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Question

In Exercises 24–26, use determinants to decide if the set of vectors is linearly independent.
$$
\left[\begin{array}{r}{7} \ {-4} \ {-6}\end{array}ight],\left[\begin{array}{r}{-8} \ {5} \ {7}\end{array}ight],\left[\begin{array}{r}{7} \ {0} \ {-5}\end{array}ight]
$$

Chris Trentman · Accepted Answer

were given a set of vectors and were asked He&#x27;s determinants to decide if this set of vectors is linearly. Independent vectors are seven negative. Four negative six negative 857 and 70 Negative five. Well, to determine if these vectors are linearly independent, consider The Matrix seven. With columns seven Negative four Negative six eight Negative 857 and 70 Negative five. So this is the Matrix whose columns are these vectors and let&#x27;s evaluate the determinant. So there are few from ways to do this. Um, I think the simplest way here is just going to be a straight up expansion. So expanding across the third column, we have our co factors seven times. Determinant of the Matrix and negative four five negative 67 plus co. Factor negative five times the determinant of the matrix. Seven. Negative eight Negative four five In evaluating. This gives us seven times this is negative. 28 plus 30 is too plus negative five times and then we have 35 minus 32 is three so 14, minus 15 or negative one. And this is, of course, non zero. And since the determinant of this matrix is non zero. And this is a matrix whose column vectors are vectors. Well, the columns of our matrix form a linearly independent set and therefore are vectors are linearly independent.

James Chok · Answer

Okay, so we will use determinants to see if these three sets of electors are the newly independence. So we just simply take the tenement 6462 minus seven. There are seven minus three minus five. So the first thing that we do is perform right operations, So we&#x27;ll take grow one and add that to grow. Three. Let&#x27;s put that in the right through. So then this is equal. This will be equal to four minus seven months. Three, 60 minus five, four plus two gives you 67 minus seven. Give you 02 minus three gives you minus one soco factor expanding along second column. We need the signs. This is positive and negative. So we negative negative seven. So that politics seven plans by its six minus 56 amorous one, which is equal to seven times with six times on modest one gives you 96 minus by six times went native file, which is needed to 30 which is seven time by 36 which is seven times by 24. Now we just know that this is not equal to zero. And so therefore the vectors are linearly independence

James Chok · Answer

okay, in this question you want to use determines to see if these sets of vectors are linearly independent. So first thing that we do is just put the vectors in a determines like this. Now we&#x27;ll just perfect expand by the lost column. So we&#x27;re taking our science positive. Negative politics. Negative, positive, negative and positive. So this is equal to negative, too. Time by 3 to 2 by minus six months. One 96 is there on three? Yeah, we&#x27;ll do some more operations. Let&#x27;s take right to and we will add three times over one. And we put that into words. So this is negative. Two times by a 3 to 2. Why? Plus the line gives you 14 negative. Six plus 60 geezer. Make it zero negative. One plus six gives you five and his unchanged 60 and three factor expanding along this color, we have our sign so positive and negative. So negative to Tuzla. Negative to four on by determined off 14 five, minus six and three. So this is four times 14 times three, which gives you for me too, minus six times negative six times by five. So that&#x27;s need to study. This is four times 42 plus 30 which we know is just the most. So therefore, the vectors thes inspectors are linearly independent.

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.

Michael Jacobsen · Answer

In this example, we have a matrix, a given here, which is of size three by four for three rows and four columns and what we want to determine, or whether or not the columns of a will be linearly, independent or dependent. To that end, consider the Matrix equation eight times X equals a zero vector. We want to determine if this equation has free variables or not. So ordinarily we might begin by row, reducing a. But this time we don&#x27;t have to because we have size three by four, meaning we have more columns and rows. It&#x27;s automatic that the columns of a are linearly dependent. Let&#x27;s also explore why that has to be the case now. I don&#x27;t know where the pivots are for this particular matrix, because I have not roll reduced it. But I do know that for the maximum amount of pivots, this would be our best case. Three columns are pivot columns, and that tells us that there is at least one free variable so automatically that implies that the columns of a forms a linearly dependent set and the solution is complete. So essentially, each time there is more columns than there are rows. It&#x27;s automatic that in that particular matrix we&#x27;ll have a linearly dependent set

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

Problem

In Exercises 24–26, use determinants to decide if…

Problem 25 Easy Difficulty

In Exercises 24–26, use determinants to decide if the set of vectors is linearly independent.
$$
\left[\begin{array}{r}{7} \\ {-4} \\ {-6}\end{array}\right],\left[\begin{array}{r}{-8} \\ {5} \\ {7}\end{array}\right],\left[\begin{array}{r}{7} \\ {0} \\ {-5}\end{array}\right]
$$

Answer

independent

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

Linear Algebra and Its Applications

Grace H.

Caleb E.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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independent

Linear Algebra and Its Applications

Grace H.

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Maria G.

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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.

Caleb E.

Maria G.

Kristen K.

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

Linear Algebra and Its Applications