mathbfm-find-a-column-of-the-matrix-in-exercise-39-that-can-be-deleted-and-yet-have-the-remaining-ma

Name: mathbfm find a column of the matrix in exercise 39 that can be deleted and yet have the remaining ma
Uploaded: 2020-07-12
Duration: 2 min 9 s
Channel: Numerade
Description: $[\mathbf{M}]$ Find a column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns still span $\mathbb{R}^{4} .$

Question

$[\mathbf{M}]$ Find a column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns still span $\mathbb{R}^{4} .$

Michael Jacobsen · Accepted Answer

for this example were considering a matrix A that&#x27;s displayed here and using a computer algebra system. We haven&#x27;t row reduced to its role, reduce echelon form. Where comes 123 and five are Pivot Collins. Here we find that all rows of a are pivot rose, and this means the following Since every row of a contains a pivot, this means that the span of the columns of A is our four itself. So take each one of these column vectors inside the Matrix. A. If you spend this set, you&#x27;ll find the set is identically are for Well, this is interesting because The Matrix A has five columns, and all we needed was a pivot and every row in order to spend all of our four. So now the question becomes, Could we delete a column and still have a pivot in every row or another? White. In other words, which column can we delete so that a still spans all of our four? To answer such a question, we have to go to the non pivot column. So this is not a pivot column and notice that if we did delete this column outright so that we have a new matrix. Let&#x27;s call it a hat where this column is missing. A hat still has, ah, pivot in every column, and so a hat still spans that target space. All of our four. So deleting the fourth column has no effect on what the span will be since we still have a pivot in every row.

Runpeng Li · Answer

Okay, so four problem for two. We need to consider the Matrix that we did before, and and we want to. I want to find a column of this matrix that it can be doing. Did it end? Have the remaining matrix columns deal? Spend our four. So instead of kind of considered the original Mitrice, I would recommend to just to consider the real reduced form off the previous matrix, which is I&#x27;ll just write again. 10 connective Seven over 13 and 00 zero juan 01 connective to over 13 and 00 and 0001 zero Andi 00001 So if we count the people columns, this is one and this is another one. And here&#x27;s to the odd one. And this the other one. So the calm that we can to delete in this case is Oh, you&#x27;re not a color. Is this column because this is another people economies. If we delete this column, then that will not affect hourly new independence off these four pictures. So that means call him. Three can deleted so that these poor rector&#x27;s steel for this four columns were still span are for. And another question is that can delete wanted one calm, because Thea answer is no. Because we have. For we only have four people columns. And if we wait, we have five columns in total way. Have deleted one column that it&#x27;s not people, columns and any any more columns. If we deleted, that will be people columns. And that will not be that will not be allowed in this case. Because if we delete one when people calm said we only have three people comes that means Sorry. Uh, we don&#x27;t have three people of columns. Three people call. So you, Stephanie not enough cannot cannot spend our are for But you can&#x27;t spend our three but not possible to spend our four

Subham Jyoti Mishra · Answer

So the question We heard this matrix here despite by seven murders married even by performing rural presents that reduced the road transform off This metrics will look like 10 13 to 050 19 street 178 to you know 01 day 202 See those Those little one minus 11. Very cool. His heroes seven his seat. Placido. Placido, Placido Mhm 11 And then we have 000 So this is the reduced public home off my taxpayer now. Even hospital construct dramatics. See, it&#x27;s there. The bottoms of C five er problems off here now that by what columns off here. So these are private elements. Is this by work? This is by book since this pilot, unless mhm So. 12 or and six. These are the problems. So they respond. Name columns off er when on this nine 53 in this 46 Honest to Unless by then we have five minutes seven by two. We have ministry by in this room in this room. And then we have six minutes, kid movement and he have another Metrics are which consists. So non zero rolls off radius Rockland home. So these are the rules. 10 13 right? His hero. 5093 You know, with them were to settle one way to spirit 0001 Unless they&#x27;re very 207 0000 See you 11 How? Performing murder pickets. You know, thes mattresses. We get seem to carry our When you do that, we hear the no tricks. Yeah, this is the final results. We often doctor off calling murdered because of CR. Let&#x27;s see, Has columns from column space here on has who&#x27;s from those base, Okay.

Runpeng Li · Answer

all right. So in problem 38 were given this matrix. Let&#x27;s say a the first row is five. Next, seven, Victor four and NYU and second roll. We will be six and activate next. 75 There were over before Nick for an extra night connective night, and the last rule will be next to 9 11 16 5 So this is a four by four matrix on. We need to determine if the columns have matrix span are for so the first thing the first thing we need to do is actually count the number off people columns in this matrix. If actually the people columns are linearly dependent. If the the number of people columns equals equals four, then that means the column so off the basic span are for so well. First, I just I&#x27;ll just write down the road, reduced to form off this matrix because it&#x27;s, uh, it&#x27;s kind of hard to calculating by hand. If you are familiar with the company&#x27;s reputation, you can just use your prime Alva or programming skills to find that reduce station form. If you&#x27;re not from here, I would recommend to just do this by hand and check your compare on your solution with me. And that will help you to, uh, get through the gate familiar with caution. Caution. Elimination. Okay, so the reduced lesion form will be one one one and 7/3 and negative 5/3 and for third. Other terms are zeros. So if we count the people positions here is one. Here&#x27;s to three. So we have three people columns, which is less than the dimension off our three. And that&#x27;s a same time you you want you can at the information that three people calls me his rank off, eh? Rank of a is three, but which is smaller, Definitely smaller than they even dimension off for. Sorry. Should be dimension off are for so the rank of A is three smaller than the dimension of our war, which is four. So that means columns off a do not spend are four

Michael Jacobsen · Answer

in this video before by five Matrix has been provided with fairly difficult numbers to work with, as we see here. So for this exercise, our goal is to use a computer algebra system. In order to reduce this matrix, we need to reduce it in order to determine where the pivot positions will be. So, using Mathematica, we find that the rule Reducto reduced form for this matrix has a pivot and calm 123 as well as in the Fifth Column. But notably if we think about this in terms of Rose, we now know that the Matrix A has a pivot in every row. It may not have a pivot never column, but certainly if we go row 123 and four, we see it. It does have a pivot and every row. But knowing that much information tells us about the columns of a since there&#x27;s a pivot and every row, this implies that the columns of a will spend are four. We use four because that&#x27;s equal to the number of rows. It would not spend our five. The number of columns

Michael Jacobsen · Answer

in this exercise, we want to check if the columns of the Matrix say can span all of the space are for first notice that this matrix A is of size four by four, so it&#x27;s possible that a could spend all of our four. But in order for that to happen, there would have to be a pivot in every single row, as indicated yellow. So if a size four by four and if a has a pivot in every row, then the columns of a will span the entire space are for so what we need to do is roll reduce this matrix, But this is a tricky matrix to work with, since the numbers are not very heavily rigged here. So this is an exercise that&#x27;s intended to be done in a computer algebra system such as Math Lab or Mathematica. Using one of those computer algebra systems ever reduced the Matrix A into the falling form. So here, after the row reduction first noticed that this would have been a lot of work if we did their arithmetic by hand. But next we find that there is a pivot and row one row to andro three So the most critical thing to notice here is that a does not have a pivot in every row. So what this means, then, is that the columns of a cannot spend Oh, off the space are for we would have to have a pivot in this position as well.

Problem

$[\mathbf{M}]$ Find a column of the matrix in Exe…

Problem 41 Hard Difficulty

$[\mathbf{M}]$ Find a column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns still span $\mathbb{R}^{4} .$

Answer

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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.

Kristen K.

Absolute Value - Example 1

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Linear Algebra and Its Applications

Grace H.

Anna Marie V.

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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.

Kayleah T.

Kristen K.

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

Linear Algebra and Its Applications