use-a-rectangular-coordinate-system-to-plot-mathbfuleftbeginarrayl5-2endarrayright-mathbfvleftbegina

Question

Use a rectangular coordinate system to plot $\mathbf{u}=\left[\begin{array}{l}{5} \ {2}\end{array}ight], \mathbf{v}=\left[\begin{array}{r}{-2} \ {4}\end{array}ight],$ and their images under the given transformation $T .$ (Make a separate and reasonably large sketch) Describe geometrically what $T$ does to each vector $\mathbf{x}$ in $\mathbb{R}^{2}$.
$T(\mathbf{x})=\left[\begin{array}{rr}{-1} & {0} \ {0} & {-1}\end{array}ight]\left[\begin{array}{l}{x_{1}} \ {x_{2}}\end{array}ight]$

Michael Jacobsen · Accepted Answer

In this example, we have a linear transformation T that takes this two by two matrix and multiplies the vector X from the domain, giving us a vector x or eight times X in the co domain. Also, we&#x27;re provide with vectors you envy. Let&#x27;s start out by sketching where u and V would be in the x one x two plane. So for X one, we go distance of five and up to So this is location of vector you the vector v goes a distance of left to then down for So here we have the vector V. Now let&#x27;s see what happens when we apply the transformation t to these vectors. For example, If we take t of you then Ray in the Matrix a first which is negative 100 negative one Amel to play by you, which is the vector five to We&#x27;ll obtain altogether negative five for the first entry and negative two for the second entry. Since the effect of this matrix here is just a multiply both entries of whatever vectors given by negative one. So for the image of you, we go to negative five. So here&#x27;s 345 and down to. So this is where t of you ends up where we started with you being here. Likewise, let&#x27;s also calculate the image of t of the So we ended up here and it is the make the vector negative 24 So this will be negative. 100 negative one times negative 24 But we also know that the effect of this matrix multiplication is to multiply the vector by negative one. So we have to negative four. So if we go down, go our distance or right to down four. This is where the vector t of V ends up. So all together, for our view, when it&#x27;s in quadrant one t of you throws it all the way to the third quadrant. But for V, that&#x27;s already in the third quadrant. The mapping sends it to the image TV in the fourth quadrant. And so this is the effect of that transformation. T on these given vectors

Michael Jacobsen · Answer

In this example, we have a transformation t that&#x27;s given and the result of tee times X is to take this particular matrix and multiply X So for the two vectors u N V Let&#x27;s determine where the vectors are and where they&#x27;re images would end up, given this transformation t So I&#x27;m going to start by plotting you, which is at a distance of five to the right and up to. So this is where we place our vector you. But if we calculate t at you, that&#x27;s equivalent to the matrix 0001 multiplying the vector five to and in this multiplication since the first rose entirely zero we obtained zero in the first entry of the image and the second entry will be too. So we had a height of two. And the effect of this transformation is to send the vector you this way until we reach the X to access. And this is where t of you ends up. Well, that should be the same effect when we look at the vector V. So it&#x27;s first plot v He is going to be negative too. That up four units on the x two direction. But we calculate TV. We know that air effect. The effect is going to be the same. The first coordinate here is wiped out, resulting in a zero. And we get the second cornet still four. So that tells us we go a distance of zero in X one direction and go to four in the X two direction. So if this is the vector V effective TFG is to send it back to the X axis or excuse me ex to access. So this is TV here. In fact, it doesn&#x27;t matter which vector. Suppose this is our vector W that we picked for this Pacific transformation. What we effectively are doing is zeroing out the X wearing Cornett and sending everything to the X to access. So this is the overall effect of the transformation T.

Ashley Boni · Answer

So we have the function you X equals the matrix 0110 times the vector X So we want a plot you be and their images. And for this transformation Ah, so efforts. Let&#x27;s find what t of you and t of e r so t of a victor you iss equal to 01 10 times our metrics you, which is five to. So we take zero times five plus one times two, which gives us two. Then we take one times five plus two times zero, which is five. So we can see our matrix has flipped. Um, first and second coordinates. So let&#x27;s do this again with T o V. So we have the same matrix times our vector V, which is negative to four. So zero times negative to post one times four gives us for and one times negative to post your terms for give us negative too. So I can see again we switched coordinates. So now let&#x27;s go ahead and plot everything we have s o. We need our why? To go up to five and down to minus two. Our excess to go Rome. Negative too. Up and 25 surgery. A little small there. That&#x27;s okay. Um, so it&#x27;s first plot you envy and red, And then we&#x27;ll plot their images and blue. So you woz 52 So we&#x27;re just gonna connect that point to the origin and ve iwas negative to four. So, again, what was Connect that back the street as possible. Eso this 1st 1 it was you on V. So now let&#x27;s plot t of you on TV. So t of you, we found was 25 Joe&#x27;s right here. And TV. We found to be four negative too s so we can see that since we had a negative coordinate envy. It changed the quadrant. Ah, but since you was both positive, it remained in the same quarter.

