in-exercises-1-10-assume-that-t-is-a-linear-transformation-find-the-standard-matrix-of-t-t-mathbbr-8

Question

In Exercises $1-10$ , assume that $T$ is a linear transformation. Find the standard matrix of $T$ .
$T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ first performs a horizontal shear that transforms $\mathbf{e}_{2}$ into $\mathbf{e}_{2}-2 \mathbf{e}_{1}$ (leaving $\mathbf{e}_{1}$ unchanged) and then reflects points through the line $x_{2}=-x_{1}$

Barsha Rana · Accepted Answer

So in the current problem, we have to find a transformation. T such that r squared too r squared. And it does too actions in particular. So the first action, if we consider, is as a t one. Okay, That is given as we change changes e to as even minus twice he took. So e tu minus twice The one bad has it leaves even as it is that is even through this transformation minute, cause it will remain even so if you think this is even and this is e to so it will change somewhere here or here somewhere even will remain same. So either there will be a sheer like this or this way. So this is nothing but horizon. Tal shit. No. How do we find out this transformation also the second action off this transformation would be then reflecting the point through the line X two is equal to minus X one. Okay, so that means what? First there will be a sheer and then x two is equal to minus X one. That means something like this. It should come here, get reflected here. So we have to find out what would be the metrics or the transformation associated with this situation. So let us go step by step. So if you see even will remain you got. So if there is a metrics which should have two columns so first column will not go through any change. That means what 10 will yield one zero. But what about he took it will see a change. What is that? That is e tu minus twice. Even so, it too is zero and one. So what will happen to eat too? This will become zero minus twice, even even first element is one again. One minus two into zero. So this will be the changing You too. That is we can again, right? Zero one will change too. Zero minus two is minus two and one minus zero is one. So that means what? This is the mattress that will change even in between this format. Right? So if it is 1001 that will get changed too. 10 minus two. What? And we clearly know that if it is a k over here, then it is a horizontal shear. Right? So we are quite going toe the courage path now what do we want after this? We want. And also we know if K is less than zero than it&#x27;s a left towards sheer right. That is this point will ship this this direction, right? If K is less than zero No, we want the reflection through X two minus x one. Right. So we now want toe, Find out the second activity. That would be reflection through X two is equals True Negative X Well, mhm x two musical through negative experts. Now can we write? Our tea is equals to t to off. Do you run off X? That means what? First we apply t even X on X. Okay. And then we will apply t two. So what will be so this is our tea even. Okay. So if we multiply this on X any variables Victor&#x27;s, then we get this final out. Now we have two similar We find t to What would we Tito Tito would be the reflection, the action, the reflection action will be defined. So through this, we will achieve the final answer now see if originally a metrics is 1001 for reflection, we get this zero minus one minus one zero. Now, this is not convincing enough. Let&#x27;s check through a diagram. This is one comma zero. This is zero comma one. Right now. What will be extra eyes equals two minus one. What would be this equation? This would be kind of her like this. So this will come here. Right? So which point is this? Here. X is zero. Why is minus one, isn&#x27;t it? So if you see 10 changes to zero minus one. That means what? This down element goes up and the up element comes down changing the site. Changing the sign. So same V. What will happen to our actual given expression? This is our final expression after we&#x27;re having tea. Even so, 10 will become zero minus one. Okay, Now what will happen to this point? This point, which is 10 we&#x27;ll also travel here. Right? So this is what here? Why is zero but X is negative. That means what all the elements changes. Sign zero becomes negative of zero is also again zero. That&#x27;s why we couldn&#x27;t see it. So if it was a point like five to that will go toe minus two, minus five. Okay, if you see a point here and we go here. So if you see if it is five comma to that will become minus two. Coma minus fine. So then we are having the second column minus 21 as this one will go up becoming negative and this minus two will come down. I&#x27;ll get taking the negative. That means plus two. So this man, this is the final expression off the transformation. T So therefore, t such that it takes a pair off. Values are square to our square says that it performs a horizon tal sheer Andi, Then reflection through extra is a construct. X x ray is equal to minus X. One would be zero minus one and then again minus one to Andi. That&#x27;s how easy it is. I hope you could understand

Ashley Boni · Answer

So we want to find the Matrix Tea for a mapping from our TUTTO are too. That leaves the one unchanged and maps to took you two plus two equals. So where do we start? We know that the Matrix has the form t of even one as the first column and t of e to as the second calm. So we want to be want to remain the same. So in other words, we want t of you want to equal one waas zero e too. And we want teeth of e too the equal. You too. Plus three you won. So in order to write this, then matrix form TV once the first column. So it&#x27;s one yuan plus zero e too. And TV too is equal to 31 plus one e two and that is our matrix t.

Michael Jacobsen · Answer

in this example, we have a transformation. T mapping are two into our to, and what it satisfies is that the image of the one will be the one minus two B two, and the image of E two is e to. Given this information, our Cole here&#x27;s to determine if we map X under the transformation. T What The Matrix A must be, in other words, were looking for the standard matrix a off the transformation t But by formula, this matrix A can be written as TV one for the first column and t of e two for the second column. So it&#x27;s determined what these values would be in this situation. First, we know that T V one equals one minus two, e two, but the one is the first column of the two by 28 any matrix? So it is a 10 My has two times e to you can think of e to as being of calling vector containing zeros. But we put a one in the two slot, so it is a 01 Then if we do the scaler, multiplication and the vector addition will have ah one and negative too altogether. Next we&#x27;re told that TV too is equal T too. But you two is 01 as before. So now we&#x27;re raid the states with the standard matrix A of t will be first go to TV one. We found that that isn&#x27;t one negative too. Then for TV to we have a 01 and this is our standard matrix A for the transformation T.

Ashley Boni · Answer

So we want to find a two by two matrix that reflects points through the horizontal access first and then reflects through Ah, the line X two equals x one. So the first rotation will call the Are one is tthe e reflection through the horizontal X one axis. Ah, so if you look at page, I believe that 74 in the section and it gives all of the EU&#x27;s general rotations and the rotation for this ISS 10 zero negative one because it slips the why coordinates. So next, the second rotation we need to reflect through the line X two equals equals one. Um, so if you think about this actually happening, um, that line is right here. So if you pick a point, let&#x27;s say this one here and you reflect over this line the X coordinate in the Y coordinate are switching. So, um, or you could just look on patient before the rotation matrix for this is 0110 So we&#x27;re switching the X on my coordinates. So the combination of the reflections is tthe e ah. Multiplication of these major cities with the 1st 1 has to be on the end because that&#x27;s the Matrix we&#x27;d would apply first. And then we would apply The Second Matrix are too. So if we multiply these two things together, let&#x27;s see what we get. So our first, uh, coordinate is zero plus zero, which is your own. Next we take second call. Sometimes the first row, so zero plus minus. One is minus one. Next second Road times first call home. One plus zero is zero. First, I want 01 And lastly, second row time, Second column. We get zero plus zero, which is zero. So this is our final, um, no tricks.

Michael Jacobsen · Answer

in this example, we have a transformation t that&#x27;s going to be linear, and what it does is it takes the first column of the two by two identity matrix, and it maps it to this element and are too. Likewise, the second element of the identity matrix is mapped to negative 5 to 00 of our two. Our goal, then, is to find a matrix. A such that t of X could be expressed as a Times X If we confined such a matrix A. Then we can use The Matrix to evaluate T at any Vector X, not just these two vectors. So the first step is to note that the standard Matrix A if we have these two elements, is given as follows. The Matrix say will be t evaluated one for the first column and t evaluate at E two for the second call. So for this situation, we know that t one is 3131 TV two is negative. Five to 00 And so this matrix A is the standard matrix for the transformation. T

Ashley Boni · Answer

So we want to figure out what the standard Matrix is for a linear transformation. Um, that first reflects the vector through the horizon&#x27;s are the vertical axis and then rotated 90 degrees counterclockwise, um, or pie over too for kittens. So let&#x27;s try to construct the Matrix. So we know that the standard matrix T is, um, given by the columns uh, TV one anti of e t. So hee oneness The vector 10 And he, too, is the vector 01 So let&#x27;s try to see what he does to these two factors. So first, let&#x27;s draw our vector E one. So here&#x27;s our vector. You won. That&#x27;s it. Ah, so this is tthe e x two axes, and this is the X one access. So first we need thio reflected through the it&#x27;s to access. So if we reflected over the vertical axis, it&#x27;s gonna become this one in blue, and then we need to rotate it 90 degrees counterclockwise. So we&#x27;re gonna rotate it, and it&#x27;s gonna be the specter here. So nothing we did has changed the length of the vector. We&#x27;ve just looked it and turned it and did these transformations. So this is gonna be the point. Negative one. So when we do these transformations, 10 becomes zero negative one. So this is t of you. So let&#x27;s do the same thing. Um, with the victory, too. So he, too, is 01 So that is the specter right here. So let&#x27;s use a different color red. And let&#x27;s do this transformations with each. So first we need to flip it over the y axis. But it&#x27;s on the wire, the X to access, so it&#x27;s going to remain in the exact same spot. Then the next step is rotated 90 degrees counterclockwise, such of the left. So we&#x27;re gonna rotate. It left 90 degrees, and it&#x27;s gonna end up right here. So our vector 01 after doing these transformations becomes the point. Negative 10 All of these are length once. So this is Artie of E, too. So we can say what our matrix to ISS first call miss TV one, so zero minus one. And the second is TV too, which is negative 10 So if we apply this matrix to any vector, it&#x27;s going to dio um, these two transformations No,

Problem

In Exercises $1-10$ , assume that $T$ is a linear…

Problem 9 Easy Difficulty

Answer

$A=\left(\begin{array}{cc}{0} & {-1} \\ {-1} & {2}\end{array}\right)$

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

Linear Algebra and Its Applications

Catherine R.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$A=\left(\begin{array}{cc}{0} & {-1} \\ {-1} & {2}\end{array}\right)$

Linear Algebra and Its Applications

Catherine R.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Caleb E.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications