find-the-values-of-h-for-which-the-vectors-are-linearly-dependent-justify-each-answer-leftbeginarray

Question

Find the value(s) of $h$ for which the vectors are linearly dependent. Justify each answer.
$\left[\begin{array}{r}{1} \ {-1} \ {4}\end{array}ight],\left[\begin{array}{r}{3} \ {-5} \ {7}\end{array}ight],\left[\begin{array}{r}{-1} \ {5} \ {h}\end{array}ight]$

Michael Jacobsen · Accepted Answer

In this example, we have three different vectors that are provided where the third vector also depends on the value of H here. One question we could ask about this set of vectors is when with the offset be linearly independent, for example, for example, with that value depend on h. Was it always be linearly dependent, or will it always be linearly dependent to determine the answer to such questions? We can start off with the Matrix A where a has calm is formed from the vectors given to be V one, V two and V three. So this would be equal to one negative 14 in the first column. Three negative 57 for calm to and then negative 15 and h so for this given matrix, we&#x27;re going to roll reduce to echelon form and that will tell us where the pivots are. First we start here and we&#x27;re going to eliminate the entry Negative one and positive for below. So let me copy row one to start out, then add row one to row to replacing Row two. That gives us zero negative two and four, then multiply row one by negative four and added to Row three will obtain zero negative five and H plus four. So our new pivot position is found on the two to entry here in the Matrix, and we&#x27;re just going to eliminate the entry below this time since we just require echelon form. So let me copy both row ones and two. So there&#x27;s our Row one and row to Let&#x27;s first Divide row to buy the quantity. Negative two will obtain 01 negative, too. Now our operation is going to be to multiply row two by positive five so that we can add it to Row three and obtain a zero here. Upon performing, that operation will obtain zero zero and H minus six. So let&#x27;s look at the situation so far for pivots. Calm one and calm to are automatically pivot columns and calm. Three. Could be a pivot column, but we require H minus six to not be zero. So if H is not equal to positive six, then A has a pivot in every column. What this would mean is then, if we consider the Matrix equation a X equals zero vector. This equation has no free variables When h is not equal to positive six. And we know that whenever we&#x27;re dealing with no free variables, we have linearly independent columns. So we&#x27;re ready to state our our conclusion will say when h is not equal to six. The columns of a which are these vectors from the start of the problem, are linearly independent.

Viswesh Uppalapati · Answer

and this video, we&#x27;re gonna be solving problem number 14 from section 1.7 of the linear algebra book. Um, so basically, we&#x27;re given three column backers. One negative one and one negative. 13 negative 578 and 11 in h and we&#x27;re has to find the value of age that would make this set of vectors linearly independent. So for for a set of vectors to be literally independent, they&#x27;re augmented matrix times a vector X has to equal zero as equals zero vector. So in this case, so the only way it, uh the only way this set of vectors could be literally independent if is if they, uh, the vector equation are the major situation X equals zero, which you learned about in section one point for has only the trivial solution, which is X equals 000 In this case, because we have three column vectors, that means we&#x27;re solving for three variables where each entry and this in X represents one variable, the solution of one variable to the system of equations. So here we set of on Augmentin matrix right here like this, where we combined the three doctors and zero vector. Um, and we basically wrote reduce it to gettinto basic echelon form where you have a staircase and the leading pivots leading entries. And each row, Um, and the pivots are the ones here in the first and second row, and the third row has, um, an H minus 26 equals zero. So for for this set for this system to be linearly independent, we need, um we needed the trivial solution. So the only way we would get the trivial solution is if h minus 26 equals zero. So H has to equal 26. Let me explain this a bit. So here we have X y Z, which are the three variables we&#x27;re solving for from this system. So the the last entry would give us Z times H minus 26 has to equal zero. And the only way, um, the only way er ze chemical zero is if it&#x27;s multiplied by zero. So therefore eight minus 26 pass equals zero and h ha sequel 26 for this, um, for this is a set of vectors. Toe be all zero. So the only way Zeke undies narrows it is If H zero if ages 26 because you subtracted from 26. And that&#x27;s the only way we get the trivial solution where all three variables are. Zero. Therefore hs equal 26 for linear independence for this set of column vector.

Viswesh Uppalapati · Answer

be solving question number 12 in section 1.7, which is based on a linear independence where it gives us three column. Victor&#x27;s and Asa&#x27;s values of H right here would make this set of vectors literally independent. So for her to be literally independent, the Matrix equation X equals zero or the concatenation of these three vectors with zero must have only the trivial solution, which in this case is X equals 000 Because, um, here three doctors, three lawmakers means that we&#x27;re gonna be selling for three variables where each entry represents a variable. So only if all the variables zero would this set of factors be literally independent. So to do that, we just set up, um, an augmented matrix like this with all three vectors contaminated with a column maker of zeros. And you perform elementary rule operations. And once you do it, I&#x27;m not gonna do all of them here. But, uh, since I did them beforehand, once you do the final reel operations, you got to negative 68 and zero for the first row. So it stays. Are that stays the same? Yeah. I&#x27;m sorry. I noticed this wrong down wrong. Says positive to their, um, the second row will become zero negative five. And, um h plus 16. It means you&#x27;re on the side. And the third row is just gonna be all zeros here. Ah, we don&#x27;t even need thio completely solve this matrix into echelon form as we know that we see a pivot here and a pivot in this columns of these two columns all the pivot. But the third column does not have a problem. Position, position. Um, so the third variable will be a free variable, which means it can be any value from the number line, which would cause the other two variables to change because there would be written in terms of the free variable when you completely solve this system. Therefore, since, um, since the system has more than just a trivial solution more solutions with just a trivial solution, there is no value of h of age that makes this system linearly independent. Therefore, it&#x27;s, um, it&#x27;s linearly dependent for all age.

