construct-a-3-times-3-matrix-a-with-nonzero-entries-and-a-vector-mathbfb-in-mathbbr3-such-that-mathb

Question

Construct a $3 	imes 3$ matrix $A,$ with nonzero entries, and a vector $\mathbf{b}$ in $\mathbb{R}^{3}$ such that $\mathbf{b}$ is not in the set spanned by the columns
of $A .$

Runpeng Li · Accepted Answer

Okay, so for problem, tell me too, when you to construct three by three matrix A with none. Zero entries and a veteran be in our three, such that he&#x27;s not set Spend by columns of a So he&#x27;s sexually just to construct a matrix A and Victor bees such that the system is inconsistent. So there are a lot of matrices and vectors like this. I&#x27;ll just give give my own matrix and you can try your own parcel So many trees high gave it so that will be one neck of three and two and or one fight and 271 And the vector B, I consider is next. If I 11 elective seven Okay, just to just to show why this, uh, why&#x27;s this system is inconsistent is you just need to consider the argumentative Matrix. So his case, it will be one left of three and two and five. Sorry, negative. Five and four or one high. And 11 271 active seven. So if we if you do the you do the couching elimination and to find out a Rory to station form, then I&#x27;ll just write it down here. So the real redo station form will be 10 17 3rd and zero second. Rory is 01 like 23 13 and zero. So that&#x27;s rule 000 And the one so clearly this term on the third row should not be one. Because we have all zero back already. There is on the left inside. And if we are given any vector, the nef s I will always be zero. But the right hand side is accounts in one, so the system is not consistent in this case.

Carson Merrill · Answer

So one way we could have a three by three matrix and a vector B Such that B is not in the calm space of a would be as if we constructed, um, a Matrix that in this case has only one pivot column. So an example of this would be 100 and then we&#x27;ll just have a bunch of zeros. So this is a simple example. Um, but here we see that there is only one pivot column. Um, and we can call this a so that way we know that the column space of a or at least the basis of the column space of a is 100 which means that it could be 200300 But there&#x27;s always going to be a number here, and then zeros are on the bottom. Um, then let&#x27;s just pick a B vector to be 010 We know that this is not possible because there&#x27;s no way that we can make that second value in the second row. There&#x27;s no way we can make that value one. So this would be a very simple example, but an effective one

Michael Jacobsen · Answer

in this example, we&#x27;re considering the homogeneous equation. X equals zero, and we&#x27;re told that we have a nontrivial solution right out of the gate. It&#x27;s X equals 111 Let&#x27;s suppose also that is, of size three by three. And the question is four. What matrix A is X equals 111 A solution to our homogeneous equation. Well, in order to determine such a matrix, say, Let&#x27;s first right out a X equals zero in Amore expanded form. Let&#x27;s put in a matrix. Say, I&#x27;m going to leave it blank For the moment, X is going to be 111 and this has to be said equal to the zero Vector, which is 000 Now we could make a completely the zero matrix, but that feels as though we&#x27;re cheating. Really, we wonder, Can we do this and get 111 a solution when a is not the zero Matrix? What one thing we can do is just because it doesn&#x27;t have to be. The Zero Matrix doesn&#x27;t prevent us from zeroing out a row or two, but I wouldn&#x27;t get too out of hand. I want to make the matrix a as simple as possible. So I&#x27;m going to make this 111 in this column. Now, if we think about what would happen if we multiplied out this much, it would look like this so far, we would take the vector entry here, which is a one, and multiply it with the first column of a So will be a column, 111 going here. We still have to decide what should go here. But we know that whatever we put will be plus one times those entries where I left question marks and for the last vector will be putting plus zero time. Excuse me. Plus one times that zeroed out column and all of this has to be equal to zero. So this is the format so far. And if we decide what can go here and then here, ultimately the vector equation to produce all the zeros will be all set will notice. That assumes we chose ah one here. If we put a negative one here, then add up, We&#x27;ll you will definitely get to zero, especially since there&#x27;s no contribution here. Some going to put negative one here and here and here in the corresponding vector equation. We then have a solution, and that tells us that this matrix A allows 111 to be a solution so effectively. If we make one column, the combination of another and zero out a column, that would be one way to determine such a matrix a.

Runpeng Li · Answer

All right. So your problem 30. We need to construct a three by three matrix, not in addition form, but whose columns do not spend our three. So that is actually a beauty, eh? Um, a matrix. That column vector. They&#x27;re the column vectors are they know any dependent? So, too there are many off them, so I&#x27;ll just try. I just try one. Say I have the column, vector y 12 connective one. And we, let&#x27;s say, uh, the second column out due to four and negative, too. And Third column. I&#x27;ll do 36 and negative three. So you can say you can see from from this Matrix, that of the second column is twice off the first column, Third column, third, the third column. He said, um, three times on the first call. So face, make sure excuses three by three matrix, which is, um, clear here. And it is not in the issue of warm, because I have not reduced it yet. And this but this mission is definitely the column. Vectors of columns definitely do not spend our three because we actually we only have one. We only have one vector, and we only have one. Meanwhile, we don&#x27;t have one people position or people call him So So this this column actors doesn&#x27;t spend our three at all.

