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$\mathbf{X}(t)=\left[ \begin{array}{ccc}{e^{t}} & {e^{-t}} & {e^{2 t}} \\ {e^{t}} & {-e^{-t}} & {2 e^{2 t}} \\ {e^{t}} & {e^{-t}} & {4 e^{2 t}}\end{array}\right]$

Chinese solution when directed with stool and avoided give some right sector own additional access this compound. So could you inform general which hard place here's the correct terms.

Room. Where else? To solve the differential equation W prime prime minus two will be prime. Plus W is equal to 60 minus two. Such that w of negative one is three and w prime of negative one is seven. Now, To solve this problem, which your eyes on slightly different initial conditions to what we want. We're gonna use a little trick by defining a function. Why of tea to be equal to W of T minus one. In this case, we then know that why I have zero physical to three and why prime of zero is equal to seven, which is the most familiar setting that we used to dealing with. Why also said this that satisfies a similar differential equation in particular. On you just have to substitute t minus one here instead of tea. So now we will solve the differential equation for why and therefore attainment solution for W. Now we start by taking the lacked last transform off beachside. We obtain that the first term is a squared Y minus. Two s plus one minus three s minus one is equal to six over X squared minus eight over SS obtained by taking the last blast transform of both sides here. Uh, rearranging gives that. Why is he for the six is equal to six minus eight s plus three x plus one times x squared. Who's that? That's minus one squared times X squared. Now, to compute the inverse transform of this, we want to compute a partial fraction decomposition. Since both s and after my ass Wanaka twice in the denominator that each contributed to terms Here we can find the square to terms most too easily by plugging in ethical toe one into this equation evening during this term and X equal to zero and ignoring this time doing this gives us the value true of s minus one squared and negative one. Uh, sorry, fool. Yeah. No, sorry. Six, um, now remains to God that the s minus one and s terms. Um, this is most easily done by Uh huh. It's easily done by subtracting two of F minus one squared from the original equation and dividing by s minus warm. Once we've done this, we can plug in s equal to one to obtain this value here, which is going to be equal to mix of one to militarily. We can subtract six over Esquire from this term, divide out a factor of s and then set s equal to zero. The resulting value will be equal to four. Once they've done this, we can now could be the inverse like last transform, which gives y equals two negative u to the T plus two um, plus two t e to the T plus full plus six t. And that's the answer. Oh, sorry. No, that's not the answer, because we nanny to substitute w of tea, which is equal to why of tea plus one. And if we do this, then we obtained that w is equal to 60 plus 10 plus two t u to the T that's one plus each of the peoples from

So for this 63 9 cancel out to seven. And I just have to subtract the value of the powers if we have the same base. So six minus four is too. So I have r squared three minus two is one. So I've asked the first power, which just s 72 6 Cancel out to 12. The art. There's no Mars on the bottom. So I'm just gonna rewrite that as squared minus over s as times as over s cancel them out. We're left with one s. So then you can see that this and this are the same. So we just have to work with the coefficients. So seven minus 12 is negative five, and I just have to tack on that.

Okay, we're given three different vectors and were asked to do things with their magnitudes. So notice the double lines mean magnitude. So for a and let's look at A. M. Because they are very similar but very different also. So notice for a we have to first add our vector U. And V. And then we find its magnitude. But for be we're gonna be finding the magnitude for each and then summing them. So for a let's go ahead and first add our vector you. So I'm just writing it and then I'm going to write my vector V underneath it. And then I can sum them together by adding my columns. So I get a to I minus two J plus two K. So now to find my magnitude I will take the square root of the square of all of the pieces. So we have a two squared a negative two squared, which would be just like a two squared. So in your calculator you can just throw that in as a two squared. And so really what we have is the square root of 12 and that's our magnitude. We can also bring out um you know four times three for 12. And so we can do a square four and bring out a two. So two square root of three. Okay, so now for part B we're going to actually find the magnitudes separately. And so for are you vector, we would be doing the square root of one squared, negative three square, but the same as three squared. And for my calculator, I don't want to get confused in with those negatives. They're not apprentices. And so we end up um um Then also doing the same thing with R. V. And it just has the two components one square plus one square. Okay, so now we have to add the two of them together. Well, our first was the square root of 14, and the second was the square root of two. No go back and get that correct. That is the square root. Okay. Okay, so let's go on to see. So first we already found our um our value for our vector U. Which was the square root of 14. So now whether the negative negative two is multiplied in at the beginning or at the end it's a property Of um are vectors that scaler piece can be pulled out front, so it's kind of like factored out of each piece. Um however the magnitude is just going to be too, because we're losing our direction and so we can take our two multiplied by our square root of 14. And then we're going to be adding that to two times are other vector, which was the square root of two. Okay, so for um d we are going to hold off and do that, we need some more space. But ian f aren't so bad. So let's go ahead and do e. So what it's asking us to do is telling us to take all of our vector W and divide each term by the magnitude. Now what this does is it produces a unit vector. So a unit vector really means that here and we'll see it. It will just show us our direction, but not our magnitude. So let's go ahead. We have all of our pieces. And then our magnitude is going to be that two square plus two squared plus negative four squared. And so that places on the bottom of square root of 24, and we can consider that's a two square root of six. So now we divide each term by two square root of six or two's cancel out our four divided by two is still just a two. And we get what's called a unit vector in a unit vector Really has the magnitude of one and they're really for comparing directions. Okay, now F looks a little bit um more different. Um But it just has the magnitude on the outside. Well we called it a unit vector because it has a length of one. And so its magnitude is one. So just by definition the magnitude of the unit vector is one. Okay, so we skipped over D. So let's now go back and look at um D. So we're gonna have to take three times are vector U. We're going to have to take not negative five times our vector V. So notice I'm just distributing these things and then we are also taking one of our vector W. And then I'm just adding up all my eyes, which gives me zero I all my jays, which is a negative 12 J. And then all my K. Is which is positive two K. Now it's asking for magnitude. So it's kind of like I have a zero squared plus a negative 12 squared plus a two squared. So in the end, that gives me the square root of 148. And if I factor out of four, I get the get Almost wrote 36 but knew this was not 12 squared, which would be 144. So it ends up that it's two times the square root of 37.


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