in-exercises-11-and-12-determine-if-mathbfb-is-a-linear-combination-of-mathbfa_1-mathbfa_2-and-mat-2

Question

In Exercises 11 and $12,$ determine if $\mathbf{b}$ is a linear combination of $\mathbf{a}_{1}, \mathbf{a}_{2},$ and $\mathbf{a}_{3} .$
$$
\mathbf{a}_{1}=\left[\begin{array}{r}{1} \ {-2} \ {2}\end{array}ight], \mathbf{a}_{2}=\left[\begin{array}{l}{0} \ {5} \ {5}\end{array}ight], \mathbf{a}_{3}=\left[\begin{array}{l}{2} \ {0} \ {8}\end{array}ight], \mathbf{b}=\left[\begin{array}{c}{-5} \ {11} \ {-7}\end{array}ight]
$$

Runpeng Li · Accepted Answer

we&#x27;ll get so far a problem. Tell work, even three vectors. They want you to a three, and we&#x27;re also keeping the victor be. So we want to determine if bees alina combination off. You wanted to be three. So that is actually to find out the Matrix a spent by the white to a three. So that is one negative, too. Two and 05 by and 208 and Victor be that is negative. Five 11 7 Elected seven. So here we only need two. The question that want to answer instead whether whether this system is consistent. So to find out what the system is consistent, we need to consider the augmented matrix. So he&#x27;s 102 negative five and negative. 250 11 and 258 Active seven. So by apply gushing elimination, we&#x27;ll find, uh, re reduced rate. You should inform off this album it&#x27;d reach matrix. I&#x27;ll just read it out here. It will be one. There are two zero and 01 for fifth and zero last role will be 0001 so you can find that the last rule it&#x27;s not possible because zero cannot be won. If we have a vector times zero vectors, then you know how it has to be there around 11 side. But on the right hand side, we have the we have the, uh that&#x27;s the return. So that means that system is not consistent. So one more do better answer our question. That is to say, he&#x27;s not a linear combination all the way to a three not linear combination.

