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Question

[M] In Exercises $37-40$ , determine if the columns of the matrix span $\mathbb{R}^{4} .$
$$
\left[\begin{array}{rrrrr}{8} & {11} & {-6} & {-7} & {13} \ {-7} & {-8} & {5} & {6} & {-9} \ {11} & {7} & {-7} & {-9} & {-6} \ {-3} & {4} & {1} & {8} & {7}\end{array}ight]
$$

Runpeng Li · Accepted Answer

All right, so four problem 40 here were given anatomy tricks, which is off four by five. Milkshakes is a form. I find matrix. Let&#x27;s say, um, it&#x27;s naming as a matrix eight. So again, we need to find out the real reduced form off a I&#x27;ll just write it down. Reduced form off, eh? And this is 10 connective seven over 13 and 00 So we have 01 connected, connected to over 13 and zero and zero. They&#x27;re pro abi&#x27;s 0001 zero. That&#x27;s Rowe ISS 000 zero and one. Okay, so if we count the number of people positions, here&#x27;s one, two, three and four. So we can. So we have people for people columns in this case, Base matrix people columns. So that means rank off, eh? Is for butt and eighties equal to dimension far for so so we can conclude that columns of a it&#x27;s been our 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

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 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.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be answering problem number six from section 1.7, which from the book with Based on Later Independence, we&#x27;re given this. Ah, augmented They&#x27;re not I&#x27;m into Matron&#x27;s. But just the 1st 3 columns of this matrix is the matrix for the problem to the problem to start. And the promise says if to justify answer and to determine whether, um, this set of our this matrix as linearly independent for a matrix to be literally independent or for its columns to be literally independent, uh, the Matrix can own the matrix equation. X equals zero can only have the solution, the trivial solution, which is just this, in this case, just a column of zeros based on depending on how many variables it has. So every single variable has to be zero, which is a travel solution. So X equals zero is a matriculation that we learned section one point for and, uh, hey, here is just the given matrix. The Matrix were given, and zero Is this Colin Dobie Koncak cabinet to the end of the matrix. Ah, and this is how you set up in X equals zero matrix and you perform row operations elementary row operations till you get the staircase of zeros and one&#x27;s in, Um, the diagonal of the matrix. Um, this is the Matrix. This is the same Matrix Elementary, uh, that this is the same matrix as the previous matrix that we have here as elementary operations results in the same matrix, but with different values. So here, essentially, we&#x27;re just holding a system of equations. Since in this matrix, we have, ah, diagonal ones. And the right side is still zero after the elementary row operations element hero operations include subtracting arose one road from another or dividing are scaling it with ah factor, um, or switching rose. So so elementary row operation results in the same matrix, but with different values. So essentially, we&#x27;re just solving the same size is Miss essentially Alexa Living system. Since we have ones in these diagonals. And let&#x27;s just say we can label each of the columns to be a variable x y z, um, here, the one here shows us that Z equals zero and one y plus our why minus four Z equals zero. So why minus are so y equals zero sends you zero and X plus three z equals zero and says zero X is also zero. So we know that this right column is just indicative of, uh, the solution set. So we know that here the, um sorry. One sec. So we know that here, um, the solution column Vector X is just 000 where each each entry represents a different variables to the 1st 1 is X the second one&#x27;s a why and what their own see. So which is exactly like the trivial solution. Therefore, this matrix are the columns of these. This matrix are literally independent. I don&#x27;t show a lot of the math behind, um, performing elementary row operations because I won did this beforehand before the video started. And it like it&#x27;s a very long process so it wouldn&#x27;t fit in one slide. So if it&#x27;s a matrix larger than, like a three by three or three by four, uh, it would be best in your times interest toe to use like a matrix calculator to get it in, uh, like basic isil on former reduce the echelon form so that it makes the problem easier if your teachers

Michael Jacobsen · Answer

for this example were considering a four by three matrix egg displayed here at what we&#x27;re trying to assess is whether or not the columns of this particular matrix are linearly, independent or dependent. One way to tell is to use row reduction, and my first step is going to be to interchange these two rows. I want my first row to be one negative 32 and last road to be zero negative. 85 That way we get a pivot position here with already a value of one toe work with. Let me copy the next rose and we only need to go to echelon form as opposed to reduce echelon form Seimei. Next goal is going to be toe wipe out these two entries. Let&#x27;s do it on this road reduction step, Saul copy Row 11 negative three and two. Then our operation to eliminate this century will be to multiply row one by negative three and added to row two, we obtain altogether 02 and negative too. Then to eliminate this century, we can just add roll one to row three and record the result. It&#x27;ll be zero to and negative to once again then copy down Bro. 30 Negative eight and five for the next row operation. I&#x27;m going to do two steps first copy. Negative one negative three to the first row, then I want to bring this row into the second row position. So it zero negative 85 I want this to be the third row 02 and negative two and now we also know once we have a duplicate of a row, it can be effectively zeroed out or visualize multiplying the old row to buy negative one and adding it to the old row three we would obtain is your of Rose, which are putting into the fourth row position. Let&#x27;s next divide Row two or excuse me row three by two to obtain a one. Let&#x27;s also make that a row too. So we&#x27;ll have 01 and negative one in her second row. First rose one negative 32 and the third row was the old second row, which is now zero negative. Eight and five and we have a row of zeros down below. So from this pivot position here, the next objective is going to be eliminate the century. Then we&#x27;ll have our matrix in echelon form. So let me copy, rose one and two first, then Multiply, wrote to by eight at the result to Row three and will obtain 00 and negative three. And now we see that there&#x27;s a pivot position in the first column, second column and third column. This that implies that that matrix equation X equals zero has no free variables. But we know in this case that means that the columns of a are linearly independent. If we had found even just a single free variable, we would have said the columns of a would be linearly dependent, and that&#x27;s because this matrix equation would have had nontrivial solutions.

Problem

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

Problem 40 Hard Difficulty

[M] In Exercises $37-40$ , determine if the columns of the matrix span $\mathbb{R}^{4} .$
$$
\left[\begin{array}{rrrrr}{8} & {11} & {-6} & {-7} & {13} \\ {-7} & {-8} & {5} & {6} & {-9} \\ {11} & {7} & {-7} & {-9} & {-6} \\ {-3} & {4} & {1} & {8} & {7}\end{array}\right]
$$

Answer

Matrix has pivot element in every row hence it spans in $R^{4}$

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

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Matrix has pivot element in every row hence it spans in $R^{4}$

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

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

Kayleah T.

Caleb E.

Maria G.

Michael J.

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

Linear Algebra and Its Applications