with-t-defined-by-tmathbfxa-mathbfx-find-a-vector-mathbfx-whose-image-under-t-is-mathbfb-and-determi

Question

With $T$ defined by $T(\mathbf{x})=A \mathbf{x},$ find a vector $\mathbf{x}$ whose image under $T$ is $\mathbf{b},$ and determine whether $\mathbf{x}$ is unique.
$A=\left[\begin{array}{rrr}{1} & {0} & {-2} \ {-2} & {1} & {6} \ {3} & {-2} & {-5}\end{array}ight], \mathbf{b}=\left[\begin{array}{r}{-1} \ {7} \ {-3}\end{array}ight]$

Michael Jacobsen · Accepted Answer

In this example, we have a linear transformation. T that maps are three back to our three. It&#x27;s going to be defined by t of X equals eight times X, where here are matrix. A is defined now for this transformation. Let&#x27;s also let B B the vector set equal to negative 17 Negative three. Now the questions are for what? X in our three is t of X equal to be so let&#x27;s determine what the solution X is in this equation now. T of X recall is equal to eight times X, so the question could be rephrased As for what X is a of X equal to be. But as soon as we write our question this form, we can get the augmented matrix and begin row reducing to solve for X. So let&#x27;s first take the Matrix A. It&#x27;s one negative to three 01 negative, too, and negative to six negative five then augments with the Vector B, which is given to be negative 17 and negative three. So now that we form the Matrix, the next step is to begin row reduction. So I&#x27;ll copy Row one and our objective is to obtain the role reduced echelon form. So for the pivot one obtained here, we need to eliminate all entries above and below. So in this case, we need to eliminate to negative two and three. So let&#x27;s multiply Row one by to add it to row two and we will get 012 and a five. Next, if we multiply row one by three or negative three, rather at two. Row three will obtain zero negative too. Ah, positive one and a two. So now we have a pivot here with zeros below. For this pivot, we have zeroes zero above, but not below. That means on the next step, we&#x27;re going to be needing to eliminate that value. Negative too. So let me copy Rose one and two. Next. The first row is 10 negative to negative one. And our new row two is your A 1 to 5. Now, to eliminate the negative to multiply row, to buy positive too. And add to row three will obtain zero zero ah, five and looks like there&#x27;s a tiny mistake here. This should have been a zero. And that means this entry here should be equal to value of 10. So now let&#x27;s check our pivots at this stage here in here, we have to pivot positions with zeros above and below. And our next pivot position will be this highlighted. Five. We need to make the 501 So for the next row operation divide Row three by five we&#x27;ll obtain 0012 and let me copy the rest. So we have 0125 and 10 negative to negative one. Now, on the next step, from this pivot position we see here, we need to eliminate the entry to and negative to found above. So this time let&#x27;s start by copying row three, which is your 012 Then multiply row three by negative to add the results to row two, we&#x27;ll obtain 01 zero and then a one for the next operation. Multiply Row three by positive to this time and added to row one will attain 100 three. Now, at this stage, we have the identity matrix here augmented with values on the right, and this tells us that X one must be equal to three. X two is one and x three is too So here&#x27;s our conclusion. Then if we set a vector X to be equal to 312 then the image of X under T is equal to the vector B, and this solves this problem.