Sean Thrasher · Answer

Okay, so we&#x27;re asked to determine if B is a linear combination of a one. A two, a three. So what does this mean? Well, it means Does there exist real numbers X one x to next three such that this is true. So can we write? B is a linear combination of a one a two and a three. Well, the best way to determine if x one x two X three exists is to try and determine what they actually are. And so we can convert this into a system of equations. And when we do that, we get the following. So minus two x one plus x two minus six x three is equal to minus six. Here. Milk minus one. Sorry, minus one. Is this guy right over here? Minus one and two ex eso. So, um two x two two X two plus eight x three is equal to six. All right. So just converted this equation to the system of equations so that the system of equations and we want to solve for x one x two and x three. Now any system of equations can either be inconsistent, so x one x two X three would not exist. What could have a unique solution? So x one x two x three In this case exists Ondas unique. So there&#x27;s only one set of one set of answers that works or have a non unique solution. If we&#x27;re in case one and then the answer would be no, we can&#x27;t write these in any a combination of a one a two a three because X one extracts through would know exist. However, if we find ourselves in case to or in case three, then the answer would be yes. Okay, so let&#x27;s try. Let&#x27;s go ahead and try to solve for X one X to an X three, okay? And the easiest way to do that is to use the augmented matrix methods. So I&#x27;ve augmented the matrix of coefficients here, and we&#x27;re going to do some road reduction. So the first thing is, I want to get rid of this entry right here where I want to make this zero. And so I take row two to be I&#x27;m going to replace that with row two plus two times, Rule One. And so when I do that, I get this. Okay, so I&#x27;ve added. So that&#x27;s minus two plus 21 plus two times zero in the minus six plus two times five minus one plus two times two. And you can see here that clearly wrote to is a multiple of row three. So row three is going to end up being pulled zeros. We can see that by replacing row three by Row three, minus two times zero. To we do this, we indeed get zeros in the weapon. And this is as far as we can go, right? I thought the reduce Rochel informed. And so and so this system of equations is equivalent to this right here. Yeah. So this matrix right here is equivalent to this system of equations, Okay. Converting attractive system of equations. And so, as you can see here, we&#x27;ve got two equations and three unknowns. And so the solution exists, but it is a known unique. All right, this is fine, because any x one x two x three that solves these two equations would work just fine. So let&#x27;s arbitrarily fix one of the variables. So I&#x27;m going to choose X three to be equal to one, right? If I do that, then X two is going to be three minus four times one, which is just minus one. And then X one is going to be two minus five times one, which is just minus three. And so I get X one is equal to minus three. X two is equal to minus one. And x three is equal to one. If I wanted the most general solution possible, I would let X three equal to tea or no tea would be just some parameter. But I don&#x27;t need to do that. This is fine. Okay, so, um right. So what we&#x27;re saying is that minus three times a one, which is this minus one climb 012 plus 56 negative eight should be equal to be. And you can check that. It indeed is right. Minus three plus five. That&#x27;s 26 minus one. Um, sorry. Six. Sorry. There&#x27;s a minus six year, So there should be a minus six here. That&#x27;s what a three is. So six minus one is minus seven and then Sorry. Uh, what am I doing? Yes. Okay. Six minus one is five minus six is negative. One and the minus two plus 86. So this this is perfectly good. Okay, so we found x one x two extra and so we&#x27;ve expressed be as a linear combination of a one a two and a three. All right, so now to the slide that I&#x27;ve been alluding to accident in the earlier. Okay, so we&#x27;ve we&#x27;ve answered the question, right? So, yes, B is a linear combination of a one A two. A three. Now I want to go a bit further and give you an additional challenge. So we found it be such that b is a linear combination of a one a two and a three. But can we find a V such? That V is not on the new combination of a 18783 So when we sold for X one extra next three, we found that there was known unique right? And in fact, the very fact that X one X tooth, the next three is not unique tells us that a one a two and a three are no linearly independent, right? Because the thing with linearly independent vectors is that if you know, so if be happens to line the span of a one a two and a three and a one a two. A three year linearly independent. Then these cool efficiency would have to be unique. Okay, Now we could have found that X one extracts. We didn&#x27;t exist at all. And if we found that, then we couldn&#x27;t say anything. Right? Then we couldn&#x27;t say anything. So a one, a two a three is no linearly independent, and we can actually see this. Right? So a three is equal to five times a one plus four times a two, right? And so a three, in fact, lies in the span of a one and a two. And so the span of a one a two a three has to be the same as the span of just a one and a two. Right. But two vectors have no chance of spending our three right. At best, they can only span a plane, okay? And so we should be able to find the V in this. So actually, the answer should not be Yeah, it should be. Yes. Sorry. So we can, Right? So we should be able to find a V, And your challenge is to do that. Now, this takes me into why? Just briefly touching on why a basis is so nice. So suppose we have a basis for our three v one b two b three and you know w is equal to x x one x two extra. And I tried to Seoul for X one extract street. Well, these coefficients here definitely would exist because by definition, since their basis for our three, they span are three. And so it is a linear combination of you want me to every three. So existence is guaranteed on. And since x one x two extra sorry since V one v two v through the newly independent, then the coefficients X one extracts, we would also be uniquely determined. And so I could represents that I can represent this w as just a column right here. Put x one x to x ray and call that the coordinates of W with respect to V two v three v one v two V three is the basis, and that&#x27;s what we do. That&#x27;s what the column vectors actually mean, right? You pick some sort of basis and then, you know the entries in the columns is what the corresponding weights of each of the basis have to be, and it&#x27;s uniquely determined because of the new independence. So I just want to point that out this bit. This last bit was unnecessary. But again, to Richard Reid, it&#x27;s right. B is a linear combination. A one a two and a three.

Runpeng Li · Answer

All right, so four problem. 14. The game we don&#x27;t given three by three matrix. That is one negative to a negative six. And And 037 and one connected to and five and or the vector we&#x27;ll give it. Here is 11 next, five and nine. So to find again to find out, um, whether whether bees Adelina combination off objectors formed by the columns of this matrix. So the thing we need to consider against the the augmented matrix hey, admitted to be And so this is actually we need to consider the linear system, Max. It would be. And we want to find whether this system is consistent and we consider such a such undocumented matrix. And we apply culture animation to find a role. Reduce station form right down the final form here. So this will be 100 2 45 over 33 and the second rule would be 010 and 41 over 33. Start negative. 41 negative. 41 over 33. And the last role will be there was everyone and negative to over 11. Okay, so, bye, Elise will reduce lesion form. We say X is x one x two Next three them from the last rule we know X tree extra A is next to over 11 and next to we&#x27;ll be on the second row. How just we write a three here. So that&#x27;s next x three will be connective till 11. X x x two will be in negative 41 over 33. So I did four over 33. So negative for 41 over 33 And next one from the first row. We know it is. Ah, Wei. Yeah, yeah, he said to 45 over 33. And so this is our vector X. So under this circumstance, we can have the Lena combination off be in terms of column self matrix.

