suppose-a-is-a-3-times-3-matrix-and-y-is-a-vector-in-mathbbr3-such-that-the-equation-a-mathbfxmathbf

Question

Suppose $A$ is a $3 	imes 3$ matrix and $y$ is a vector in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{y}$ does not have a solution. Does there exist a vector $\mathbf{z}$ in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{z}$ has a unique solution? Discuss.

Runpeng Li · Accepted Answer

problem. We&#x27;re given the Matrix three by three matrix and the vector y and such data a eight times acts. It was why has no solution. And meanwhile, we want to We want to figure out that whether there exists a vector Z in our three such that the equation the X people see has a unique social. All right, so first, let&#x27;s consider the each of the reduce station of form of eight. So let&#x27;s say a first. It has to be three by three matrix. Now consider reduced agent for reduce station form should be one the erosion of you needed by one and with some stuff over here and zero over here, so Oh, I would just leave this part blank. All right, so this is our eight, and we&#x27;re considering system a axe. It was why. So this is Tom. This system here must have a solution right here. Because no matter which, why we&#x27;re taking here, we have already writing a new page. So consider Hey, X, it was why No. Put a here, all zeros and some stuff and, uh, yeah, no bearable. Here. Victor ags X Y x two x three And why want y two by three. So I should be white too. And the white three. Now, remember that, uh, this system must have a have a solution, because we we have right here in the meadow, no matter which, which, which is why we&#x27;re taking We can always take X three equals flight three and next to can be determined by X three and X one can be determined by x two and next three. So we&#x27;re done. So we have to make he ordered to make this system none consistent. We have to make it, at least warned. Roll all zeros. So let&#x27;s say in this case, we have the last rule, all zeros, and this should be one. This is zero, and we have some stuff over here. Uh, say I&#x27;ll just draw like this. All right, so right now we consider a x equal. See, So we need to make this system consistent, but you see from here that we can it&#x27;s impossible to have a unique access here because we have who are taking. We&#x27;re taking a Z. That is, that makes this system consistent. We have to make the last entry zero right now. We have again. We have to, uh, three unknowns, which is X one plus some cooperation. That&#x27;s a a 13 times x three because Z y and next two plus a 23 times eight x three will see to. Now we have here x one, the person 1st 1st on noon on X to the second unknown Onda X three, the third unknown. And we only have two you wishes here. So that means this this system has infinite many solutions. So it it&#x27;s just impossible to have a unique, unique solution in this case. So that means that means we cannot. That means we cannot have.

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.

Runpeng Li · Answer

All right, so in problem 34 suppose A&#x27;s of three by three matrix and A&#x27;s B&#x27;s a vector in our three with the property. That Exodus p D has a unique solution. And we want to explain why the column Self Bay must spend our three. So Okay, so in this case, since we know the leaner system they X equals B is has a unique solution. So that means if we consider the or commended matrix and furthermore, if we considered the rule reduced form off this matrix, it must has has to ruin form like this 81 1 a two to something on the diagonal, diagnose and have some stuff there some stuff here. But it doesn&#x27;t matter. What are they? And we have zeros here, and I&#x27;m on the right hand side. We have B one b two B three as a constant. Okay, So this is the only impossible for this system to have a unique solution because only in this way we can solve a unique by the By the last row, we can solve the unique, unique x three, and furthermore that from the second road that X two will be determined by X three and x one would be determined by X two and x three. So this the only impossible way that we can have a unique solution for this system. So that means by observing this matrix, we have three people, two columns and three people. Columns means, uh, means axes. Um, yeah. So three people columns actually gives me, gives us the information that all the columns because A said three by three matrix And since every every single column is, is that people column So these columns are linearly independent, very independent, and furthermore, they must spend our three because in this case, we actually have three vectors and our three estimation three and those three Rector&#x27;s Arlene Early independent so thes three directors must spend our free so

Michael Jacobsen · Answer

in this example we&#x27;re dealing with the three by 30 matrix A that&#x27;s provided here for such a matrix. We can ask interesting questions such as if we were to solve a matrix equation eight times X equals, say, zero vector. What would be the solution set to such a system? Well, this is kind of a strange deal, since our Matrix A has nothing but zero entries. So if we go to the augmented matrix for this system, then it would have this form. First. I&#x27;m inserting a as green. So there&#x27;s a three by 30 matrix and I&#x27;ll use blue for the augmented portion. So there it is. This is the augmented matrix for eight times X equals zero. Interestingly, the variables are x one x two x three, and we have that x one x two x three are all free variables, since if a is zero matrix, there is just no pivot columns, and that makes all variables free in this situation. Well, in this case, if we were to describe the solution sets to eight times X equals zero matrix, the solution would be X, which is equal to x one x two x three and if we write this in Parametric Victor form, it takes on the form of X one times 100 plus x two times 010 plus x three The last three free variable time +001 where x one x two and x three are in our the set of real numbers. So this is the solution set in Parametric Vector form. But possibly this is just a descriptive f x one x two x three could be any real number whatsoever. And this is the solution, then the solution set to our matrix equation is Theo. Entire set are three. So that&#x27;s the solution. Sets to a X equals zero when we&#x27;re dealing with the zero matrix.

Michael Jacobsen · Answer

in this video, we start out with the four by three matrix, and we&#x27;re going to consider the equation. X equals B. Let&#x27;s start off by making an assumption. Let&#x27;s suppose that the Matrix Equation A X equals B has a unique solution for some vector be in our for So let&#x27;s next. Consider what that would specifically mean for this matrix. Well, it would imply that in the equation, X equals B If we have a unique solution, there are no free variables, so start there. So that&#x27;s last line implies that there are no free variables. How do we know this? Well, if there was a single free variable, we would have infinitely many solutions, not a unique one. So next, knowing that there are no free variables that implies the following. Ah, free variable will always come from a non pivot column of The Matrix, say, so we know that A has a pivot in all Collins, and there&#x27;s three com specifically. So let&#x27;s say all three columns squeezing the three in there. Sophie has a pivot in all three columns, but it is of size four by three. This is our next immediate conclusion. I will say that a must be equal to or rather not equal but row equivalents to a four by three matrix. That would look like this so we can have 100 now. Calm One is a pivot column for row to go 010 Now we have to pivot columns, but we require three. So the third row will be 001 and we&#x27;re looking at the reduced row echelon form, so their final row will be all zeros. So given just this minimal set of information, we now know that the role reduced echelon form of a takes on this form.

Problem

Let $A$ be an $m \times n$ matrix and let $\mathb…

Problem 38 Hard Difficulty

Suppose $A$ is a $3 \times 3$ matrix and $y$ is a vector in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{y}$ does not have a solution. Does there exist a vector $\mathbf{z}$ in $\mathbb{R}^{3}$ such that the equation $A \mathbf{x}=\mathbf{z}$ has a unique solution? Discuss.

Answer

No

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

Linear Algebra and Its Applications

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Maria G.

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Michael J.

Absolute Value - Example 1

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Linear Algebra and Its Applications

Heather Z.

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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

Heather Z.

Maria G.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications