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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 reflects points through the vertical $x_{2}$ -axis and then rotates points $\pi / 2$ radians.

$\left[\begin{array}{cc}{0} & {-1} \\ {-1} & {0}\end{array}\right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 9

The Matrix of a Linear Transformation

Introduction to Matrices

Oregon State University

McMaster University

Lectures

01:32

In mathematics, the absolu…

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In Exercises $1-10$ , assu…

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The linear transformation …

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For the following exercise…

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So we want to figure out what the standard Matrix is for a linear transformation. Um, that first reflects the vector through the horizon's are the vertical axis and then rotated 90 degrees counterclockwise, um, or pie over too for kittens. So let'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's try to see what he does to these two factors. So first, let's draw our vector E one. So here's our vector. You won. That'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's to access. So if we reflected over the vertical axis, it's gonna become this one in blue, and then we need to rotate it 90 degrees counterclockwise. So we're gonna rotate it, and it's gonna be the specter here. So nothing we did has changed the length of the vector. We'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's do the same thing. Um, with the victory, too. So he, too, is 01 So that is the specter right here. So let's use a different color red. And let's do this transformations with each. So first we need to flip it over the y axis. But it's on the wire, the X to access, so it's going to remain in the exact same spot. Then the next step is rotated 90 degrees counterclockwise, such of the left. So we're gonna rotate. It left 90 degrees, and it'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's going to dio um, these two transformations No,

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