let-mathbfuleftbeginarrayr2-3-2endarrayright-and-aleftbeginarrayrrr5-8-7-0-1-1-1-3-0endarrayright-is

Question

Let $\mathbf{u}=\left[\begin{array}{r}{2} \ {-3} \ {2}\end{array}ight]$ and $A=\left[\begin{array}{rrr}{5} & {8} & {7} \ {0} & {1} & {-1} \ {1} & {3} & {0}\end{array}ight] .$ Is u in the subset of $\mathbb{R}^{3}$ spanned by the columns of $A ?$ Why or why not?

Bobby Barnes · Accepted Answer

now to determine if something is in the span of the columns of some matrix. Um, that&#x27;s just a fancy way of saying, Is there some solution or some vector X? When we multiply this, we end up, um, getting you out of it. And one way we can do this is by signing up in Augmented Matrix and Row reducing it. So let&#x27;s go ahead and do that. 58701 negative 1130 And then over here on the right, this will be to negative 32 Now, if you can use a calculator to help you out here, um and you can just plug this in and do the road reduced echelon form of it. You actually end up with this matrix. So 1030 01 negative 100001 And from this notice that we end up with this saying zero is equal toe one, which is definitely not true. So but this implies is we have and inconsistent system, which means there is no solution. So there is no vector X where if we multiply this by our matrix A, we end up getting you And so this also implies further is that you is not in the span of the column of a. So this is going to be what our conclusion ends up big. Um, but if for whatever reason you&#x27;re not allowed to just use a calculator on, you actually have to do the road reduction by hand. We&#x27;ll do that as well. Now, normally, we would try to get it. Kind of like what they have here. They have, like, this upper triangular form. Um, you could also put into the lower triangular form without any trouble. Um, and the reason why I&#x27;m going to actually do this in the upper I mean, the lower versus the upper. Because notice that we have a one here and we have a one here so we could just clear everything above it. And then it gives us that nice lower triangular matrix. You could do it Either way. It doesn&#x27;t really matter. You&#x27;ll get the same answer regardless. I also just don&#x27;t want to divide that first row by five to get fractions because I try to stay as far away from fractions, if I can. Yeah, so let&#x27;s go ahead and clear this five out here, so we would need to do, uh, negative five row three plus Rogue one. And so we&#x27;re only changing broad one. So five become zero negative. 15 plus eight is going to be, um, negative. Seven zero times sank to five. That&#x27;s just zero. So that&#x27;s seven. And then two times negative five is negative. 10. Add that to to that would be negative eight. And then the rest of the matrix stays the same. So 01 negative one negative. Three, 30 030101032 And now notice that if we were to multiply this second row, we can cancel this seven. Um, if we multiply the second row by seven, I think didn&#x27;t say what I was doing s 07 row two plus row one. So this is going to become zero that 10 in the negative one time. Seven that zero. And when we add that to seven and then we get negative 21 plus negative eight, which is going to be negative 29 and then the rest of the matrix stays the same. So you&#x27;ll see we get something slightly different from the reduced form. But we still end up with our contradiction here because this says zero is equal to negative 29 which again, inconsistent system implies you is not in the span of the columns of a because remember, Like I was saying before, this is just a fancy way of saying, Is there a solution to this matrix equation here? And if you get a inconsistent system, that means there is no solution.

Heather Zimmers · Answer

in this problem were asked to determine if be compliment is a subset of set A And in two previous problems, we found all the subsets of a and we found be compliment. So here we are, looking at me compliment or the compliment of be notice all the elements it has in it are all of those elements. Also elements of a well, one is and three is and five is and seven is. However, to is not an element of a and 15 is not an element of a and since those air not elements of a then B compliment, is not a subset of a

Michael Jacobsen · Answer

In this exercise, we start out with the Matrix A and we note about this matrix that the columns are not multiples of each other. And because of that, if we were to span the columns of A, they would form a plane in our three. What we&#x27;re trying to determine then is for this provide vector you we see here. Will you be in that span? Well, to determine that we conform either the vector equation or a matrix equation. But in either case, we&#x27;re going to be augmenting as follows. Take the matrix a augment with the vector, you in question. If we have that, there&#x27;s only pivots that are not in the right. Most column, then the following equation A X equals you would be consistent. And if that&#x27;s the case, you would be in the span of the columns of a. So it&#x27;s right out what this matrix is going to look like. First, I&#x27;m going to copy and a which is three negative 21 negative 561 and zero for four. After we augment with the vector you Now for my first row operation, I&#x27;m going to just do a full swap of row one and three so that 114 is above, which gives us a pivot position here, which is easier, easier to work with and places three negative 50 below. Next, let me copy negative to six and four. Well, instead of using the negative to six and four, let&#x27;s just go ahead and divide row two by two and we have instead negative 13 and to a slightly easier to work with. Now Oracle is to just determine whether the system is consistent here, here, Rather. And so our goal is going to be to eliminate this entry in this century working towards just echelon form. So let me copy the first row 114 then Visualize adding row one and Row two together and recording the results here and the new row equivalent matrix would be 04 and six. When we do that operation next, multiply row one by negative three added to row three and copy. The result here will obtain zero negative eight and negative 12. Now we recognize that Row three is a multiple of row two because if we multiply this by negative to, we do produce negative eight and negative 12 so immediately, Row three could be zeroed out for our next row equivalent matrix. So we have 114 046 and 000 I don&#x27;t recall that. Don&#x27;t for get to recall that we&#x27;ve been augmenting with the vector you. And now we have a pivot position here and here and none other. This tells us that the equation a X equals you is consistent, since there is no pivot in the right most column. But if a X equals you is consistent than that implies you is in the span of the columns of a and geometrically the span of the columns of a forms a plane in our three. So you lies in this plane as well.

