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(1 pt) Let T be an injective linear transformation from R" to R* . Let A be the matrix associated to T and let B be the rOw- echelon reduction of ADetermine wh...

Question

(1 pt) Let T be an injective linear transformation from R" to R* . Let A be the matrix associated to T and let B be the rOw- echelon reduction of ADetermine which of the following conditions can hold:A.T = 7,8 = 5 and Ehas pivots: B.r and Bhas pivots 5,8 and Bhas pivots. and Ehas pivots E None of the above

(1 pt) Let T be an injective linear transformation from R" to R* . Let A be the matrix associated to T and let B be the rOw- echelon reduction of A Determine which of the following conditions can hold: A.T = 7,8 = 5 and Ehas pivots: B.r and Bhas pivots 5,8 and Bhas pivots. and Ehas pivots E None of the above



Answers

Let $T_{1}$ be the linear transformation from Problem 10 and let $T_{2}$ be the linear transformation from Problem 5 (a) Find the matrix representation of $T_{2} T_{1}$ relative to the standard bases. (b) Verify Theorem 6.5 .7 by comparing part (a) with the product of the matrices in Problems 5 and $10 .$ (c) Use the matrix representation found in (a) to determine $\left(T_{2} T_{1}\right)(-3+8 x) .$ Verify your answer by computing this directly.

Were given real matrices that define a linear transformation of our two. And for each matrix were asked to find all the Eigen values and a maximum set s of linearly independent Eigen vectors were then asked which of these linear operators are diagonal Izabal, in other words, which can be represented by a diagonal matrix in part A. We are given the real matrix A with components 563 negative two. Well, first we'll find the characteristic polynomial. This matrix delta T apartment he's been riding. It's a two by two real matrix. This is simply T squared minus this trace of of a which is three times T plus the determinant today, which is negative 10 minus 18, which is negative 28. Oh, is that what happen? Which we can factor as t minus seven times T plus four. The roots lambda equals seven and lambda equals negative four. Are the Eigen values of this matrix A. You want to find the corresponding Eigen vectors is the Eigen values helping to find the corresponding Eigen vectors. Well, first we'll subtract seven down the diagonal of a. So the matrix and this is a minus seven I. And this gives us the two by two matrix negative 263 negative nine. Which corresponds to the homogeneous system negative two X plus six Y equals zero. And three X minus nine. Y equals zero, which corresponds to the system, X minus three Y equals zero. Now we see that Vector V one with components 3, 1 is a solution non zero solution. In fact, it's an Eigen vector. Now, on the other hand, will subtract lambda equals negative four down the diagonal of a. Find our matrix M. So M is a plus for I Which is the Matrix 9632. And this corresponds to the modernist system. Nine X plus six Y equals +03 X plus two Y equals zero. Which just corresponds to a single equation. Three X plus two Y equals zero. And so we obtained the solution the two with coordinates to -3. Yeah. And these are our Eigen vectors. And therefore the set S. Which is vectors V. One and V. Two for 31 and two -3. Is a maximal set of uh linearly independent Eigen vectors. Each is from a different item space since S. Is the basis of our two. Yeah. We should get him. Dude. We should threaten uh D. O. D. Officials. Dude. Yeah. We got the power of podcasting on our side. Yeah. Yeah. We got combination. Stop. Come on. Do someone's gonna get pinkeye one of these days? No, you shut up. Well, S is a basis for our two. And therefore it follows that. Our Matrix A. Sorry, one second. Because you're Ronald Mcnair the cloud. Yeah. Matrix A. Is diagonal. Izabal. Well be careful because she's made out of balloons. Yeah. And she might pop on the way over to your new place. That's good timing. Yeah. You got that in writing. Hey buddy. You want hey I got to make sure a white face paint. In fact using the basis S. A. Is represented by the diagonal matrix D. Yeah. Which is the diagonal matrix with entries 7 -4. Yeah. Uh Best friends, All of us. Okay. We're so close. Um When you guys you guys uniforms Then in part b. were given the Matrix 1, -1 2 -1. No that's it. Which we call B. But Jack in the box. That's not a clown. No, it's literally a jack. Once again, we'll find the characteristic polynomial. This is a two by two. Matrix. The characteristic point of the old delta of T is t squared minus the trace of B which is zero tense T plus the determinant of B which is mm Positive one. But notice that there are no real roots. Think of. Yeah, I guess so. But maybe technically besides you. Therefore it follows that B is a real matrix, no Eigen values and no I don vectors. It's funny, it's funny that that's the worst one and therefore it follows in particular that B is not diagonal. Izabal. Alright finally, in part C were given the matrix five negative 113 first or find the characteristic polynomial delta T. This is a two by two matrix. So this is T squared minus the trace of C, which is eight times T plus the determinant of C which is 15 minus negative one or plus 16, which we can factor as t minus four squared. Therefore it follows that the roots lambda equals four is the only Eigen value of our matrix C. Now to find Eigen vectors will subtract forward down the diagnose or C. To get our matrix M. So M is C minus four I. Which is the matrix one negative 11 negative one corresponding to the linear system. X minus Y equals zero. Homogeneous system. Yes, has only one independent solution. For example, we could take The equals 1 1 as a solution and therefore this is an Eigen vector. Especially stop, you should be looking right now but I'm gonna you're listening to Adams deal corner of C. There are no other Eigen values. Therefore it follows that the set S. Which is the Singleton set 1. 1 is a maximum set. Yeah, a linearly independent Eigen vectors of C. But notice that S. Is only one vector and therefore is not a basis of our two. Therefore it follows that our matrix C is not diagonal. Izabal luxuriously, now that I have the walls, you should, I think. Do you know what kind of mattress you fucking with? I have a full already have a full. You should clean. You're an adult, you know?

Problem questioning. So do you warn Be to Our Is big toe are serene. It's Amy. It was seen equally some staple one, perhaps and next. Sprayer and being. And that's slim. Something born X. It's weird and it's you don't so you know you one for one. It's equal to into, Oh, you want one, which is a pretty toe team Through or X plus one. Find one when she's April Brexit. Let's run it. Just exposed one man. So we put forward. Suddenly we're really lead to war. One. It went to It's squared, my sex, their industry toe X. All right, that's one. Me to you. One. Well, that's where x you not thanks Were they restaurant meals? Were there must lights there were trying to write these transportation using at the linguine, combining in or vector from scene So this can be right now one times one waas your plane. Plus, you find that script. This can be a little one times one Wellstone times Must your fun. It's spring when those can be written as your plans. One most two times. That's last three. Find it's It's play so accordingly. Using these confessions, you can say that need to be more off CNN day. It's secret to 10 wins here. I want to rent to you. So you know, for questions being or question being using the fear in our 6345 saying that Do you want on seeing a in tow ups? Evening you are. Oh, me. Hey, from in society training any? Thinks that aren't you? Want a Weren't one year deal. You won. What would you You know you want one in here? It's you. You You run this year on you. Number one You know you know you and you using the application for these two men tree Susan's We can not seem to want to be a It's important to zero euro foreign tour. You you question and seeing can you seems free for fun. And we're saying that into do you want off the s is different to you to work steaming well being. Oh, yeah. Remember of our yeezys? Well, you're in Really worry 12 And this is we're heading from here again. It's runnings. New ribbons in your own want to end you You know the application for a system nine races is able to seeps toe and since, But I want you to remember we were all the same the same so, uh, need to be worn. Oh, slur in one of those letters. Plus where once a squirrel. 2 to 1 of seven ones. That's us. We're in tow. It's a sport. Someone. Warren was lost for words. We're plus speaks. Thanks, Brian, must you? Jesus. What Must no seats X. That's where this Iran's nuisance suits to instance. What's your name? You, you know? Well, I need to do you work Seating. It's different of one times Two times Two times two minus warmly times U minus. Nearly times nice meal times. You your level, you know, thumbs. He usually sleeps. This is Mark in front of you, sir. Sense that runs nothing for you. Uh, it's you news about to you want It's not. You work too

So our first question is Is the matrix in row Ashlan form in this case? Yes, it is because we see we have leading one. So our first non zero digit is a one in each row in part B. We're looking for reduced row echelon form, which means that we need to have zeros above and below those ones. In the second column, we have three. Above are one which means No, this is not in reduced echelon for finally, we will write our system of equations represented by this matrix. Remember, the first column represents a coefficient for acts are Second Column. Why, if we had four columns would consider Z, but we don't. And so our final number is our constant. So in the first row we have X plus three y equal to negative three and then the second column we have. Why equal to five. So that is our system of equations.

So using these definitions to look at the matrix in numbers step first of all, we see that the first number in each row is a one. So in part, A Yes, this matrix is in row echelon form. So the first number non zero number is one. Then in part B, we need to look for those zeros above and below our once in this case, there are zeros above and below the ones. So yes, this is in reduced row echelon form Finally, in part C. We want to write this system of equations given by this matrix. And remember, we have x followed by why followed by the number that that is equal to so looking across the first row, we have one X plus zero y equal to negative three and then in the second row we have zero x plus one y equal to five. So we have X equals negative three And why equals five as our solution


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