Ashley Boni · Answer

Okay, so we have a function t of X that takes our vector and ah, right multiplies it by a matrix with points lives on the diagonal. So we want to see graphically what this does to vectors. So first we want a plot you be t of you and TV on a graph. So first, let&#x27;s find out what t of you and TV is so t of you iss our matrix times the vector u which is given to be five to. So we do this. Ah, we get 0.5 times five, which is two and 1/2 plus zero times too, which is zero. So two and 1/2 plus zero is two and 1/2. Um, second row 0.5 or zero times 50 plus 0.5 times two, which is one and now t A V. Let&#x27;s do over here. V is minus two for so 0.5 times. Negative too is negative. One plus zero was negative one zero plus 00.5 times. Or which gives us two. So now let&#x27;s, um, photograph this. So we need thio Go up to five um, X axis and we didn&#x27;t go to four on why? And go to negative too. Okay, so first, let&#x27;s plot our original vectors. U and V in red. So you is 52 So we&#x27;re going to the right five and up to and we&#x27;ll start from the origin. So this is you. He is. Ah. What? Negative to four Negatives to four is right here. Connect that to the origin. So this is our rector V&#x27;s. And now let&#x27;s plot their transformations in blood. So t of you and TV t of you is two and 1/2 comma one. So here&#x27;s two and 1/2 one is right here. So here is our t of you. TV is negative one too, which is right here so we can see that by applying, um, this matrix to any vector. It cuts the vector in half. Eso contracts it by a factor of two

Carson Merrill · Answer

So for this problem, we, um, see that we have different values of U and V were given starter equations. They are X equals U squared minus V squared. And why equals to UV So these equations will be extremely helpful and moving forward. We want to substitute using either U or V. So if we use you, then what we&#x27;ll get for the substitution is that you is equal to the square root of X plus V squared. So with that, we get that y is equal to to the times the square root of X plus B squared. So now we just substituted you. We&#x27;re not able Thio Express V as an equation of X and Y, um so we have accomplished the goal of getting it in terms of just X, y and V. So now we&#x27;re gonna substitute the constant V Well, let V equals zero. So if we equals zero thin, that yields y equals zero. If we let V equal one, then that&#x27;s going to yield. Why Equalling to root X plus one. Um and then we could repeat the same thing with you. So when we do that, we get that V is equal to the square root of you squared minus x which gives us why being equal to to you you squared minus X So when you equals zero, why equals zero and when U equals one, why equals, um, the two times the square root of one minus X. So based on that, we could look at the graph and it&#x27;s gonna look something like this. How which tells us that we have a solution right here on that&#x27;s going to be at the value of two.

Bobby Barnes · Answer

So we want to use some kind of graphing utility to help us sketch this function here. So if you don&#x27;t have a program that allow you to do this website I really like to use is called dust most. You can see the Ural up here and the way we&#x27;re going to write this in decimus for To understand that this is ah, vector valued function is we&#x27;re gonna put a pair of front sees as opposed to a pair of brackets for a vector at Bid or the vector components. We&#x27;re just gonna put them into the X and y like we would for a normal point. So this is gonna be too go side of t swear. And then we put a comma and then to minus so that we could come down here to the square root AT T. And then once we have this in terms of tea, it&#x27;ll let us pick for what? Bounce a t. It wants us to graph it. So I don&#x27;t know, Let&#x27;s just do like, 100 or something, zoom in a little and then well, looking at the domain, if we were even to plug in like negative 100. You could see it doesn&#x27;t really change anything because that square root there exit. So it&#x27;s undefined for any of the negative values. So the only think that desk most doesn&#x27;t write do it doesn&#x27;t tell you the orientation of it. So if we were to go ahead and plug in zero into this, though, so we&#x27;d have co sign of zero. So actually, it&#x27;s just go ahead and plot that point so would have to host side of zero and then to minus the square root of zero. And then you can see how it labels this point right here. And so then, if we plug in another value, we can see that it I would just be oriented going like this here so I can&#x27;t really draw on desk most. But the orientation should start from that point, and you could just draw your arrows going this way forever

Problem

Use a rectangular coordinate system to plot $\mat…

Problem 13 Easy Difficulty

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

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Linear Algebra and Its Applications

Alayna H.

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Absolute Value - Example 1

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

Alayna H.

Caleb E.

Maria G.

Michael J.

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

Linear Algebra and Its Applications