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Question

In Exercises $1-4,$ determine if the vectors are linearly independent. Justify each answer.
$\left[\begin{array}{l}{0} \ {0} \ {2}\end{array}ight],\left[\begin{array}{r}{0} \ {5} \ {-8}\end{array}ight],\left[\begin{array}{r}{-3} \ {4} \ {1}\end{array}ight]$

Viswesh Uppalapati · Accepted Answer

in this video, we&#x27;re gonna be solving problem number two of section seven off. Um, chapter one, which is on dinner. Independence. Um, so we&#x27;re given three column vector 00205 native eight negative 34 and one which I, um which I combine into an argument in matrix for an X equals zero equation. What we&#x27;re doing here is basically, just rewriting, um, a system of equations into an augmented matrix where we get zero x plus zero y plus native three equals zero and zero x plus five wipe lists for Z equals zero. And so on, Um, and that system of creations history wrote into this argument matrix can only have the trivial solution for it to be a linearly independent. Um, which means that all the variables, uh, can have to be zero. So in this case, X has the equal 000 since, um, there&#x27;s three columns, which means there&#x27;s gonna be we&#x27;re solving for three variables, um, and each each entry in this column represents a solution to one of these variables. So to get that to see if, um, this solution has a has on Lee the trivial solution um, we have toe reduce it. So to do that, I will flip a road three under 01 because it makes it easier. Um, because we already are. Get a staircase pattern, which we want and which is an integral part of the echelon form. Um, and then we want ones at this diagonal. So I just divide the first row by two. Can I get a negative one or one negative? Four and one, huh? The second row by five. So I get zero one and 4/5 in the third row by negative three sawyer 001 And on the right side. Hey, um, So since this matrix is now in basic Ashland form, weaken. And since they&#x27;re ones at every leading entry, which is also the position, we can conclude that there there can be no free variables possible for this for this augmented matrix. So are for this system. So they&#x27;re no free variables in the system. That means, um that means this right. Call them the right most home must be the solution to the system, which is X equals 000 And we don&#x27;t have to solve this any further because we started with zeros on the right and ended with zeros on the right. Um, and we didn&#x27;t performing your operations where we were. The values of these zeros changed. So we can conclude that the final solution of this side of system of this system would be X equals 000 in a column vector. Ah, And since this set of vectors on Lee has, ah, the trivial solution, that means this set of vectors is linearly independent on, and that is the answer to this.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number four from section 1.7 of the book, which is on linen independence. So we&#x27;re giving these to Colin vectors. Ah, negative. 14 and negative. Too. Negative. Eight. And the only way these this set of Colin vectors is linearly independent, as if X equals zero. The of matrix equation has only the trivial solution, which is, um, X equals a column vector of zeros, depending on the number of um, column vector is your given. So the number of variables depends on the number of column Decker&#x27;s There are so exit polls zero is just another way of writing a system of linear equations where you have to variables, um, with negative one negative to four and negative eight on something called zero. So you want to set up a matrix like this where you combine the two, uh, column Victor&#x27;s, and then I like to draw a line to distinguish it and then combine it with zero. That&#x27;s just a way to set up the major situation X equals zero. To solve this, you just have todo, uh, put this Convert this into echelon form where you perform row operations to reduce it so that there is the ones in this diagonal or, um, or you get a row with zero equals zero on both sides, which means they&#x27;re free variable. So to do that, you basically just, uh first, I would like Thio and divide the second row by four. So are too times one for it. So you have one here. Negative two year and then 00 And I like to multiply the first row by negative warm so that I could change these two positive, Um, and the next you would, uh, would subtract row one from row two. So to minus row one. So the rolands stays the same and row two becomes zero. Uh, negative 40 And then last year, I would divide rowe, um, two by four. Again are negative for this time. So row two times negative. 1/4 is one too. 00 was your one. We have ones in this diagonal, so we have, like, a staircase pattern. So we know that this, um, augmented matrix is now in basic salon for on the right side. Hasn&#x27;t changed its toll 00 That means, um since It&#x27;s just zeros on the side. We can disregard this, too. And because here, let&#x27;s say we have two variables X and y, where each called column represents a variable in the system of equations. We know that Y equals zero because there&#x27;s no other entry in this, uh, in the second row. So we know y equals zero, and the second equation would be X plus two y equals zero. And since why a zero x plus two times zero equals zero X equals zero. So this, uh so basically the right side is a representative of the, uh, solution to the system equations. So you got X equals 00 which seems trivial solution and therefore this set of vectors Negative 14 and negative. Too negative. It was Your column vectors are linearly independent. That&#x27;s it.

