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Points) Determine whether the Vectorslinearly independent in R'. Fully justify your aneter...

Question

Points) Determine whether the Vectorslinearly independent in R'. Fully justify your aneter

points) Determine whether the Vectors linearly independent in R'. Fully justify your aneter



Answers

Determine by inspection whether the vectors are linearly independent. Justify each answer.
$\left[\begin{array}{r}{1} \\ {4} \\ {-7}\end{array}\right],\left[\begin{array}{r}{-2} \\ {5} \\ {3}\end{array}\right],\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right]$

In this example, we have four different vectors provided, and the goal here is to determine whether this set will be linearly, independent or dependent. Well, what we could do is use the normal situation or solution. Where we take a Times X equals a zero vector were A's. The matrix, formed from calm is being be one be to V three, and before then, we would roar dose this matrix into echelon form to determine if we have free variables. But actually by inspection, we can tell that this set of vectors would always give free variables to this matrix equation. Let's explain why when we look at this particular matrix eight that we formed, it has two rows because every vector we placed has to Rosa's well. It was also of size to buy four because there are four columns now. That means this matrix a has, at least to free variables, and we know this without even using row reduction. The reason it would have at least two free variables is because with the two by four matrix, we look at the smaller dimensionality that is the most amount of pivots. It could possibly have just two leaving us at least two free variables. That means there is non trivial solutions to this matrix equation, so it's right below a X equals zero. This has nontrivial solutions, and this implies immediately that the given set of vectors is linearly dependent. So what's the shortcut on this particular situation when we analyze a particular set of vectors that has two entries but four vectors? As soon as we detected that there is more vectors than entries, it's automatic. That linear dependence is the conclusion for the set of vectors and all the material we did in between here and here is the argument of why this must be the case.

Okay, so to explain is if when all these vectors air literally independent. First we had to see what we have. So you have the 35 negative Juan V one all zeros and V two in negative 654 and V three. So to be linearly independent, that means that a combination off to these vectors should not equal the third vector. So let's start off with V one. What we want equal B two plus V three. Okay, as you can tell, that there would be no combination of the two Cosby three even if we set our factor like three times 33 Nothing. Making equal toe the one we cannot change me to, since that's just a zero vector and we can just ignore that. Basically, no, let's look at me too. Sequitur. Be born close p three. So this would be called a homogeneous system since this zero Equalling t V or impulse the three. But as you can tell, that the only solution to this is multiplying both by zero because we can never multiplied. But I might multiply any coefficient one of these to make it cancel each other out. But let's just say that if this no, I want to be negative, then they would make it a step easier to negate it. And this would have to be different number. So if it were when we dependent, I'll just say we have Vito has the same zero equal to Oh, 123 plus you too. I have four native six. So, as you can tell, we can put coefficient of two from this. When we add these up, it would cancel each other out and make it zero equals zero. And that is a non zero coefficients that I would make it. You gonna read dependent, sister. So all these vectors are literally independent, even with homogeneous vector in there.

This video, we're gonna be solving a problem. Number 18 from section 1.7 of a book was on video independence. So here were given four column Becker's 44 negative 13 to 5 and 81 and were asked to determine just by looking at collectors. So no work if ah, these columns are if these columns of vectors are literally independent. So linear independence is if the the matrix equation X equals zero. So if we combine all these vectors and tremble in matrix and concoct, contaminate them with zero vector, Um, and soft as the moment solve using element here operations. If that has the trivial solution, which is, um, in this case is X equals 0000 Since we have four column backers, so we're gonna be selling for four variables here. Um, so an easy way to solve this problem just by looking at it is if the number of vectors, which is four we have four column Beckers is greater than the number of entries in Easter Vector, which is to each vector, has two entries. So for Israel into then the system is linearly dependent, so by theory, made this system is linearly dependent. Here's why. Um So this system is literally dependent because, um, if there are more column backers than entries were gonna be solving for four variables, but you don't have enough equation set up toe solve for these foreign variables. For all these four variables tohave what a singular value. Therefore, this will result in two of the variables being free. Variables were the other two variables that are not free are written in terms of these 23 variables. So these 23 variables could be any number on the infinitely vast number line and which would cause the other two non three variables to change. So all these. So this set of actors has an infinitely many, uh, set of solutions to the Matrix equation. X equals zero, and not just the trivial solution, therefore,

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'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.


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