Michael Jacobsen · Answer

In this exercise, we have a transformation t that maps are three into our to buy the royalty of X equals E X. Where are specific matrix A. Is this two by three matrix? Let&#x27;s suppose next that a new vector here be, which is equal to negative to negative two was given. Then we might wonder for what vector X will we obtain? Be so in other words, for what X in the domain are three is t of X equal to this vector B in the CO domain or two. To solve this recall that t of X is equal to eight times X. That allows us to rewrite the question t of X equals B as eight times X equals B. This is helpful because it reminds us that now that we&#x27;re considering the system of equations, we can solve X by going to an augmented matrix. First place A, which is one negative three negative 57 and 75 and augment with the vector B that&#x27;s given here. So we&#x27;ll have negative two and negative to inserted in here. Now we begin row reduction copy Row one, which is one negative 57 negative too If we multiply Row one by three Editor Row two will attain altogether zero negative eight negative 16 and a negative eight. Now we need to make this position a one. So it&#x27;s divide row to buy negative eight. We&#x27;ll have that. This is row equivalents 20121 Then copy in row one, which is one negative. 57 negative too. Now, for this pivot position, we need to eliminate the entry negative five found above. So if we multiply row two by five at Euro three will be able to do just that. First, let me copy the road to it. 01 21 Then multiply it by five and row one will be one zero. I&#x27;m noticing that this entry should have been a negative seven. So we should have negative seven here, here and here. Now, when we multiply by five, we get 10 minus seven for a value of three multiplied by five here Well of five minus two, 43 as well. Now we can express what the solutions that is. Recall thes air columns, corresponding to x one x two x three and we&#x27;ve augmented for the right hand side of the equation. So we can say X one is equal to negative three x three plus three since we&#x27;re treating the augmented portion as an equal sign and solving for X one. Likewise, in Row two, the coefficients 1 to 1 tell us that X two is equal to a negative two x three plus one. From this information, we&#x27;re ready to express X So if X is equal to and X three times the coefficients for X three, which are negative three and two negative three and negative to rather plus this constant matrix +31 or constant vector where X three is in our then will satisfy that the image of X under T is equal to be. Now we have a free variable here. So since x three is free, there are in fine nightly many such x we found one such that t of X will be equal to be, for example, if we let X ray be one, then add the two vectors here together and we have one solution. But X ray could also be zero. So three and one is also a solution, and so on

Ashley Boni · Answer

So we want to find about your ex such that a X equals B. Um, or this is our matrix A. And this is our vector B. So we want to row reduce the augmented matrix. So first, um, I want to make these two value zero using the first equation. So first and third equation are gonna remain the same. So now let&#x27;s change the second equation. I&#x27;m gonna take negative three times the first equation and add it to the second. So we get a zero here. Negative three times negative to a six. Minus four. Gives us too negative three times wanted negative. Three plus five is, too. And then we have negative three times. One, which is negative. Three plus nine gives us six. And we&#x27;re gonna do a similar thing for the last equation. Take three times the first equation and add it to the fourth. So one times 33 minus 30 Uh, times three is minus six plus five is minus. One times there is three. Minus four is minus one on time. Serious, three minus six is negative. Three. So now our first columns taken care of. Um, next, we have In the second rule, we have 02 to 6. So we can, um, cut everything in half. That&#x27;s one of the rules were allowed to do with, uh, producing matrices. So instead, I&#x27;m gonna write 0113 So now I&#x27;m going to use this new row two thio, get rid of these two values. So I&#x27;m gonna multiply this new row to buy negative one and add it to this room surgery here. So I have a zero. I have zero, um, times negative one. And at one, a zero three times naked of Juan is negative. Three plus three is zero. And then I&#x27;m gonna do the same thing for the last equation. Um, I&#x27;m gonna multiply this row two we have in our matrix here. Bye. Positive one this time and add it to the Lord. So that makes this negative one a zero. Um, again makes a negative on a zero. And ah, we have three postnegative three, which is zero. So since we have rows of zeros, um, we can expect to have a free variable, So let&#x27;s see. Um, what these equations say? First equation says X minus two. Why? Plus z equals one second equation says why plus C equals three, and that&#x27;s all we have. So the second equation let&#x27;s rewrite it as why equals three minus C and substituted in the first equation. So we have X minus, too. Why so over not replace that with three minus C plus Z equals one facility, right? X minus six plus two Z plus Z equals one. So finally we have X plus, they&#x27;re easy equals seven. So, uh, one of these variables has to be free, so it&#x27;s juicy to be free. Ah, so our general solution iss going to be our vector X iss equal to seven minus three z three minus c access the expression we had for why and Z is free so we can rewrite this as, um 730 plus some vector times are free variable. So first equation has negative three z Certain put minus three. Here. Second has a negative one z. So negative one and last is one time. See? So here&#x27;s our expression for our General Vector X. If we want a particular solution, um, we can pick Zee to be some regular value s. So let&#x27;s just take the equals one. So if Z equals one, um, we just have 730 plus negative three negative 11 which gives us 4 to 1, so that&#x27;s one exit would work.

