in-exercises-5-8-determine-if-the-columns-of-the-matrix-form-a-linearly-independent-set-justify-ea-3

Question

In Exercises $5-8,$ determine if the columns of the matrix form a linearly independent set. Justify each answer.
$\left[\begin{array}{rrrr}{1} & {4} & {-3} & {0} \ {-2} & {-7} & {5} & {1} \ {-4} & {-5} & {7} & {5}\end{array}ight]$

Michael Jacobsen · Accepted 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

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.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving question number eight from section 1.7, which is based on linear independence here. Were given a matrix that is three by four matrix initially and were asked, uh, were asked if it is new. If the columns of this matrix are literally independent, Um, to find that out, we need to learn about Leonard. Defendants. Linear independence is basically a set of vectors is linearly independent? If, um, if if the matrix equation X equals zero only has the triple solution, which, in this case is this right here a column of four zeros. Because here, uh, we know that if you split this matrix into individual column vectors, you have four column vectors, which means that that&#x27;s how many variables we&#x27;re gonna be solving for if we were to set this question up as a system of equations Ah, and setting them equal to it like, uh, zero. Hence X equals zero. So this is how you set up in augmented matrix of X equals zero right here. Um, and to solve this ah problem. In an easy step, you just consider the roommate from the book and I&#x27;d recommend reading that chapters of the book or a section of the book as they&#x27;re very concise. And the book does a really good job of explaining how to do every proper, like the concept of how to do every problem so they&#x27;re made of. The book states that less. Let&#x27;s say we have n representing the number of row vectors and appear representing the number of column vectors. So here N is three as their three row vectors and, um, so n equals three and P equals four as there&#x27;s four, um, Colin vectors as we&#x27;re selling for four variables. Ah, so four is greater than three. Therefore, this system is literally independent, because solving this system would result in a free variable where one of the variables, uh, is called a free variable, where you can choose any value so there on the number line. So there&#x27;s infinitely possible values on, and that free variable causes the values of the rest of the variables to change, as the rest of the variables will be in terms of that variable. So once you perform elementary row operations on this, um, this augmented matrix you&#x27;ll end up with this matrix where um there&#x27;s leading entries in three of the four columns, but the four ah, one of the columns does not have a leading entry, which means not call them. If he labeled this x y, uh x one x two x three and x four x three would be a free variable. This means that, um, this means that this system has more solutions than just the trivial solution. Therefore, um, the system is linearly dependent. That&#x27;s what Jeremy it is really saying.

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

Sean Thrasher · Answer

Okay, so we&#x27;re asked to figure out if be this problem back to be is a linear combination of the columns of A And I&#x27;ve written down the columns of a is the factors v one b two, b three. So the question is, is be a linear combination of B one b two, b three. And if you seen Problem 11 is the same exact process behind it. So we want to find x one x two x three Such that we can write be as this some right here. That&#x27;s what it means for B two b a linear combination of you want me to be three. And so this equivalently is if you&#x27;ve if you substitute and what b one b two B three years and write it out was a system of equations. You get this. And so the question is, does x one x two x three exists So to ask that we&#x27;re going to try and find the solutions x one x two X c We&#x27;re trying. We&#x27;re going to try and solve the system of equations. Now a linear system of equations could either be inconsistent, so solution might not exist. It could have a solution, and the solution is unique. So on Lee one you know, triplet of X one x two x three would work, or you could have non unique solutions of a solution would exist. And there be also yet another one. Hey! And so if we&#x27;re in case one, then we know that B is normal linear combination of a 18 23. And it turns out that this is the case that this system of equations boils down to. So to solve it, we augment the Matrix like this. We have no events matrix, the matrix of coefficients here, and the vector B, which is the right hand side of the equation. And so we&#x27;re going to use role reduction. And first step is to take row three. I want to get rid of sentry. So I ads two times, roll one to row three. And when I do that, I get this. So I get zeros. Except for this last entry here, and it&#x27;s a negative three. So 000 Negative three. Sorry. You can&#x27;t quite see that. Uh, sorry. No, it will be plus three. So +000 plus three and we can see that there&#x27;s a problem. So if you convert this back too, a simpler an equivalent system of equations, you have this. And the third equation is just zero is equal to three. And certainly this is not true. No matter what you know, this is not true. Regardless of what x one x two extra you choose and so boring. Case. One system of equations is inconsistent, and so B is not linear combination of the columns of a and we&#x27;re done.

Michael Jacobsen · Answer

in this example, we have three different vectors provided, and what we want to determine is whether or not these vectors will be linearly, independent or literally dependent. Well to check for linear independence. One method is to consider the Matrix equation eight times X equals zero vector, where a is going to be formed as follows a will be V one v two V three for each of its columns. Now what we&#x27;re mainly concerned about when we formed this formal genius equation is whether or not it has trivial, non trivial solutions. So to determine whether we would have nontrivial solutions, let&#x27;s begin reducing the matrix A itself so calm. One. We indicate us 500 then seven to negative six for column, too, and 94 negative eight for the Third column. Now, in order to determine if there will be nontrivial solutions, we just need to put this matrix into echelon form. So for echelon form, the major pivot we&#x27;re concerned about is this one, and we just need to produce a zero here. So let&#x27;s re copy this matrix say its role equivalent to first copy 579 their first row copy in the second row 0 to 4. Now what we&#x27;re going to do is multiply row two by three, add the results to Row three that would give us 004 Then, once we get to hear, we see that there&#x27;s a pivot in this column, this column and this column, which implies that this system has no free variables, which means X equals zero has on Lee the trivial solution. But if you have only the trivial solution to the Matrix equation, it would mean that X one times v one plus x two times v two plus x three times V three equals a zero. Vector also only has the trivial solution, and in this case, we know by the definition of linear independence and dependence that the vectors V one V to V three are linearly independent. And it all came from the fact that there was a pivot in every column, which means there&#x27;s only the trivial solution then, by definition, these vectors air linearly, independent

Problem

In Exercises $5-8,$ determine if the columns of t…

Problem 7 Easy Difficulty

In Exercises $5-8,$ determine if the columns of the matrix form a linearly independent set. Justify each answer.
$\left[\begin{array}{rrrr}{1} & {4} & {-3} & {0} \\ {-2} & {-7} & {5} & {1} \\ {-4} & {-5} & {7} & {5}\end{array}\right]$

Answer

The given vectors are linearly dependent.

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Grace H.

Anna Marie V.

Kayleah T.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

The given vectors are linearly dependent.

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Anna Marie V.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications