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Let band let A be the matrixin the range of (he linear transfoxmaton X-Ax? Why Or Wiy nar?Is b in the range 0f the Iinear ~transtormation? why or why nol? Yes; b I...

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Let band let A be the matrixin the range of (he linear transfoxmaton X-Ax? Why Or Wiy nar?Is b in the range 0f the Iinear ~transtormation? why or why nol? Yes; b Is In the range ofthe linear transformation 0 B. No; b is not in the bacause the syster represented by the aporopriate augmentcd [ ranqe 0f Ine hneal- transtoration because Mator & consistant 0 C not in Ihe Tange ol Ine the system represented by Ihe eppropriate aupmentad matrs I5 linear trans(ormaton because tle system cotsrleni 0

Let b and let A be the matrix in the range of (he linear transfoxmaton X-Ax? Why Or Wiy nar? Is b in the range 0f the Iinear ~transtormation? why or why nol? Yes; b Is In the range ofthe linear transformation 0 B. No; b is not in the bacause the syster represented by the aporopriate augmentcd [ ranqe 0f Ine hneal- transtoration because Mator & consistant 0 C not in Ihe Tange ol Ine the system represented by Ihe eppropriate aupmentad matrs I5 linear trans(ormaton because tle system cotsrleni 0 D Yes, bi5 the range of Ihe Iineal represerted by tne appropriate augnented matru [s nconsistent transtormation because [hve systcm represented Ine aopropral? apmentud maffu %5 nconsisient



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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

In this example were provided with the transformation T that maps defector ex into a X. Our Matrix A is defined here as a four by form matrix, and we're going to be working with this particular vector B. So the question is, Is there some X in the domain are four for which t of X is equal to be? Well, we know that t of X equals B as an equation is equivalent to a X equals B, so we can rephrase the question is the system A X equals B as a matrix equation consistent to determine that we would do the following Take the Matrix A and augment it with the vector B. This would be a four by five matrix, which takes quite a while to reduce by hand. And the trouble with doing all those reductions is your learning less linear algebra when you get that bogged down. So for a matrix of this size, we're going to use either a computer algebra system or a calculator to perform this role reduction in a calculator. I obtained the role reduced echelon form of 1000 0100 then 0010 Now the fourth column gets a little bit more interesting. It's negative. Seven halves negative nine halves and 00 Finally, this is the moment of truth. The fourth column from where we augmented using B With all these air operations, we get 000 and a one that tells us for the potential or the actual pivots of theirs. Reduced matrix will be in columns 123 and four. But that tells us that the Matrix equation A X equals B is inconsistent, since the right most calm is a pivot column. Or viewed another way, the right most column is a short hand. Since we've augmented for the equation, zero equals one, which is a contradiction. So that tells us that four all X in our four t of X is never equal to be sold. Our solution is complete. B is not the image of some vector X. Let's take a moment to also explain what we've shown here, since we cannot find a vector X from the domain that results in the Vector B and the co domain, we've also shown the following B is not in the range of the transformation, t, but just to be specific. This is extra information below, and the solution is provided here

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.