Ashley Boni · Answer

So we want to find a vector X such that eight times X is equal to be. And to do this, we row reduce augmented matrix, Um, of a would be So we have this matrix. And first I want this to be a zero. I&#x27;m gonna take negative three times the first row and added to the third row. So that&#x27;s going to give me no change in the first row. No change in the second row and in the third row, Negative. Three times 30 nine minus five is four negative. Six minus nine is negative. 15 and negative 18 minus nine is minus 27. So next up, I want this to be a zero. So I&#x27;m gonna take negative one times the second equation and add it to the third equation. So first row doesn&#x27;t change. Second row doesn&#x27;t change on. And the state zero, this becomes a zero. Because what we want, um, we have negative 16 plus. Ah, 16 miles 15 is one and negative. Four times negative. Seven is 28 minus 27 is one. So, um, last step. I&#x27;m gonna take this third equation, and I&#x27;m gonna use it to make these two zeros just to simplify things for us. So last year I was going to stay the same and let&#x27;s do the second equation first. So we&#x27;re gonna take four times the third equation and add it to the second. So that gives us zero here and four times on his four minus seven is negative three. And now I&#x27;m gonna take negative two times a second equation or the third equation and added to the first to make this guy zero. So he becomes zero. Nothing happens to the 1st 2 and negative two times. One is negative, too. Plus six gives us four. So here&#x27;s our system. We need us off. So this tells us that X minus three Y is equal to four. Why equals negative three and C equals one. So our general or vector X, that we want to find We already know Y and Z and X is equal to four plus three times. Why wise negative three. So this is four minus nine, which gives us negative five and we could see one solving these equations. There is no other option. There&#x27;s only one solution. So the solution is unique.

Jack Chen · Answer

in you miss question. We want to verify the function X, um, self, the, um, system for the so the system for the east given by ex are the vectors xs security vehicles to a Times X Crosby and the weekends. So in this case, X equals two. Um, takes one x to a column vector. You can just write it out. What&#x27;s the system for? The so that the idea of a perfect X one? Because toe eight times eggs. So it&#x27;s one. So x one minus four x two plus the first entrance and be so it&#x27;s minus two times call sign off P process sigh of tea and the second ODI will be X to prime you close to minus three x one cross, um two x two in the plus seven. Signed p my, um, press to consistently. So this is the system for these. Now we want to verify that the following you ah function actually soft this system for these. So we want to verify, uh, this given function. Um, reaches are four times e to the minus two T crossed twice off sine of t and then the 2nd 1 is ah three e to the minus two t miners Society and kill yourself this system for these. Okay, so just verified both equations for the 1st 1 LF inside because toe x one prime sex one primed by the, um Rose off the the relative, especially the general Ah, first term is given by minus eight. You to remind us to tea The next time is two times called softy and the right inside because to excellent minus four x two miners to times co sign he crossed sigh of tea and the parking x one x two Um, so once we plug in X Y extra, we have four meat to the miners to a t press uh, twice off sign key minus 12 e two reminders to t press for Kasai t and the miners twice off Kasai t in the minors. Twice off society we kicks which is exactly you close to 11 side after this canceling right and we do the same thing for the 2nd 1 left inside because the because to the derogative off x two with respect to t, which is miners six you to the miners to t pross sigh of tea, right, so the right hand side. You just tow minus three x one press two x two class seven Safety press to costs I of tea. Then just replace x one x two by the functions we have. So it&#x27;s a minus 12 e to the minus two key minus six side off the This is minus three x one press six e to the minus two T minus two co safety. This is, uh, twice off x two and process seven side T plus to society and, uh, after the cancellation. Mom. So this consults this and they&#x27;re combined the on the same terms. So we have the right hand side, close to 11 side and the way verified. X one extra indeed solves this system of equations.

