construct-a-nonzero-3-times-3-matrix-a-and-a-nonzero-vector-mathbfb-such-that-mathbfb-is-in-mathrmco

Question

Construct a nonzero $3 	imes 3$ matrix $A$ and a nonzero vector $\mathbf{b}$ such that $\mathbf{b}$ is in $\mathrm{Col} A,$ but $\mathbf{b}$ is not the same as any one of the columns of $A .$

Melvin Adkins · Accepted Answer

in this problem, we want to use a three by three matrix A. So that be is in the column space of a but be is not the same as any of the columns of our Matrix. A. So one way to go about this would be to make a the identity matrix, and we know that the columns of a are linearly independent and the column space of a is the span of the column vectors of a And the span of these three column vectors is actually equal to our three, because any vector in our three can be written as a linear combination of these three vectors thes three column vectors. So now we can choose any B that&#x27;s in our three as long as it&#x27;s non zero and also a long as it doesn&#x27;t match any of the columns of a so we could let be, be equal to the Vector 111 And that linear combination is just the first column vector of a plus. The second column Vector of a plus. The third column vector of a. So this shows that B is in a column space of a and clearly be is not equal to, uh, any of the column vectors of a although it can be. But in this problem, we wanted to choose B, so that s o that it&#x27;s not. And that completes the problem.

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

Runpeng Li · 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.

Mike Verwer · Answer

Okay, So this question is asking us to construct a matrix B so that the product a Times B will equal to zero and so far first represent the products eight times. Be using the column vector definition so that this matrix will be a time of the first column of B one. Uh, not plus sorry. And then the other second column will simply be eight times the second column Vector of B. You want this equal zero. And so maybe one then will be equal to three minus six minus one two, the 11 e 12 And this new sequel zero. And so this resulting multiplication will provide us with the following two equations. Three B 11 minus six. Be one, too, is equal to zero and two minus B 11 plus to be one too equal zero. And so what we do here is we&#x27;re going to solve one of these equations for one of the variables, and I&#x27;m going to solve the second equation for the variable B 11 We&#x27;ll isolate that. I&#x27;ll do that over here. So what we get is be 11 is equal to to be one too. So now what I&#x27;ll do is I&#x27;ll sub this value in for B 11 into the other equation into equation one here. Every time I see a B 11 and here I&#x27;ll substitute this value to be one, too. And so what we get is that three times to be one, too is equal or minus. Uh, 61 too, is equal to zero, and I&#x27;m going to factor it to be one two&#x27;s here so that we get be one, too is equal to six. Minus six equals zero. Our time. Six months, six equals zero. But this is always true. It&#x27;s kind of trivial. E true. And so what we say is that there are infinitely many solutions for this equation, and what we&#x27;ll do then is we&#x27;ll introduce a dummy variable will say, Let&#x27;s be one too equal T He&#x27;s just a dummy variable that we create For the purpose of this question. It could be any number and then since be want to is equal to tea and be 11 is equal to two times beauty. One, too. It follows them that be 11 is equal to two tea, and this calculation will be exactly the same for eight times the column. Vector be, too. And so by symmetry, we&#x27;ll introduce a different dummy variable. But we&#x27;ll get a similar result. We&#x27;ll find that B 22 will be equal to the dummy variables s and then by symmetry again, beats who won will be equal that to us. And so are matrix be here Then will be the following to t t in the first column two s as in the second column. Now, to get an explicit answer, uh, that the question wants weaken. Just so been any value that we want for S and T. So I&#x27;m going to let t equal one and ass equal to so that be then becomes 21 in the first column and for two in the second column. And you can confirm that this matrix will indeed produce zero when you multiply it with a just as you asked

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

Okay, um, for this problem work even. Ah, zero matrix. Say three by three matrix. Let&#x27;s say, is a 11 A 12 a 13 and a 22 8 to 1. Uh, sorry. Should be. Should be a 8 to 1 here. And 8 to 2. And a 23 and a 31 A 32 and a 33 Now we take the matrix multiplication, which is given Vector. Which one? Negative. Two and one. So that is a 11 minus 2 82 A 12 plus a +13 And the second entry is a 21 minus two. A 22 plus a +23 And the third entry will be similar is a 31 minus two a three to us A 33 And it has to be zero. So that means these three entries has to be all the same. And it has to be zero. So we can just take a 11 and just take all the entries A i g for where i and j our speaker or equal to one smaller equal to three. And of course they are N j has to be intruders. So it&#x27;s a I j t take all a i g to be one. So now a 11 This one minus two plus one So that zero and we should shower for centuries. You&#x27;re here and the same is the one minus two. What? Plus one and zero on their entry will be one minus two plus wine list again. Zero. So here a I j t a i g to be one that It&#x27;s all one matrix. This is our, uh this is our nontrivial solution here.

Problem

Construct a nonzero $3 \times 3$ matrix $A$ and a…

Problem 27 Easy Difficulty

Construct a nonzero $3 \times 3$ matrix $A$ and a nonzero vector $\mathbf{b}$ such that $\mathbf{b}$ is in $\mathrm{Col} A,$ but $\mathbf{b}$ is not the same as any one of the columns of $A .$

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Linear Algebra and Its Applications

Grace H.

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Absolute Value - Example 1

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

Grace H.

Anna Marie V.

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Michael J.

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

Linear Algebra and Its Applications