Viswesh Uppalapati · Answer

in this video only solving problem. Number 36 of section 4.1 Grand word. Uh, we&#x27;re asked to determine if why is in the subspace of our four spend by the columns of a where why is, um, why is already in our for a zit has four entries given here, and but we need to see if it&#x27;s spend by these specific columns in our four. So in a better part of the matrix, a first we need to set up in augmented matrix here, um, to solve this system. So we just combine the two matrices are to, uh, the Matrix and the vector. So here we have 38 negative five to as the first column. Negative five seven. Negative eight. And negative, too, was the second. Call them negative. Nine Negative 63 Negative nine. Is that their call? And then a line in between. Negative four. Negative eight six and negative five. As the last column as this is the B column in the Matrix equation, X equals B in resolving for a set of scale er&#x27;s X, where any where x one multiplied by the first column and x two multiplied by the second call in the next three multiplied. But third call them would equal why so this season are you have calculator and to make your life easier sense Doing this by hand would take a long time. But, um, so you get if you saw this, you should get and put it in our area. We&#x27;re reduced to ash lawn for me. Should get negative one on DDE 3/5 on zero. So, um, we solved for three values your negative 1 52 53 5th where? Negative 1/5 times. What&#x27;s called this? The one or a one 81 83. So the night of 1/5 time&#x27;s a one plus negative to fifth times a two and was 3 50 times a three equals. Why? So this, um since there is a solution to this system, you know that we know that, um, Why is spanned by these columns the columns of this matrix, and are for

Sean Thrasher · Answer

Okay, so we have this matrix a here, Inspector B and the columns of a we call a one a two, a three we call the span of a one a two a three w. Wright McCall That set W The first question is, is be in the set a one a two, a three. Well, of course it isn&#x27;t because, you know, what does it mean for B to be in this set? It means that it is either equal to a one or equal to a two are equal to a three. But it is neither of these guys, so it doesn&#x27;t belong in this set. How many vectors are in the set? A one. A two a three. Well, it&#x27;s three. That&#x27;s because a one a two a three distinct vectors. If it just so happened that a one a two a three world, the same, for example, then this would actually have one vector. Because, remember, sets do not have redundant elements. Okay, there no repeats allowed. So the answer to part is no. And three for this one is being w how many vectors Aaron W. So let&#x27;s tackle this one first. So W is the span of a one. A two, a three. So what we&#x27;re really asked is does be this vector line the span of a one. A two, a three. And we&#x27;ve done a question like questions like these before. So what does it mean for me to be in the span, Will The span of this is the set of all in your combinations of a one a two and a three, right. So for beats being so to ask, if he&#x27;s in the span is the same thing to ask. Does there exist? X one x two x three All real numbers such that b is equal to this. So can I write? B is a linear combination of a one A two a three. Well, this right here is just this. I was there 36 with three. And then this was minus four minus 23 Equal to 41 minus four. Right. This is the same thing as the following system of equations. Next to someone is to x three him over here, then the third component gives us this equation. Okay, so you should know how to do everything up till now. Um okay, so this system of equations can either be inconsistent on if it isn&#x27;t consistent. Then x one x two x three does not exist, right? There is no such x one x two x tree that can satisfy all these three equations. If it is the case that the that the system of equations is inconsistent and then we&#x27;d find that B is no in the span. However, if it is consistent, if we confined x one x two x tree, then we conclude that, yes, B is in the span of a one a two, a three. So, really, what we&#x27;ve gotta do is just solve for what x one x two x three is So we do this using augmented matrices. Okay, so this is the augmented matrix. I&#x27;m checking it. That&#x27;s right. So the first thing we can do is well, I want to get rid of this pivot right here. This entry. So I do. Row three goes to Row three minus. Nope, B plus plus two times roll one. And when we do that, we get this. Okay, So minus two plus 26 The 6.3 minus eight that&#x27;s minus five minus four plus eight. That&#x27;s plus four. Now, I want to get rid of this entry right here. This is the next pivot. So that&#x27;s gonna be Row three goes to row three minus two times Wrote to roll one is untouched. Road to is untouched but rose six. Now becomes this and I want to make this a plus one on right. And so I let row three go to minus row three. Obviously rode and rode to are not touched. And now I want to get rid of this entry. Right? So wrote to goes to row two plus two times, row three And so when I do this, I get this right because I&#x27;m trying to get this in reduced. Rochel informs I&#x27;m trying to get 111 here in some numbers over here, right? That&#x27;s what I&#x27;m aiming for. So the second wrote, obviously, we can divide by three. So let&#x27;s do that for the second row. Don&#x27;t touch the other two rows. Now I need to get rid of this guy right here yet. So roll one goes to row one plus four times. Row three, right. So I get this. And so yes, I found x one x two x three x one x two x three is this And so you know B is equal to minus four times a one minus a to minus two times a three. Let&#x27;s quickly verify this. So minus four times a one. So that&#x27;s 10 negative, too, in the minus 03 six and then minus two terms minus four minus 23 So what do we get in? The first component is negative for plus eight. That&#x27;s four looking good, minus three plus four. That&#x27;s one eight minus six. That&#x27;s too, um, minus six. That&#x27;s minus four. And yes, that&#x27;s indeed be so. Yes, Be is be can be written as a linear combination of a one. A two, a three. And so yes, B is in W. Now for the final question. How many well, from the final bit of part B. How many vectors? Aaron W. Well, w contains the span of a one a two a three so it contains. Well, it contains all the actors, so it contains all vectors where X one extracts through, you&#x27;re free to choose whatever that is. So, for example, I can let X won&#x27;t be one x to be zero an extra e B zero. Right? And so I&#x27;d see that clearly a one is in w Okay, that answers part. See, so a one is in W. But so is Lambda Times. A one for any lambda for any lambda in our is in W right. And in fact, there are even more factors than that. But this is already an infinite amount of actors, right? In fact, uncountable. So the answer is infinity. Hey, infinite number of infinite amount of actors w is not w is not a finite set part. See, show that anyone is in w So this is very straightforward, right? You know, we just need to show that we can find we confined x one x two x three x one extracts through such that This is true, right? This is our new B. We&#x27;ll take X won to be won on X to an extra to zero and you&#x27;re done. Okay, The point is a one. A two a three previously lies in the span, right? But then so does a one plus a two a one plus a three, a one minus eight tunes on an old and your combinations. That&#x27;s what the span is. Okay. All right, So that&#x27;s it for this question.

