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Canbe written as linear combination of V,and V If so. which linearcombination? If not; explain why not:...

Question

Canbe written as linear combination of V,and V If so. which linearcombination? If not; explain why not:

Can be written as linear combination of V,and V If so. which linear combination? If not; explain why not:



Answers

Linear combination Let $\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}+\mathbf{j},$ and $\mathbf{w}=$ $\mathbf{i}-\mathbf{j} .$ Find scalars $a$ and $b$ such that $\mathbf{u}=a \mathbf{v}+b \mathbf{w}$

We want to find some linear combination. Uh, A and B scale er's where you is. You go to a Times B plus B plus w. So first, let's go ahead and write in what a b m p w we're gonna be. So we're gonna have a time. So be is I plus J be Is I minus J Now we can go ahead and distribute the A would get a I plus a J and then distribute be will get be I minus e j Now you It's supposed to be two i plus j so they don't notice if we were to go ahead and think about the equation that only adding the AI and the B I will give us two I Well, we can just set that equal to each other. So end up with a equation of two. I is equal to a I plus b I. And now we can also do the same for Jase. We have Jay is even too a j minus b j. And if we were to go ahead and divide the first equation by I in the second equation by J, that's going to give us that too is eager to a plus B. And our second cordage is gonna be one, is he goto a minus bi. Now we can go ahead and add thes two equations just like we would in Algebra and Saul for A and B. So that's gonna be three. Is he going to weigh when we divide the two over? So that's going to give us that a Is he going to three halves? And to get B, we can just plug it into one of our equations up here. So I'm just gonna plug it into this top equation. So, too, is he Go to three halfs plus B. So we subtract three halfs over so we'd get 1/2 is equal to be so a is three halfs b is 1/2. So if you were to come over here and plug in three house and 1/2 so you multiply be by three house W by 1/2. Add them up. You should get to I plus J

So we went to the trimming. Whether or not we can read an earlier question, this is any system of linear equations, a documented mind tricks matrix and wheels one to the German. How to write this Met matrix. So first of all, we can read any system of linear equations as a matrix ascena make metric serious. Any system of linear question can be written. Listen, no, l commented Matrix, and the reason for these is because any leaner question combat readiness a one Tim Specs was richer. Tim's wife plus 8th 83 Sorry, this trip. Money two times 83 him C plus. So this can continue to all the way to infinity is equal to a constant. And so for easy stand off to a question we have the X Bless me, too. Why Husby three z plus descriptions all the way to Infinity Musical Chair Constant Be so we continue all the way of writing different system of questions vertically, horizontally, And so for an argument of The Matrix, we can break all the coefficients for a particular equation. Let's hit this equation. One. Just a question, too, for Question one will have a row with all of the corporations for a love or a X terms in white terms, A C terms all the way to infinity. So we have a one. A two in three for second equation will have to be one. Me too. Maybe three. So now we're gonna place a vertical line over here to separate their constants. It's will have pain. P all the witches commuting this how you write in undocumented metrics matrix from a system of linear equations. So remember that yes, you can break any system of linear questions. It's undocumented metrics because you can write them in the center form A One Times X was a two times one plus a three time scene all the way to infinity encircle Chuck Johnson. And because you can write this all of the questions in this form, we can place all of their coefficients in this set of matrix That's a Sichuan represents one of questions with the coefficient of X, y and Z lying up in his way and on the side we placed a vertical line two separate that Constance from our coefficients

I think this problem We're given a set off the nearly dependent vectors. We want me to be free and report. We have to solve this. We're gonna use the humans. Let's try it back to w uh, using an arbitrary number K. So W's came one of me One plus get to be too plus k for before. What we know is this We know that this be Zarley nearly depend by so they depend on each other so well, let's assume not be one is equal to, uh, some constant A to B two plus a three d three plus a Ford We pour hear Keyser arbitrating And here a czar Arbitrary. So let's write to be one in 31 in terms of B to B B m w will be, um, a one a two we two plus a three d three plus a four before plus K to be two plus three v three plus cape or we fall. So then w will be equal to a one a two plus que tu we to plus one a three plus three Well, three plus J won a four plus four me for So since case and a CZ arbitrary. And since we can, right, let's say V two in terms of We want me three Emmy for orchid Correct. Be three in terms of other valuables are directors. It means that there's more than one way Jax Prospector read, So Dan W. Can be expressed as a linear combination off their vectors, we warn me to briefly and in numerous ways.

Prove that the span friend right here Ban of V one V two V three equals the span of V one V two if and only if the three can be written as linear combination of V one and V two. So I'm gonna write out what these spans are. The span of V one v two V three is V and elements of the such that V is C one B one plus C two v two plus C theory V three with Ah c one, C two and C three being real numbers. But if we write out the spin of V one and V two, it would be all V all vectors in the vector space. Such that V is some D one rial number times view one plus some real number D two times V two. No, I don't have space. But ah, I did wonder, right? That um c one c two, c three d one and d to are all rial numbers. So if we equate the two sides, then we see that every vector could be written as C one v one plus C to v Teoh plus C three v three. But that same vector could also be written as D one v one plus de to the to. If we now solve this, we could write it as see one minus D one times Vector one plus C three minus not three Tu minus D to times Vector to equals Negative C three Vector three Now if we divide by negative C three we see that vector The re is D one minus C one over C three times Vector one plus C to minus D to whoops de tu minus C to oversee three times vector to, um, this is a real number and this is a real number. So, um, Vector three must be able to be written as a linear combination of vectors one and vector to


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