John Irizar · Answer

the best way to see if a point is a solution to a system of equations is toe. Actually, plug in the point here. I will go ahead and plug in our X value six times negative too minus two times, and then plug in our why value negative five. And does this actually equal negative, too? Six times negative two is negative. 12 negative, two times a negative. Five is positive. 10. Negative 12 plus positive. 10 is in fact, negative, too. So our point works for our first equation. Now let&#x27;s see if it works in our second equation. Three times negative, too. Plus in negative five. Does that equal negative 11 three times negative two is negative. Six minus the five that does, in fact, equal, Negative 11. So our point is a solution.

Runpeng Li · Answer

Okay, So in this problem, we only need to apply our the&#x27;re, um your mate, you know, textbook. So it says if we have the Matrix P Well, in this case, our matrix peas. Just a span, not B one B two. So Matrix B. We know this matrix. Then our bi matrix will be peeing burst. Time say times P. So we just need to find out the product. So, Pete, members, by my calculation, I won&#x27;t do this step by step. Just a house. Just write down this of embers. It&#x27;s 11 I think the 23 now Matrix A is next to one for next 2 to 3 and matrix P. It&#x27;s just a three negative 1 to 1. Okay, So as first find, it was too part of the first product. So first entry, Nick one times. Like to warm. Plus two two times one. So that&#x27;s negative. Three and one plus one times four plus one times three. So that&#x27;s seven and two minus six. So it will be negative for And the last entry we have connective aid. That&#x27;s nice. So it&#x27;s one now, three negative. 1 to 1. Okay. Now for this to matrix. We&#x27;ll search for these two missions, since we have next of three times three active and I last 14. So it will be five and three. Ah, seven. So he&#x27;s 10 and next to 12. Us, too. So selective. 10. And we have four plus one. So he&#x27;s five. No, just put the skater into our matrix. So we have one too connected to and one So this is

Sean Thrasher · Answer

Okay, so we&#x27;re asked to figure out if be this problem back to be is a linear combination of the columns of A And I&#x27;ve written down the columns of a is the factors v one b two, b three. So the question is, is be a linear combination of B one b two, b three. And if you seen Problem 11 is the same exact process behind it. So we want to find x one x two x three Such that we can write be as this some right here. That&#x27;s what it means for B two b a linear combination of you want me to be three. And so this equivalently is if you&#x27;ve if you substitute and what b one b two B three years and write it out was a system of equations. You get this. And so the question is, does x one x two x three exists So to ask that we&#x27;re going to try and find the solutions x one x two X c We&#x27;re trying. We&#x27;re going to try and solve the system of equations. Now a linear system of equations could either be inconsistent, so solution might not exist. It could have a solution, and the solution is unique. So on Lee one you know, triplet of X one x two x three would work, or you could have non unique solutions of a solution would exist. And there be also yet another one. Hey! And so if we&#x27;re in case one, then we know that B is normal linear combination of a 18 23. And it turns out that this is the case that this system of equations boils down to. So to solve it, we augment the Matrix like this. We have no events matrix, the matrix of coefficients here, and the vector B, which is the right hand side of the equation. And so we&#x27;re going to use role reduction. And first step is to take row three. I want to get rid of sentry. So I ads two times, roll one to row three. And when I do that, I get this. So I get zeros. Except for this last entry here, and it&#x27;s a negative three. So 000 Negative three. Sorry. You can&#x27;t quite see that. Uh, sorry. No, it will be plus three. So +000 plus three and we can see that there&#x27;s a problem. So if you convert this back too, a simpler an equivalent system of equations, you have this. And the third equation is just zero is equal to three. And certainly this is not true. No matter what you know, this is not true. Regardless of what x one x two extra you choose and so boring. Case. One system of equations is inconsistent, and so B is not linear combination of the columns of a and we&#x27;re done.

Problem

In Exercises 13 and $14,$ determine if $\mathbf{b…

Problem 12 Easy Difficulty

Answer

$\mathrm { B }$ is not a linear combination of $a _ { 1 } , a _ { 2 } ,$ and $a _ { 3 }$

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$\mathrm { B }$ is not a linear combination of $a _ { 1 } , a _ { 2 } ,$ and $a _ { 3 }$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Heather Z.

Alayna H.

Caleb E.

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

Linear Algebra and Its Applications