Bobby Barnes · Answer

So you want to find these three vectors at this point over here? Uh, as you can see, we just really need to fill out these formulas. Let me throw these off side, and we&#x27;ll start with the tangent unit. So we need to find the derivative of our So there&#x27;s going to be our prime of t is equal to So this is just going to be well, we take the drove of each component wise would be to t and then to t squared and then one, and then we need to take the magnitude of this magnitude is going to be what we just square each of them and then add everything up. So it would be four t squared was for t to the fourth, uh, and then plus one square, which just one. And then we take the square root of this. So we plug that all in over here we get our unit tangent is going to be two t, uh, to t squared one and then all divided by the square root of four t squared plus or T to the fourth plus one. Yeah. Now, um, well, we need to figure out what is T in this case. And if we come up here and look well, our last position is just t. And that says it&#x27;s just equal one. So we&#x27;ll just come over here and plug in T 0 to 1 to get it at this point. Should be t of one. So that is going to be 2 to 1. All over would be four plus four plus one. So that would be eight plus one, which is nine. Route nine is three. So we just divide three into everything. So that would be two thirds. Two thirds, one third. So this is our unit Tangent at one now to find the normal what we need to take tea and take the derivative of it. So I&#x27;ll first need to distribute this each of the positions in order for us to be able to take this derivative. So it&#x27;s going to look a little messy now, unfortunately, so we&#x27;ll have four t square plus four t to the fourth plus one all over two teeth. Then we have two t squared all over that magnitude again. And then one over this magnitude. Okay, so now we can attempt to take the derivative of this. And hopefully we don&#x27;t mess that up because that is kind of involved. Especially let me just this up. Scoot it down here. So, the derivative, we&#x27;re just gonna take it component wise for each of these. So we&#x27;d end up with t prime of t being equal to. So for the first one, we&#x27;re going to use quotient rule. So it would be Whoa, Do you hide the derivative of two t is just too, um, minus high. So to t take the derivative of this. So actually, let me write that off on the side. So D by D t of square root of 40 square plus 14, 4 plus one. Um, so this would be to a one half power sort of be one half, and then the new power, since it&#x27;s one half would become negative one half, which is the same thing is just one over that square rooted. So four t square plus four t to the fourth plus one. And now we can go ahead. And, um, what do we want to do? Take the drift on the inside? I don&#x27;t know why it took me so long. to figure out, but it would be eight t and then was 16 t cubed. And so now we can get this to to counsel with that, that becomes a four that becomes an eight. So the derivative of this looks like it will be, um So I&#x27;ll factor that four out So it would be or t was to t cubed all over the square root of Fourty Square plus 42 the fourth plus one, um, art, but now will plug that in over here. So this would be four t plus two t cubed all over the square root for t squared plus four t to the fourth plus one and then all over, um, the denominator squared. So that would just be four t squared plus four t to the fourth plus one. Okay, um, so let&#x27;s see if we can simplify this at all before we go into that second position. Well, I could multiply the top and bottom to rationalize this, And in doing that, we would end up with so this would be squared. So it be four t squared plus for T to the fourth plus one. And there would be times two, and then we&#x27;d have minus two t. Or actually, if I distribute the two T, that would give us 18. Uh, and then that would be eight t squared, plus 16 t to the force and then all over. And then we&#x27;d have to multiply the denominator by that route. So now would be to the three halfs power. Mhm. Okay, uh, and then if I distribute this to 30 of us eight t squared plus 82 the fourth plus two, and then the negative would be minus 80 square in minus 16, T to the fourth, and then we can go ahead and combined are like terms here. So the T squares cancel out, and then we just have that, too. So it&#x27;ll be too two minus eight t to the fourth, because we can combine those two t to the force, and, uh, I don&#x27;t see anything else we can do. So then we just go ahead, and but what we have in the denominator again, So, yeah, that was pretty annoying to write out. And we only have to do that, um, two more times. So I&#x27;ll actually do is take this and I&#x27;ll do it off on the side again. So let&#x27;s write that one up top here. So we have d by d t of two t squared all over the square root of fourty square plus or T to the fourth plus one. So again, we have to do quotient rule. And we already know what the Denon the derivative of the denominator, is going to be. So, um, we first strike this outside shall actually write it down here. 40 square plus war t to the fourth plus one. Then the derivative of the top is for tea, then minus those split. And the derivative of this is going