Michael Jacobsen · Answer

In this example, we have a three by three matrix which has been constructed. This particular matrix is not initial inform, and that&#x27;s strictly due to this negative six quantity, which is represented here, highlighted in yellow. Let&#x27;s analyze this matrix further by putting it in echelon form. Once it&#x27;s a natural inform, we will be able to analyze its pivots. So for my first row operation, we&#x27;re going to eliminate this negative six. So let me copy Row One and grow to next multiply row two by three so that we can add it to Row three, the result of B zero zero and one. Now that the Matrix is in its echelon form, we have a pivot here that we can analyze here and here. In other words, this matrix. If we decide to call it a allows us toe show that A has a pivot in every row having a pivot and every row is now very special for this matrix. Because we have a pivot in every row, that&#x27;s logically equivalent. It&#x27;s all right to the bi directional implication, simple to the statement that the columns of a spans are three where the three comes from specifically the fact that we have three rows. So for this particular matrix, we have an example where there&#x27;s a pivot in every row and therefore the columns of this matrix will spend its space are three.

Donald Albin · Answer

Let s be the subspace of three by three matrices with riel elements, Um such that the elements in each row and each column add up to zero. Interesting. Well, I&#x27;m going to write a be negative a minus, bi, because a plus B plus negative a minus bi is zero. I&#x27;m gonna write c de negative C minus D and I&#x27;m going to write e f negative e minus f. However, all the columns after Adam zero also so a plus c plus e has to equal zero and ah b plus d plus F has to equal zero. So e is going to be negative. A minus C and AF is going to be negative. B minus d. So I&#x27;m going to rewrite this a b negative a minus B C d negative C minus d negative a minus C negative B minus d His last one&#x27;s interesting. First of all this. Plus this has to add up to zero, But then we got some substitution as well. Let&#x27;s put in the substitution and we&#x27;ll see what happens. Um, it&#x27;s negative. E minus. F negative e minus f is negative. A minus c. It&#x27;s just a plus. C plus a plus B A plus C plus a plus B. There&#x27;s gonna be two A plus B plus c. However, I&#x27;m not confident that that gets all of the restrictions, because we&#x27;re further saying that Negative A minus B minus see, mine is D plus to a plus be plus C equals zero. So I&#x27;m thinking maybe I should have started this a different way and just did a B c d e f g h I, um But let&#x27;s keep going. Let me let me think so more here using with rose e f negative u minus f I used the rose to figure this out and then, um, I figured out that negative u minus f We used the two columns there so kind of used rows and columns. We didn&#x27;t use this last column for sure, and that&#x27;s what I&#x27;m using right now. So, um, to a minus is a negative B plus B that cancels out negative C plus C cancels out. Mine is D is here. So a equals D. That&#x27;s what I was thinking. So if a equals d, then we can rewrite this. And ah, I wish I had more space. I&#x27;m gonna write it over here. A is the same is D So it&#x27;s a be a minus bi. That&#x27;s still good. C d negative C minus a but again, de is a Okay, um negative A minus C negative B minus a to a plus B plus C and so there&#x27;s no longer a d. I think this is correct, but let&#x27;s just go across the bottom row and make sure negative a minus C minus B minus a plus to a plus B plus c past equal zero. This this and all this Cancel out this and this. Cancel out this and this Cancel out Very good. So I can make my ah matrices using and let me start by writing this using First of all a is one b zero See is zero that&#x27;s going to give me one zero one zero a. Is one is negative. One negative, one negative one to So I&#x27;m just getting all of those with a equals zero comma. Let&#x27;s try another one This time A is zero B is one and C is zero 01 negative. 100 zero. Interesting. Um, b I&#x27;m looking for be Vero negative 11 cool. And the last one. This time I&#x27;m going to use a zero. B zero c is one 000 one zero negative one negative 10 negative one. And there we have a subset and there are more. All right, so this is not the only set of vectors. Um, I would say, though, that this is the minimum number of vectors. Three Ah, vectors. You can see when you look in the back of the book that Ah, it lists four vectors. Um, and that&#x27;s okay. Um, at first I was thinking, that doesn&#x27;t seem right to me. However, just because the minimum number of vectors that you need is three doesn&#x27;t mean that if you&#x27;re just asked for a set of actors, it doesn&#x27;t mean that you can&#x27;t write MAWR. So, um, we could just keep writing vectors if we want to dio um, nevertheless, this answer is correct, and you could also get the answer in the back of the book. And that is another set of vectors. I do notice one mistake in my work. Um, up here, remember, it was negative. A minus, bi. So negative a minus bi should go there um, so that&#x27;s actually gonna put a negative here, so they&#x27;re all gonna add up to zero. Otherwise, this is correct.

Problem

In Exercises 23 and $24,$ mark each statement Tru…

Problem 22 Medium Difficulty

Construct a $3 \times 3$ matrix $A,$ with nonzero entries, and a vector $\mathbf{b}$ in $\mathbb{R}^{3}$ such that $\mathbf{b}$ is not in the set spanned by the columns
of $A .$

Answer

$A = \left[ \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { 1 } & { 1 } \\ { 1 } & { 1 } & { 1 } \end{array} \right]$ and $b = \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 3 } \end{array} \right]$

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$A = \left[ \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { 1 } & { 1 } \\ { 1 } & { 1 } & { 1 } \end{array} \right]$ and $b = \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 3 } \end{array} \right]$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

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.

Heather Z.

Michael J.

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

Linear Algebra and Its Applications