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11_et b =and let A be the matrix in Exercise 9.Is b inthe range of the linear transformation X F > Ax? Why or why not?12. Let b =and let A be the matrix in Exerc...

Question

11_et b =and let A be the matrix in Exercise 9.Is b inthe range of the linear transformation X F > Ax? Why or why not?12. Let b =and let A be the matrix in Exercise 10. Isb in the range Of the linear transformation X F > Ax? Why Or why not? In Exercises 13_16, use a rectangular coordinate system to plot [2], = -2 u = and their images under the given transfor- mation T . (Make a separate and reasonably large sketch for each exercise.) Describe geometrically what T does to each vector X in R

11_ et b = and let A be the matrix in Exercise 9.Is b in the range of the linear transformation X F > Ax? Why or why not? 12. Let b = and let A be the matrix in Exercise 10. Is b in the range Of the linear transformation X F > Ax? Why Or why not? In Exercises 13_16, use a rectangular coordinate system to plot [2], = -2 u = and their images under the given transfor- mation T . (Make a separate and reasonably large sketch for each exercise.) Describe geometrically what T does to each vector X in R? . 13 Jro) = [- _8][:] 14. T(x) =[6 S][~_ 15 To) = /0 'I#~



Answers

Let $\mathbf{b}=\left[\begin{array}{r}{-1} \\ {3} \\ {-1} \\ {4}\end{array}\right],$ and let $A$ be the matrix in Exercise $10 .$ Is the range of the linear transformation $\mathbf{x} \mapsto A \mathbf{x} ?$ Why or why not?

Okay, in this caution that given metrics is one my next city five minus five. Zito one minus two D Fight. Tu minus four for minus four. Okay. And the solution B is one. Sorry. Minus 11 and zero. And we have to solve the equation. X equals to be. In other words, try to find X such that X equals toe Be so we will write documented metrics for the given system. And this will be one zero to minus 314 53 negative four minus 55 minus four. And the solution is minus 110 Okay, now we will do some operations in here. We will multiply, are one with minus two and and tow and to Artie. And this will be okay. May one minus three, five minus five. Minus one and 01 My necessary five one. Andi, this will be zero, 296 six and two. And now? Yeah. Next operation. Multiply second roll with minus door. And at the all three on. This will be when play in this city five minus five, minus one. Tzeitel one minus three, five one on 000 My next 40 So from the last metrics, we can say from the third row, we can see minus four x four, it was 20 That is X for request 00 And from the second row, we can say x two minus three x three was 21 that is X two equals to one last three x three. Okay. And from the first rule, we can see x one minus three x two last five X three equals minus one. And now we will put the value off x two. In the terms of X three, that is X one minus three minus nine X three plus five X three equals to minus one. Okay. And now, No, it was funny for the store. Excellent calls to minus one plus three plus four x three this x 21st two plus four x three. Okay, so there is infinite number of solutions. For example, to be so be is in the range or a X. So we can say our answer is there is there is in finite number off solutions for it was Toby. So B is in There is a change or X S X. This is our final. Honestly, thank you

So we're off to use the same matrix is in number 38 again, Um, we're encouraged to use Matt Lab two softies. So I, uh, typed in the augmented matrix off a in the factor B, which is here on the right and took the reduced row echelon formed. So Matt lab did all the reduction for me. And we get this matrix here, which, um, I called our So this is the right hand side so you could throw in a little hot in here, so Ah, we need to see if this matrix is in the range of a So in order to do that, we take the reduced matrix and see if it gives a consistent system. So let's read out the equations we have. So this gives us that x one is equal to negative Three forts x for minus 5/4 x two is equal to negative 5/4 x four minus 7/4 herself. 11 forts, um, lastly, Way have x three is equal to one and 3/4 which is 7/4 x for plus during 1/4 which is 30 over four and X for it's free. So this is a consistent system. Um, we can choose any X four and get a corresponding ex well, next to x three. So if we want to write the general form of, uh ah, vector, that is ah, solves a system. We can do so. So we would have negative three force negative. 5/4 7/4 and one. So we have this matrix multiplying x four. And then we would add to that the matrix or the vector Negative five force negative. 11 4th 13 4th and you're so this is the form of a matrix that is under that image.

Were given a Matrix A and column vectors B one B two, which we're told me when in B two form the basis Be and a Is the Matrix for a transformation from X to a X and were asked to find the bee matrix for this transformation. So to do this, we're going to take p to be the column vectors. Be one matrix of column vectors. B one b two. It's The Matrix with Colin Victor's to negative one in 12 Then we have following the example in the book Be Matrix is given by P in verse a p. We can calculate p inverse pretty easily using p, using our formula for the inverse of the two by two matrix. So the determinant of P is four minus negative one, which is five. So we have 1/5 times I'm sorry times and then we have Well, we're going to want to switch D and A so we have to and then negative be so negative. One negative sees a positive one, and then a so too times. And then we have our matrix A which were given us matrix of column vectors. Three negative 14 negative one and finally, p again. So to negative. 112 And so I get 1/5 times to negative 112 and then calculating a P. I get three times two is six minus four is too. Three times one is three plus eight is 11 and negative too. Plus one is negative. One negative one minus two His negative. Three 1/5 times multiplying again. I get four plus one is five. 22 plus three is 25 tu minus two a zero and 11 minus six is five. And so distributing the scaler I get the matrix one five 01 And so this is Thebe matrix for this Trans.

Okay, So in this problem, we only need to apply our the're, um your mate, you know, textbook. So it says if we have the Matrix P Well, in this case, our matrix peas. Just a span, not B one B two. So Matrix B. We know this matrix. Then our bi matrix will be peeing burst. Time say times P. So we just need to find out the product. So, Pete, members, by my calculation, I won't do this step by step. Just a house. Just write down this of embers. It's 11 I think the 23 now Matrix A is next to one for next 2 to 3 and matrix P. It's just a three negative 1 to 1. Okay, So as first find, it was too part of the first product. So first entry, Nick one times. Like to warm. Plus two two times one. So that's negative. Three and one plus one times four plus one times three. So that's seven and two minus six. So it will be negative for And the last entry we have connective aid. That's nice. So it's one now, three negative. 1 to 1. Okay. Now for this to matrix. We'll search for these two missions, since we have next of three times three active and I last 14. So it will be five and three. Ah, seven. So he's 10 and next to 12. Us, too. So selective. 10. And we have four plus one. So he's five. No, just put the skater into our matrix. So we have one too connected to and one So this is


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