to be for T plus two t cute all over the square Root for t Square plus four t to the fourth plus one all over what we have and the denominator. Uh, and so that would be for T Square plus four t to the fourth plus one. Okay, let me down Song. Um, So again, I&#x27;ll go ahead and get rid of that square root like I did the last time. So then it&#x27;s going to be or t square plus four t to the four plus one times for teeth and then minus. So that is going to be eight t squared. That would be eight t cute. Uh, plus 16 t to the fifth and then all over again. That denominator is now going to be, uh, squared. So it will be four t square plus. Or actually, uh, this is where Why did I after before? Hold on. Oh, because I was for something else. I don&#x27;t know why I Rufete so this should have nothing down there because I should have squared it 40 square plus four t to the fourth plus one because this should have the same denominator as the other one. So this should be to the three house power. Now catch. Let me pick this up. Then we can scoot that down, so that should go here. Um, and then I don&#x27;t think I have enough room, so I&#x27;ll just do dot, dot, dot for that. And then we can go ahead and combined are like terms. So let&#x27;s go ahead and distribute that fourty sewed beef 16 t cute and then plus 16 t to the fifth plus 40 and then minus 80 cute minus 16 t to the fifth. Um so these will cancel out, and then we can go ahead and combined those cubes. So that would be to my s a t to the fourth all over Fourty Square plus 42 4 plus one, raised to three half and then at 14, just days 14, and then it would be positive 80 cubes would be 80 cubed plus or t all over for Tea Square plus or T to the fourth plus one raise to the three halfs. Okay. And we&#x27;re almost done because we just have to take the derivative of this last part here. So I guess I should just do this here because I don&#x27;t see me actually coming and filling these in. Well, the only real difference between this is it&#x27;s just reciprocated. Um, so we could go ahead and just use chain rules. So let me go ahead and do that down here. Mm hmm. My home. So, yeah, I just kind of scroll down a bit and write it out. So d by d. T of one and actually all right it as or t square plus or T to the fourth plus one. Raised to the negative one half power So now this is going to be so We use power organs would be negative one half and then that would just become negative. Three house to be fourty square plus four t to the fourth plus one to the negative three halfs. Then we take the derivative on the inside, which is going to be eight t Well, not 80 square, but just 18. And then plus 16 t cute. And then we go ahead and cancel this too with, uh, and that again. Then we can just go ahead and collect all those terms. It would be negative or T was 80 cubed all over now. Four t squared plus four t to the fourth plus one raised to the three house. Okay, so now let&#x27;s go ahead and pick this up. Now we can put it right here, okay? And that is only the derivative of our unit tangent. That is joyful. And at this point, we can go ahead and find the magnitude of this. So if we get magnitude, we&#x27;re just going to square all of these animal together, so this would be to minus 82 4th square. So if I square everything in the denominator is just gonna become what we have there. Cubed. So before t Square plus for T to the fourth plus one. Cute. Um, actually, they all have the same denominator, so I&#x27;m just going to write that once, so I don&#x27;t have to write it out 1000 times. So mere race this mhm looks like it&#x27;s flag in the little. So yeah, I&#x27;m just going to make this as one big fraction since again they&#x27;ll have all the same denominator. So would be to my s 84th squared plus eight t ah, cubed plus 40 squared And then plus right, So I do something weird. Well, both of these should have been negative here, so that&#x27;s a negative. Also, because the negative on the outside should have distributed to everything there. Uh, and then that would be negative for T minus 80 Cube squared, and then it would be all over again. The denominator squared, since I&#x27;d have to square each of those when I was adding all this up so it would be four t squared plus four t to the fourth plus one. Cute. And then we have to take the square root of all of this. Okay, Now we need to just expand everything there and actually, let me just get rid of this. We really don&#x27;t need that anymore. And, um, if we expand all of this out in the numerator, um, this is just going to be perfect square and yeah, so it should be or minus 32 t to the fourth plus 64 t to the eighth and then plus 64 t to the sixth loss. So it would be 60 or t to the fourth, right? And then plus 16 t squared and then here, Actually, since I am squaring this, um, I can pull that negative out, and it would essentially just become positive, so we don&#x27;t need to worry about that anymore. And then we would get 16. He squared, plus, um 32 64 t to the fourth, and then plus 64 t to the six. And then again, all over this denominator, meet the or plus one cubed. Um, actually, since the by normal vector, um, we don&#x27;t necessarily need to get that exactly, because it doesn&#x27;t depend on anything else. We could just go ahead and plug everything