suppose-the-first-two-columns-by-and-mathbfb_2-of-b-are-equal-what-can-you-say-about-the-columns-of-

Name: suppose the first two columns by and mathbfb_2 of b are equal what can you say about the columns of
Uploaded: 2019-07-31
Duration: 3 min 19 s
Channel: Numerade
Description: Suppose the first two columns, by and $\mathbf{b}_{2},$ of $B$ are equal. What can you say about the columns of $A B$ (if $A B$ is defined)? Why?

Question

Suppose the first two columns, by and $\mathbf{b}_{2},$ of $B$ are equal. What can you say about the columns of $A B$ (if $A B$ is defined)? Why?

Mike Verwer · Accepted Answer

Okay, so this question is asking us. Suppose the 1st 2 columns of B R. The equal are the same. What can you say about the columns of a Times B? And we&#x27;re going to use the fact that each column of a Times B is a linear combination of the columns of a using the corresponding column of B as weights. So what that means is, let&#x27;s say that a be sub one is the first column of the Matrix a B, and that&#x27;s gonna be equal to first column of a plus second column of a plus third column of a and so on using the entries of the first call of a B. So be 11 be to one be 31 and so on. So the the column vectors of a here are acting as sort of the variables that we&#x27;re including in our linear combination, and then the the weights of the linear combination are the entries of the corresponding column of ah of be, like notice how this column, the number representing the column matches the column where we&#x27;re working with the whole way through. And so the 2nd 1 will be the same. So the second column of a B will be equal to well, well, Instead, we&#x27;ll be using the second column entries of Be So this will be Be one, too. Ah, a one plus B 22 a two plus B +32 a three. Right and so on. But if the first column of B is equal to the second call, maybe then, um, be I One is equal to B I to is equal to just be so. I will just call it. We&#x27;ll name it will name them all be so bye for I want to end if he is an n by n majors or as an collins at least. But that means then that these two mate, uh, the two columns a B one in 82 are equal, right? Let me just write them both out, all right? They&#x27;ll just be able to be one times the first column Vector of a plus B two times The first column vector of a second column vector of a plus B three times the third column vector of a and so on. So these two columns will be equal, and that&#x27;s what we can say. But the columns of baby

Lucas Finney · Answer

19. Here is a pretty quick one. So we are given are so we have, um, Matrix B, where we know that its third column is the sum of its first two columns. And we want to know What can we say about matrix A B So matrix a B going to be a Times B one b two b one plus B two. That&#x27;s the third call with the way that matrix multiplication works, we would get each column vector by multiplying each column by Matrix A. So we get a B one a B two than a Times B one plus B two. Now, if we look at it, we should be able to recognize that for the Matrix A B just like Matrix B. Its third column is the sum of its 1st and 2nd Collins. So pretty quick and straightforward questions

Mike Verwer · Answer

Okay, so here were asked to comment on the product. A Times B If the second column of B is entirely zero, and we&#x27;re gonna use the fact that each column of the product A B is a linear combination of the columns of a using the corresponding column of B as weights. So what that means is, if I write a be sub to here, meaning the second column of the Matrix AIDS has B that&#x27;s gonna be equal to the first column. Obey. Plus, um, the second column of a plus, the third going away and so on using be, ah, throw one calling, too, and then be to to and be 33 year 232 as weights. So the call in vectors of a here are what we&#x27;re taking the letter combination or the linear combination of and the elements of the second column of B R. What we use is the weights in that linear combination, and what that means is, if b I, too, is equal to zero for ah, I going from one way to end, then that means that this a B two will also be entirely zero right, because we&#x27;re multiplying it by zero each time for each. Each column vector then will be multiplied by zero, and we&#x27;ll get Aaaah this year a vector.

Patrick Burns · Answer

so we know with matrix multiplication that the number of columns and our first Matrix must equal to must be equal to the number of rows in our second matrix. So you just wayto explain this is to give two matrices so we&#x27;ll just do a two by two matrix and we&#x27;ll say we&#x27;re multiplying it by another. Or let&#x27;s do a matrix with one column but to rose. So we should still be able to multiply these matrices because we have does it same number of rows and be as we do columns and a so just to run through it, that look like again, even multiply it. This the first row times this column and be our only column, and then that will give us our first position in our resulting matrix. Then we do the next row, same column and B that will give us the next one. So that word. Now let&#x27;s say that be was instead a matrix with two columns in a single row. Now, if we try to do matrix multiplication, we&#x27;re going to start in our first row of a You multiply it by the first column, so we&#x27;re going to have two times one is one we&#x27;re sorry to plus to from Matrix a times. But then we have nothing to multiply it by because we&#x27;re still in the first column of B. However, we only have one value here and so we can see that matrix multiplication is now impossible. So that is why we must have an equal number of count Earned are a number of columns in Rome matrix a must equal or number of rows in matrix B.

Michael Jacobsen · Answer

in this example we have two major sees were considering a is of size and buy em B is of size and by P. We also have set the notation for the Matrix B so that columns are B one B two all the way through Bisa p With this information, if we were to say multiply a Times B, then by the definition of matrix multiplication, we would have a Times B one for the first column of the product. The second column is a Times B two all the way through a times B sub p for the last column of the product here. Let&#x27;s suppose that a Times B s A P equals the zero vector and BCP itself is not the zero vector. This is an interesting statement, since a BCP is not a zero vector and we have a homogeneous equation we&#x27;re considering. We can make an immediate conclusion here. This implies directly that the columns of the Matrix A are linearly dependent, and this is really through the definition of linear dependence. We have here an equation that could be converted into a vector equation, and we have a set of scale er&#x27;s that come from Bisa P Not all equal to zero since we said this is not the zero vector And in this multiplication we have a linear combination set equal to zero with a nontrivial solution immediately calm survey are linearly dependent.

Ankit Gupta · Answer

here we have a metrics, a off the order three close to and a metrics. Be off the order. Do a cross to and we need to find that. Is it possible that AEG was to be than a equals? Toby is not possible. It is only possible when the mattresses are off. Same dimensions. So here, because A and B are off different dimensions, so it is not possible. Secondly, we are being asked that we have toe little mind whether a plus B is defined. So a plus B is also not defined because they are off different dimensions to aid two mattresses, both the mattresses should be off same on dimensions. Next, we are being asked whether a B is defined. So we have a off the order t close to on be off the order to close to. And we can observe that a number of rules in a number of columns in a question number off rules in B, we have a number off columns in a It was too number off rules and be and hands A B is a defined on dimensions. Off a B will be equal to three cross to

Problem

Suppose the third column of $B$ is the sum of the…

Problem 18 Medium Difficulty

Suppose the first two columns, by and $\mathbf{b}_{2},$ of $B$ are equal. What can you say about the columns of $A B$ (if $A B$ is defined)? Why?

Answer

if the first two columns of matrix $B$ are equal then the columns of $A B$ are equal.

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

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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if the first two columns of matrix $B$ are equal then the columns of $A B$ are equal.

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Alayna H.

Maria G.

Michael J.

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

Linear Algebra and Its Applications