Angelo Rendina · Answer

we have these metrics? A. We want to find the metrics Dean, which is four by two. So to calling vectors of land for such that a time Z is there entity offsides, too. Now these requests is equivalent to the system. A B one is equal to the first column of I two, so 10 and at the same time, a V two is equal to the second column over that entity. So 01 So let&#x27;s solve both lines the same time. So let&#x27;s say that you Won is given by a B. C. D, where we want to find what is a B, c D E r, and let&#x27;s write a V one Now, because of a weak computers as a plus, people see and people suppose the and we want this to be equal to 10 now because a B, C D. R. By request on Lee zeroes or ones, there&#x27;s only one possible solution here, which is a, B, C and e. R. All zeros and therefore a must be one. So everyone is 1000 Similarly, we do. Do we call a B C d e N t. That we want to find. And again a V two is given by the very same product, and this time we request these be called to 01 and again because ABC and the are either zeroes or once. There&#x27;s only one possible solution here, which is a B and C are old zeros nd is one. So we, too must be given by 0001 and we&#x27;ve constructed our medics D. Now let&#x27;s say we want to instead find See which again is four by two. Such that C A is the identity off size for Let&#x27;s say that, see, he&#x27;s given by the first roll, the second row, the third row and the fourth row and each of these rose are of length to so now to solve the system C A is equal to did any detail size four. We actually right the metric system, see I for here and then proceed to reduce by rose. But now scenes C has four rows and only two columns, the best we can hope to obtain. He is did entity offsides, two up here and then two rows of zeros at the bottom and then some mad tricks here on the right left college you obtained by the elementary operations on I. But now the system can no have solutions because you is not goingto have zeros in the bottom row because use obtained by elementary operations on their entity, and so we have an impossible solution.

Problem

Let $A=\left[\begin{array}{rr}{2} & {-1} \\ {-6} …

Problem 14 Easy Difficulty

Let $\mathbf{u}=\left[\begin{array}{r}{2} \\ {-3} \\ {2}\end{array}\right]$ and $A=\left[\begin{array}{rrr}{5} & {8} & {7} \\ {0} & {1} & {-1} \\ {1} & {3} & {0}\end{array}\right] .$ Is u in the subset of $\mathbb{R}^{3}$ spanned by the columns of $A ?$ Why or why not?

Answer

$u$ is not a subset of $\mathbb{R^{3}}$ spanned by the column of $A$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Grace H.

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

Grace H.

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

$u$ is not a subset of $\mathbb{R^{3}}$ spanned by the column of $A$

Linear Algebra and Its Applications

Grace H.

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

Grace H.

Anna Marie V.

Kayleah T.

Maria G.

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

Linear Algebra and Its Applications