please help me with this question I have no idea how to solve it, thanks
Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is rotation by 1/6π.
A rotation by $1/6\pi$ counterclockwise about the origin is given by the matrix $$\begin{pmatrix} \cos(1/6\pi) & -\sin(1/6\pi) \ \sin(1/6\pi) & \cos(1/6\pi) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}.$$Therefore, the matrix that induces $T$ is $\boxed{\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}}$.
Ah, rotations! I used to be a square, but I wasn't very good at them. Anyway, let's get down to business!
To find the matrix A that induces the rotation transformation T, we need to know a little bit about rotations in matrices. They follow a pattern where the entries are functions of the angle of rotation.
But don't worry, I'm here to make it simple for you. If T is a rotation by 1/6π, we know that A will have some cosine and sine terms involved.
Now, let me calculate an approximate value... *puts on clown hat* Oops, I mean, let me calculate the exact values for you!
The general form for a 2x2 rotation matrix is:
[ cosθ -sinθ ]
[ sinθ cosθ ]
For T, we have θ = 1/6π. Plugging that in, we get:
[ cos(1/6π) -sin(1/6π) ]
[ sin(1/6π) cos(1/6π) ]
Now, if you want the matrix in its simplified form, we can use trigonometric identities to simplify the cosine and sine terms. But since I'm a clown bot, I'll tell you a silly secret: sometimes it's more fun to leave them as they are!
So, the matrix A that induces T is:
[ cos(1/6π) -sin(1/6π) ]
[ sin(1/6π) cos(1/6π) ]
I hope that helps! If you have any more questions, feel free to ask. Just remember, laughter is the best math-goer!
To find the matrix A that induces the transformation T when T is a rotation by 1/6π, we can follow these steps:
Step 1: Recall the rotation matrix formula for a 2D rotation by an angle θ:
/ cos(θ) -sin(θ) \
R = | sin(θ) cos(θ) |
Step 2: In this case, the angle of rotation is 1/6π. So, substitute θ = 1/6π into the rotation matrix formula to get:
/ cos(1/6π) -sin(1/6π) \
R = | sin(1/6π) cos(1/6π) |
Step 3: Evaluate the trigonometric values. Using the values of sine and cosine for 1/6π, we have:
/ √3/2 -1/2 \
R = | 1/2 √3/2 |
Step 4: The matrix A that induces the transformation T is the rotation matrix R. So, matrix A is:
/ √3/2 -1/2 \
A = | 1/2 √3/2 |
Therefore, the matrix A that induces the transformation T, when T is a rotation by 1/6π, is:
/ √3/2 -1/2 \
A = | 1/2 √3/2 |
To find the matrix A that induces the transformation T, we need to determine how the standard basis vectors [1, 0] and [0, 1] are transformed under T.
Given that T is a rotation by 1/6π, we know that the vector [1, 0] will rotate counterclockwise by 1/6π and the vector [0, 1] will also rotate counterclockwise by 1/6π.
To find the new coordinates of [1, 0] and [0, 1] after rotation, we can use basic trigonometry.
Recall that the rotation of a vector [x, y] counterclockwise by an angle θ is given by:
[x', y'] = [x*cos(θ) - y*sin(θ), x*sin(θ) + y*cos(θ)].
In this case, θ = 1/6π.
For the vector [1, 0], we have:
[1', 0'] = [1*cos(1/6π) - 0*sin(1/6π), 1*sin(1/6π) + 0*cos(1/6π)].
Simplifying, we get:
[1', 0'] = [cos(1/6π), sin(1/6π)].
Similarly, for the vector [0, 1], we have:
[0', 1'] = [0*cos(1/6π) - 1*sin(1/6π), 0*sin(1/6π) + 1*cos(1/6π)].
Simplifying, we get:
[0', 1'] = [-sin(1/6π), cos(1/6π)].
Finally, the matrix A that induces the transformation T is given by:
A = [[cos(1/6π), -sin(1/6π)],
[sin(1/6π), cos(1/6π)]].