in-exercises-1-10-assume-that-t-is-a-linear-transformation-find-the-standard-matrix-of-t-t-mathbbr-2

Question

In Exercises $1-10$ , assume that $T$ is a linear transformation. Find the standard matrix of $T$ .
$T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ rotates points (about the origin) through 3$\pi / 2$ radians (counterclockwise).

Michael Jacobsen · Accepted Answer

in this example, we have a transformation. T mapping are two into our to and what T is going to dio is take points in the plane and rotate them by an angle of three pi over two in the counter clockwise direction. Our goal here is to find the standard matrix A of T so that he can be written when evaluating X as eight times X. Her goal is to find the standard matrix A. We know that we can calculate t one and th e to. Then our work is going to be complete where e one e two are the first and second columns of the two by two identity matrix. But we have a great formula for when we&#x27;re ever were rotating points by an angle in the counter clockwise direction. The First Inc. The first element is going to be co sign at three pi over two, and the next element will be signed at three Pie divide by two. And if we change the angle, we just have to change what goes in here. Likewise, for the second set of points we have that t at e two is going to map into negative sign at three pie Divide by two for the first century and co sign of three pi over two for the second entry. So let&#x27;s evaluate each one of these components or will have all together. In the first case that co signed three Pi over two is a zero and signed three. Pi over two is negative. One for the second case Sign a three Piper to is negative one. So with the extra negative, we get a positive one here and co sign is zero at three pie or two. This tells us our standard matrix A can be written as TV one TV, too. But now it&#x27;s equal to the falling matrix. We put in zero negative one and 10 and this causes the appropriate rotation by this ankle.

Ashley Boni · Answer

so in the text for the section were given an example three the general rotation matrix for, um, the rotation about the origin through fate of radiance. So in this example, we want to find the specific matrix when data is negative. Pi over four and the negative just means where instead of going counterclockwise, we&#x27;re going, uh, clockwise. So let&#x27;s just plug in our data into this formula, So, co sign of negative pi over four gives us one over square to two. So this is gonna be our value in a diagonal. And if we take sign of negative pyre before we get, uh, negative one over square two. So that&#x27;s what we&#x27;re going to get in this entry in the bottom left and negative sign of data would be positive. One of us were todo. So here&#x27;s our, um, General Matrix

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,

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.

Ashley Boni · Answer

So we have the formula for the standard matrix of a transformation t and that is given by t of E one TV too, and t of a three. So these are the columns off our matrix. And you wanted me to any three artists the standard basics vectors with the one in the, um, 1st 2nd or third position. So all we need to dio is plugging. The vectors were given because we&#x27;re already given t of a 20 to 33. So TV one is 13 T v two is four negative seven and TV three is negative five.

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

Problem

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

Problem 3 Easy Difficulty

In Exercises $1-10$ , assume that $T$ is a linear transformation. Find the standard matrix of $T$ .
$T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ rotates points (about the origin) through 3$\pi / 2$ radians (counterclockwise).

Answer

$A=\left[\begin{array}{ll}{-e_{2}} & {e_{1}}\end{array}\right]=\left[\begin{array}{cc}{0} & {1} \\ {-1} & {0}\end{array}\right]$

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Kayleah T.

Caleb E.

Absolute Value - Example 1

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$A=\left[\begin{array}{ll}{-e_{2}} & {e_{1}}\end{array}\right]=\left[\begin{array}{cc}{0} & {1} \\ {-1} & {0}\end{array}\right]$

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Kayleah T.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Anna Marie V.

Kayleah T.

Caleb E.

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

Linear Algebra and Its Applications