Michael Jacobsen · Answer

for this exercise. We are provided with three different vectors, and the vector V three we see here also depends on the value of age. So questions we might ask about this particular set of vectors are for what values of age would this set be linearly dependent? Well, to answer that kind of question, we can start by making a Matrix A, which is formed using the columns V. One, V two and V three are provided vectors. Then our metric say would be 15 negative three for the first column. Negative to negative 96 and call him, too, and three h negative nine for the third column. Let&#x27;s re reduce this toe echelon form. That way we can analyze the pivots and now determined linear dependence. So let&#x27;s begin then with Row one or entry here our first pivot position and will take out the five and the negative three below. So let me copy one negative two and three for a first step. Multiply row one by negative five. Added to row two, we obtained 01 and H minus 15 for the next step. If we triple row one multiplied by three and after Row three. Then we&#x27;ll produce a zero zero and zero as well. So that&#x27;s kind of convenient. With one row operation, we are in echelon form now. The pivots will be here and here and nowhere else. And the value of H does not change where the pivots will be. Let&#x27;s consider that matrix equation eight times X equals a zero vector. This equation has nontrivial solutions. The reason we know this is that there is a free variable that corresponds to the third column in this particular matrix. So since we have non trivial solutions, we know nontrivial solutions to a homogeneous matrix equation implies immediately that the set of vectors V one, V two and V three is linearly dependent, and the linear dependence we found again does not depend on H whatsoever. So V one v two V three is linearly dependent for all values h in the set of Real&#x27;s. So this is how we can determine whether or not we have linear, independent or dependent vectors in this situation,

Michael Jacobsen · Answer

in this video, we have three different vectors and the Third Vector as an interesting situation where the third entry depends upon H and H is allowed to be any real number. Now. Suppose we were interested in the span of V one M V two. The question might be for what value of age would be three being that particular span. To check this, let&#x27;s consider the equation A matrix A with calmness V one and V two times a vector X equals B three. If this matrix equation is consistent, then be three is inside the span. Otherwise, it will not be in the span. So let&#x27;s augment this into a matrix containing be 11 negative three to be, too, which is native 39 negative six and V three, which is five negative seven and h. So we have a three by three matrix and to reduce, we&#x27;ll start by pivoting off this entry and eliminating these two entries. So first let me copy Row one. It&#x27;s one negative three and five. Now multiply row one by three added to row two and we will get 00 and positive eight. If we multiply row one by, say, negative two. At the results of Row three, we will this time get 00 and H minus 10. So now note that we&#x27;re not quite in echelon form, but we do have a pivot here and a pivot here. And those pivots do not depend on the value of age. Since we have a pivot in the right Most column. It follows that this system is inconsistent, But this then implies immediately that V three is not in the span of V one and B two, and this is for all values of age. Next, let&#x27;s consider the set B one b two, b three once again, and determine whether this set could potentially be linearly independent, as exposed as opposed to V three being the span of the prior two vectors. Well, we did. Most of those were essentially all of the lake work for linear independence. What we would want to do is take all three vectors, place them into a matrix, as we&#x27;ve done here, and were reduced to echelon form, which is exactly what we have here now in that echelon form. Notice that this column here column two is a non pivot column. That means the System V one, the two V three tens of Vector X, equals a zero. Vector has non trivial solutions, since there is a free variable, so the free variable gives us non trivial solutions to the homogeneous equation. But now we know once we have non trivial solutions. This then implies immediately that the set of vectors V one to and V three is linearly dependent. So nontrivial solutions is take tied together and linear dependence if and only if we&#x27;re looking at the homo genus equation here.

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.

Problem

Find the value(s) of $h$ for which the vectors ar…

Problem 11 Medium Difficulty

Find the value(s) of $h$ for which the vectors are linearly dependent. Justify each answer.
$\left[\begin{array}{r}{1} \\ {-1} \\ {4}\end{array}\right],\left[\begin{array}{r}{3} \\ {-5} \\ {7}\end{array}\right],\left[\begin{array}{r}{-1} \\ {5} \\ {h}\end{array}\right]$

Answer

Thus, the vectors are linearly dependent if and only if $h = 6 .$

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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Thus, the vectors are linearly dependent if and only if $h = 6 .$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Heather Z.

Alayna H.

Caleb E.

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

Linear Algebra and Its Applications