into this. Um, I guess we actually didn&#x27;t need to find this exactly. I was thinking of how, like, we couldn&#x27;t just use what we had before, um, to do this, which I mean, I wish I would have thought of that earlier, but, uh, yeah, let&#x27;s actually get rid of all of this because we don&#x27;t really need it. Um, and I&#x27;m just going to plug in one, and then we use that to find the magnitude. Uh, so yeah, sorry for attempting to go through that, but Yeah, I just kind of thought about how we could do this. Um, because again, I, for some reason, thought we needed that. Uh, yeah. So let&#x27;s plug the sense that would be just to minus eight over and then be four plus four plus one raise to the three halves over here. It would be eight plus four over. Well, that would just be nine raised to the three halfs. And then over here, it&#x27;s going to be negative or minus eight over nine to the three halfs. This should be TF one. So that would give negative six over. Uh, well, nine to the three Halfs should be 27 because it would be, like three cubed. Uh, this is going to be 12/27 then this is going to be negative. 12/27. All right? And now this is something that I would much rather find the magnitude of. And so now we can square each of these components to be negative. Six squared plus 12 squared, plus negative. 12 squared. And then this is all over. 27 squared, and we square root the whole thing. So this is, uh, 36 plus 1, 44 plus 1 44. And, um, so 3. 20 for over 27 squared. And if we square both of those 3 24 is actually 18. Would be 18/27. Well, okay, So our normal sector in T again was supposed to be t prime A t over t prime of T magnitude. But since we just want this at one, just go ahead and plug in one. So let&#x27;s come up here, so I&#x27;m just going to pick this up. I don&#x27;t have to write it out again. So this is going to be this. And then we divide by 18/27. Okay, um, so this would just become negative. 6/18. 12/18. And negative. 12/18. Um, and then I can just divide the top bottom of all these by six. So that is going to be negative. One third. Two third, negative two thirds. And so that&#x27;s in a one. And lastly. So actually, let me come up here and hold this down so we can try to collect everything together. It&#x27;s all need this or the next part. So let&#x27;s come up here. So to get the last one we need to do our by normal, we need to do the unit tangent with the normal cross product. So, um, we want b of t. And this is going to be almost order again. Uh, t then in. Okay, So it would be t of t crossed within of t. So to do this cross product, we have i, j and K. And then we put t first. So it would be two thirds. Two thirds, one third, and then we would do negative one third, Two third negative two thirds. And actually, this is going to be at e one, all right. And Now we can go ahead and just take our determinant. However, I normally do, um, this method for it. So I rewrite the first two, the V i two thirds negative. One third j two thirds. Two thirds. So I do my downs here. First I multiply those together, Adama. So it would be, um, two thirds times negative two thirds, I plus one third times negative. One third jay, and then plus two thirds times two thirds. Okay. And then I subtract going up the column. So let me scoot the silver. So minus. So first, we&#x27;ll just write negative one third, two thirds k so you can see how I&#x27;m going up the column, then Plus two thirds. One third, uh, I So two thirds, one third eye. And then it would be negative. Two thirds. Two thirds j. And that&#x27;s going up like that. Uh, Now we just multiply. All these, uh, combined are like terms. Let&#x27;s see, um, this here would be for or negative four nights, I That&#x27;s going to be minus minus 1/9 J and then plus 4/9 K. And then this would be minus so two nights negative two nights. Okay, um, then plus two nights I and then minus four nights. J uh, actually, let me just go ahead and make sure I didn&#x27;t do anything weird with this first. Yeah, everything looks fine. Uh, I think I wrote everything down. Right now, we just go ahead and add everything up. So distributing this negative, so that would be positive. This becomes negative. That becomes positive. Um, so eyes are going to become negative. 69 the chaise are going to become 3/9 and then the K&#x27;s are going to become 69 which we can go ahead and simplify it down to negative two thirds, one third and two thirds. And so then this is going to be our by normal at one. Mm. Um, so let&#x27;s put all of these in one place now. Yeah, Mhm. Because we&#x27;ve been doing this for a while. We actually even hit the page length. We have just how much work we had to do. Yeah. So these are going to be those three vectors that they asked for Just a ton of work. Um, wish it wasn&#x27;t as difficult of a process, but unfortunately, that&#x27;s just kind of what we have to do.

Problem

With $T$ defined by $T(\mathbf{x})=A \mathbf{x},$…

Problem 3 Medium Difficulty

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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Linear Algebra and Its Applications

Grace H.

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Anna Marie V.

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Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications