33 and 000 Is you all right? At last, we can use the first roll minus three times the second rule. So that is 104 and negative five. And the second road. Still the same. 01 negative. Three on three and all zeros in the natural. All right, so let&#x27;s check out. Check out the equations here. So that&#x27;s that is X one minus Sorry, it should be a plus. That&#x27;s four x for X three. He caused negative five and x two minus three x three equals three. So that means X three is a free bearable here, so that&#x27;s X Juan Can be could be represented as a negative five minus four x three and x two can be written as three plus three x three. So our solutions our solution is connected. Five minus four x three and three class three x three and extremely and we separate. First, inform off ex threes of connective four x 33 x three and that&#x27;s three plus accounts in terms will be negative. Five. So active, negative five and three and a zero. So that&#x27;s we take out the X three. So that&#x27;s negative. Four to be one times X three clothes, clothes negative. Five, three and zero. So we&#x27;re done. Now remember, we can compare our solutions with our previous solution about about exercise Six. That is, uh our previous solution is next. Four, three and one times x three. This is this is our previous solutions. And compared to this, can you notice that we have a constant term? You are a new solution. So that&#x27;s the thing to take care of

Runpeng Li · 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.

Problem

In Exercises 5 and $6,$ follow the method of Exam…

Problem 5 Easy Difficulty

In Exercises 5 and $6,$ follow the method of Examples 1 and 2 to write the solution set of the given homogeneous system in parametric vector form.
$$
\begin{aligned} x_{1}+3 x_{2}+x_{3} &=0 \\-4 x_{1}-9 x_{2}+2 x_{3} &=0 \\-3 x_{2}-6 x_{3} &=0 \end{aligned}
$$

Answer

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

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

Linear Algebra and Its Applications