Alright for this problem, we are given a linear system. A ax equals B where the reduced row echelon form of the augmented matrix a augment B equals now going down each row than across each column. 100 All right, 0 to 000 0100 3200 1400 Augment Negative 2500 So one thing that we can note is that the first column the third column have leading ones there and there, whereas the 2nd, 3rd and 4th do not. So what that tells us is that we're going to end up having three parameters in our solution. So translating that first expression into a equation give us that X one plus to actually what I'll do is, since we know that we have three parameters, we'll say that off the bat X two equals R X four equals S and X five equals t, then translating the two non zero rose that we have there we have that X one plus two are plus three s plus t equals negative too. And we have that X three plus two s plus fourty equals five now what we can dio So we can rearrange these expressions and turn them into matrices. So the meat or not, just matrices rather vectors instead, or a sum of vectors. So what we would do is we would want to go through. And essentially we are first rearranging and figuring out what x one and x three are in terms of S and T as well as their Constance. So we would have that X one equals negative too. Minus two are minus three s minus. T X three equals five minus two s minus 40. So that gives us the vector formed. First of all, we'll have our vector holding all of our constants. So we have that the constant part of X one is negative two. Then there's no constant part of X two. Should be a four by one or sorry, five by one. Rather, um so x two has no constant part. X three has constant. Part five x four has no constant part, and x five has no constant part. So is the constant part of our vector. Then we'll have our our component so we can go through and figure out the our parts so x one has negative to our we said that X two equals our lips. I should just write negative to their X four are started. Rather X two equals are so we have a one in that place and then zero for the other. Actually, you have in X three. Yes, we have zero ours in x three. None in x four x five. So we get negative. 21000 Then we would have plus s. Now we go through and find our s terms. We have three in, um, rather negative three in X one. We'll have zero and x two is x two is just are you'll have negative two in x three. That guy, we have positive one in X four from the definition and we'll have zero in x five. And lastly, we have our tea component, which will have negative one from x one zero from X to negative four from X three zero from x four and one from X five. So that gives us our solution or all possible solutions. Yeah, to that system of equations where we have further specify that r s n t are real numbers next we want to consider if the first column was replaced by 1134 and the third column was replaced by two negative 113 And we were still doing a X times be just with this modified matrix. Then what is B s? So we have, you know, a one a two a three a four a five where a one and a three have been changed. So right, a little prime on them, then A to a four and a five are the same and then we're multiplying them by the same X factor. What's that going to be? So first, what we'll do is write out what that matrix looks like. I'll pause so our modified matrix looks like 1134 2000 to negative. 113 32001400 And then we're multiplying by the X that we found in the previous part. So it's multiplied by negative two minus two ar minus three s minus T are five minus two s minus 40 s t turning our some of vectors the way that we expressed s or ex rather in part a turning that some into a single vector here. Then we have to do our matrix multiplication. So keep in mind that the way that the matrix multiplication works is that we go across, we take the cross product of each row vector with the column vector. And in this case, since we're doing a multiplication of a matrix by a vector, we go through and each component of the resulting vector is going to be the cross product of one row with the vector. So we would have the first or sorry. I said cross product, I meant dot product rather s. So we have a one dot Exe the first element a two x the second a three got x who, actually important thing to be careful about here. In terms of the notation, I believe that with the notation the textbook uses, we would right a one a two a three a four like this to refer to the row vectors. Whereas the arrows overtop are being used to refer to column vectors. I'm just going to confirm that yes, So there is a convention that the arrow overtop is used to designate that something's a vector, but in this case, when we're saying a one a two, a three and a four were saying the, um we're saying the row vectors, whereas using the arrows when referring to Indices of the Matrix, is referring to the correspondent column vectors. So that being said, we have a one dot exe a three x a four x So a one X is three dot product of 12231 one with negative to negative two ar minus three s minus t are five minus two minus 40 an s than tea. Which do you excuse My sloppy writing here going to equal s. So we have one times the first row. So we'll have It's not going to be a vector anymore. It's gonna be a scaler. We'll have one times The first row is going to be just negative. Two minus two are minus three s minus T. We have two times a second row which is just going to give us plus two are Then we have two times the third row, which is going to give us 10 minus four s minus 80 and we have three times the fourth row which three times the fourth row, which is going to give us plus three s. And then we have plus T for a to not ex and so on. What I'll do is, you know, I've gone through this in specific breaking it down and for the other elements I won't be as I won't be showing it as step by step like that. We can copy down that first element here this copy and paste that up there for the second. Um, for the second row of the resulting vector, we should get again. Negative. Two minus two ar minus three s minus T. And no, ours from the zero times are and we'll have negative one times five minus two s minus 14. We'll have negative five plus two s plus 40. We'll have two times s or plus two times s. Rather, we'll have plus four times t. Yeah. Then we'll have three times negative. Two minus two ar minus three s minus t. So we'll get negative. Six minus six R minus nine s minus three t. Then plus zero times are then plus one times five minus two s minus 40. Then we'll have for the next row we'll have four times Negative. Two minus two ar minus three s minus t. So we'll give us us will give us negative eight minus eight are minus 12 s minus 40 than plus three times five minus two s minus 14. So it will be 15 minus six s minus 12 t which when we simplify that down, we should get moment here. Lee, pull it up on neither screen. We should get that. The most simplified form of what we got there. It's negative. Two plus two times five minus two s minus 40 then negative seven. It's too are actually. One moment I realized, as I was starting to copy it down that I could have simplified it further. So the most simplified form actually be eight minus running alone. Battery eight minus four s minus 80 than negative seven minus two are plus s plus 70. Negative one minus six are minus 11 s minus 70 and seven minus eight are minus 18 s minus 16 T. Where again, R S and T can be any real number


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