let-a-be-a-3-times-2-matrix-explain-why-the-equation-a-mathbfxmathbfb-cannot-be-consistent-for-all-m

Question

Let $A$ be a $3 	imes 2$ matrix. Explain why the equation $A \mathbf{x}=\mathbf{b}$ cannot be consistent for all $\mathbf{b}$ in $\mathbb{R}^{3}$ . Generalize your argument to the case of an arbitrary $A$ with more rows than columns.

Michael Jacobsen · Accepted Answer

in this video, we&#x27;re starting off with a matrix. A. It&#x27;s going to be of size three by two. So what we want to do here is to consider the Matrix equation eight times X equals a Vector B. Now he has chosen arbitrarily on the right hand side. So a big question about this equation is whether it is consistent or not. That&#x27;s hard to do when we don&#x27;t know what The Matrix A is specifically but will use the size of the matrix toe. Analyze the pivots. Let&#x27;s start by imagining what a could look like. It&#x27;s going to be of size three by two. So I want to write something that has three columns and or excuse me three rows and two columns, so they&#x27;ll be entry here, here and here. I&#x27;m just going to put little dots here to give them blink arbitrary entries as well as entries here, here and here. Now, the best case for this Matrix is that we have a pivot in the first column, then in the second column. But to get any more pivots, we would require 1/3 column. So there is always going to be a row if we&#x27;re dealing with three by two where there is no pivots. So this matrix a has cannot have a pivot in every row, so a cannot have a pivot in every row. What this tells us, then, is that a X equals B is not consistent for all vectors. Be in our be sure to use the three here. So for all vectors, be in our three. It doesn&#x27;t tell us that a X equals B is always inconsistent. What it does tell us is that we can find a vector B for which it will not be consistent. No, we can ask ourselves why did this happen? Suppose a is and m by N matrix. Our matrix was a size three by two. And if m is bigger than n so that we have more roses than columns just like in our case, then we will know immediately that there cannot be a pivot in every row and we can generalize even further to say that for such a situation, a X equals B will not be consistent. These air s is sorry for every vector Be in are now We have to be careful. It would be always this dimension First, just like we picked this dimension. So in our M and that generalizes the three by two situation to an end by N one where specifically their arm or rose than columns.

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

Mike Verwer · Answer

Okay, So this question we have the assumption that eight times D is equal to the M by M identity matrix, right? A D is equal to ay so them. You need to show that for every our size column, vector be a X equals B has a solution. Okay, so we&#x27;re gonna do this. We&#x27;re gonna start with a tee time eight times equals I am, and we&#x27;re going to multiply on the left. Ah, a column Vector B that will give a d Times be is equal to be And we&#x27;re gonna do a little bit of size. And Alex is here. So a is, um m by n matrix, which means that de must be on end by m matrix since their products is results in m by m okay, and if be, since B is a ah, and by one matrix, right, it&#x27;s a column vector. It&#x27;s an M size column vector. That means that we can multiply with D or Deacon, be multiplied by be right. The product, um d b is defined, and it will be on n by one vector. Okay, so that means that we can use the associate of law on this equation and we can turn it into a times d B equals B. And now all we do is we say, Let&#x27;s just let de times be the end by one column vector. Well, just give it the name X, and then a X equals B as desired. And this column vector here d b will be the solution. Okay, So the second question is, why can&#x27;t we have Why can&#x27;t a have more rose than columns? And to show this or to explain this, I&#x27;m going to I think about the equation X equals B in terms of an actual, um, collection of linear equations you&#x27;re tryingto like Solve a win your system of equations And if a A&#x27;s and M by n right, if mm is larger than and if there are more rows and columns than in our linear system of equations, that means that their arm or equations and there are unknowns, right? It would be something of a form like ah, like X minus y equals two. And then your other equation could be two x plus y equals one, and then 1/3. It could be, you know, five x minus. For why is equal to minus four or whatever, right? Just ah, um a system of linear equations. And to solve this, what you do right is you would just do it by hand without even So you have matrices. You would you would pick on equation, right? Maybe this equation here, I&#x27;ll do a different color. Yeah. You say you pick equation here, you isolate you, pick, you pick an equation, you pick an unknown. You isolate this equation for that unknown, and this becomes, you know, X equals two. Plus why, right? And then you sub this, you pick another equation, right? Maybe you picked this equation and you sub this new value of X into that equation, right? And then you are resulting with just two times two plus why, plus y equals one. Right? So this is just a one variable equation which you can just solve. He can isolate for why? And you get a number out, right? You get why equal some some number. And then you said this number back into into this equation here, right? And then you get your solution right. X is equal to two. Plus, you know, whatever Y is equal to and there&#x27;s your solution. What the issue is, we didn&#x27;t use this equation at all. So the solution that we get, you know, this X Y solution that we get we have no guarantee that the solution will be consistent with this equation, because we never used it to find the solution in the first place, right? It was never touched. So that&#x27;s why we can&#x27;t have more rose than there are unknowns. Because we can&#x27;t be sure that that a solution for it would be consistent with every equation involved in the linear system. Okay, hopeful that makes sense.