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

Michael Jacobsen · Answer

in this video, we have two vectors, B one and B two, and we want the test. If they&#x27;re linearly independent. Orly nearly dependent now What we could do is throw these both into a matrix and begin row reduction. But let&#x27;s take a different method. This method works with just two vectors, which is our situation here. Notice that if we take and consider the first elements, they&#x27;re only off by multiplication. By negative three. That makes us suspicious. Let&#x27;s suppose we take negative three times V one. We would obtain negative three and positive nine. But that means negative three times V one is equal to V two, and in this case, negative three times V one minus V two is equal to the zero Vector, in other words, were able to find scale er&#x27;s C one and C two not both. Zero such that C one B one plus C two V two equals zero vector by taking see one to be negative three and see to to be equal to negative one. This, then implies that the vectors or the set of vectors B one B two is linearly dependent. We have to be careful if we use this method because we&#x27;re using just two vectors. And in this case, two vectors are always linearly dependent if they&#x27;re multiples of each other like we just found.

Michael Jacobsen · Answer

In this example, we have a set of two vectors that were considering B one and B two. We want to determine whether they are linearly, independent or dependent. First notice that since we have two vectors, we have a specific method that we can use If we look at, say, the entry here on V to comparing with this entry in V one. If we multiply negative four times V two, then the result would provide us with a negative eight in this position here. Then multiply it by Negative. For here we get a positive 12 multiplied by negative for here, and we get a positive four. Now, If this was equal to be one, we would have said that these vectors are linearly independent because they would be multiples of each other. But that&#x27;s not the case. We have one deviation here and here. So this allows us to say that the vectors V one and V two are not multiples of each other. And in this specific case where we have just two vectors, we can immediately make the following conclusion. Since they are not multiples of each other. It follows that B one and v two are linearly independent. We need to be careful if we use this technique considering the multiples of vectors, because if we had three vectors, this would not work. It might be more effective for three vectors to use row reduction instead in many cases.

Viswesh Uppalapati · Answer

in this, uh, in this video, we&#x27;re gonna be solving problem number 20 from section 1.7, which is based on their independence. So here were given three back. Here&#x27;s one for negative seven negative. 253000 On these there are column workers and were asked to determine just by looking at homes and no work. Uh, if they&#x27;re literally independent or dependent so and easier to solve. This is by looking at the Book of the theorems and the serum. Nine states that if a set of column acres contains the zero vector, then the set is clearly linearly dependent. So we have the answer. The set is linearly dependent. And here&#x27;s why. Um so for a set to be literally dependent are literally independent, the vector equation or the major situation X equals or that we learned back in section one point for Has equal has to have only the triple solution was in this case is 000 because we have three vectors. That means there&#x27;s gonna be three variables that we&#x27;re gonna be stalling for if you were to set this up as a system of equations. So if you were to set up a matrix augmented matrix to solve this. Well, uh, for X equals zero, we would set it up like this. I&#x27;m sorry. Sorry for my handwriting, but these these two are zero backers. Um, so here, right off the bat, we know that this zero accurate causes a free variable. For that, they&#x27;re variable to be free. So the other two variables will be written in, uh, terms of the third variable Where, uh, that they&#x27;re variable. It can be anything on the number line can be any value number line. And, uh, the other two variables will be written in terms of the three. Very also be able change. Based on what value The free variable is, uh, chosen to be so, since this has way more solutions with just a trivial solution. Uh, this set of factors is only really

Problem

In Exercises $1-4,$ determine if the vectors are …

Problem 2 Medium Difficulty

In Exercises $1-4,$ determine if the vectors are linearly independent. Justify each answer.
$\left[\begin{array}{l}{0} \\ {0} \\ {2}\end{array}\right],\left[\begin{array}{r}{0} \\ {5} \\ {-8}\end{array}\right],\left[\begin{array}{r}{-3} \\ {4} \\ {1}\end{array}\right]$

Answer

Linearly Independent

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

Linear Algebra and Its Applications

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Linearly Independent

Linear Algebra and Its Applications

Grace H.

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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.

Catherine R.

Heather Z.

Kayleah T.

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

Linear Algebra and Its Applications