The question reads, if vector one in vector to our vectors in a vector space V and you want you to and you three are each linear combinations of them. So I'm going to write Vector you one is a linear combination Vector V one and vector V two. Likewise Vector. You too is a linear combination of the two V vectors and vector you three is a linear combination of the two vectors. Prove that this set is linear ley de pendent. Um, So what I'm gonna do is I'm going Teoh solve you the first equation for V two. So it's going to say V two equals You won minus a V one, Overbey. And now I'm gonna substitute that into the second equation. You, too, equals C V one plus de V two, which is you won minus a the one Overbey. All right, so now I have you to related to you one and V one. So, no, I'm going to take the 3rd 1 and solve it for V two. You three minus E V one, you three minus eveyone over F. But now these two are the same. So I can write. You won minus a V one over B equals you three minus E V one over f. And so I'm gonna keep going with the red one now. And before I continue, I'm going to think about Does this make sense? I believe so. So I'm gonna solve this for V one. So first. Well, I'm gonna cross multiply, so it's gonna be f you won minus F A V one equals B. You three minus b e view one. Okay, so let's great this again. I gotta be careful saying that 1st 1 f vector you, um minus B. You three glows F a V one minus b E V one. All right. And now we get f times vector. You being careful there minus bu three over f A minus. B e equals V one. All right. Now I can take this and substitute it into this. So it would say you to equal c V one plus de Overbey. You one minus a V one. But now we know what V one is. It's f vector. You won a minus. B, you three over f A minus, bi. Hey. Okay, let's keep going. Vector. You too. Uh, constant times V one. I gotta put that in there. Also vector you hips CV. Okay. Plus de Overbey. You minus D a Overbey times f A minus B f vector. You one minus b. You three already getting there? This is supposed to be a one here, making sure that I've got ah, everything written correctly. Okay, Vector. You too equals C over f A minus bi e f vector You one minus see Be over f A minus bi e vector You three plus de Overbey vector You minus de a a half over b times f a minus bi you times vector you one thus d a b o dab over be Oh, the B cancels out So we're not gonna dab anymore. Uh oh. Was gonna say we could dab at the end of this, but no, the dad is gone. A que Such is going to be d a over f a minus bi Ah, the a over f A minus bi e you three. So you two is just combining like terms C f over f a minus bi, do you? Plus do you Overbey minus D a f over B f A minus. Be name you want mine s or you know what I'm gonna do? Plus, um, notice that the U three terms have a common denominator already. So this is just gonna be de a minus C b Cool. So you to is a linear combination of you want and you three. Ah, because this ends up just being an unreal number, and this ends up just being a real number, and so they are linearly dependent. Um, I'm just interested to see this whole thing here. What does it become? Because, um, we could write it as c f be over f a minus bi me be minus no plus de f a minus b d e over f A minus bi e be minus d a f over b f A minus bi e. Okay, so if we group this together, the denominator is be times f a minus bi e and the numerator cfb plus d f A minus b d e. I would have thought a B would cancel out here minus d efs. Ah, interesting C f b plus de if a minus B d e mine is d A f and so we cant cancel out to be. Can we cancel out an F A minus bi? No. Because that first term, neither has an FAA in it, nor a be in it. So I guess that's what the first term would be in its not very neat. All right, It is possible. Maybe even likely, that I made a mistake in here somewhere. But the point remains, even if somehow I did make a mistake. Um, they are a linear combination of each other You to is a linear combination of U one and U three, and so that set is linearly dependent.