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.

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.

Michael Jacobsen · Answer

in this video, we have a three by four matrix, and we have certain assumptions that are going to be placed for this matrix. First we have that the Matrix equations that eight times X one equals white one, and at the same time a Times X two equals White, too, are consistent. Were we picked vectors white, one in white to to come from our for the number of columns in our Matrix. So now that we know that these air to given to be consistent systems, let&#x27;s also let a new vector w b equal to why one plus white, too. Then let&#x27;s also consider a new system of equations a X equals W. So what are the questions here? The question is to determine whether or not a X equals W is consistent or inconsistent. Well, by what we have so far, it turns out that the system is consistent, but we need to next explain just how we know it&#x27;s consistent. Well, we can prove most directly that X equals W is a consistent system. The moment we come up with a solution for X, let&#x27;s do the following next for our next step, let&#x27;s let X be equal to X one plus x two. If we let the Vector X represent that, then my claim is we have now a solution. Since if we multiply a with this vector X, we should be able to get to this vector w But we have to show it. So it&#x27;s right that as our next step I will say then eight times X will be equal to. And here&#x27;s where we make our first substitution. We know that X is given to be x one plus x two. So it&#x27;s right that in they&#x27;ll be eight times x one plus x two. Then this turns out to be the step of our work here. That is the key to showing we have a consistent system and that we even have a solution given by Ex. That key is that a matrix Times a vector satisfies a distributive law where we&#x27;ll have here eight times x one plus eight times x two Notice that now that we&#x27;ve made this matrix multiplication through the group distributing, we now have two places where we can substitute eight times x one and eight times x two R Y one and y two, respectively. So this equals Why one plus. Why too? Well, we&#x27;re not there, Remember, we need to get to the vector W to prove that we have a solution and the system is consistent. But that&#x27;s where we go to this substitution. Why? One plus white to is W. And so we have shown that this is system is consistent and even mawr. We&#x27;ve shown that this vector X is the solution. And really, the whole key to this was this step right here that a matrix can distribute into a vector some

Problem

Could a set of three vectors in $\mathbb{R}^{4}$ …

Problem 31 Hard Difficulty

Let $A$ be a $3 \times 2$ matrix. Explain why the equation $A \mathbf{x}=\mathbf{b}$ cannot be consistent for all $\mathbf{b}$ in $\mathbb{R}^{3}$ . Generalize your argument to the case of an arbitrary $A$ with more rows than columns.

Answer

In general, when matrix is $m \times n$ matrix and $m>n$ (more rows than columns) it can span $\mathbb{R}^{n}$ and not $\mathbb{R}^{m}$ because it can have at most $n$ pivot columns.

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

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

Absolute Value - Example 1

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In general, when matrix is $m \times n$ matrix and $m>n$ (more rows than columns) it can span $\mathbb{R}^{n}$ and not $\mathbb{R}^{m}$ because it can have at most $n$ pivot columns.

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications