Question

Consider the equilateral triangle with vertices 1,2,3 initially placed, respectively, at \((-1/2, \sqrt3/2)\), (1,0) and \((-1/2, -\sqrt3/2)\)in the x-y plane.
Give 6 2 x 2 matrices implementing all of the symmetries of this equilateral triangle.
Please, help

Answer 1

I got just
\[f_1=\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]\] it reflects across x-axis (or around perpendicular bisection segment 1,3 through 2)
\(f_1\) sends \(\left(\begin{matrix}1&2&3\\3&2&1\end{matrix}\right)\) I mean sends 1 to 3, 2 to 2 and 3 to 1

\[f_2=\left[\begin{matrix}-1/2&\sqrt3/2\\\sqrt3/2&-1/2\end{matrix}\right]\]counterclockwise rotates around center an angle \(2\pi/3\)
\(f_2\) sends \(\left(\begin{matrix}1&2&3\\3&1&2\end{matrix}\right)\) I mean sends 1 to 3, 2 to 1 and 3 to 2
\[f_3=\left[\begin{matrix}-1/2&\sqrt3/2\\-\sqrt3/2&-1/2\end{matrix}\right]\]counterclockwise rotates around center an angle \(4\pi/3\)
\(f_3\) sends \(\left(\begin{matrix}1&2&3\\2&3&1\end{matrix}\right)\) I mean sends 1 to 2, 2 to 3 and 3 to 1

Answer 2

\(f_4\) denotes symmetry across the perpendicular bisector from 1 to segment 23. But I don't know how to set it in matrix form, the equation of the line is \(y= \sqrt3 x\)
Please, help

Answer 3

@SithsAndGiggles

Answer 4

Sorry for taking so long, I was playing around with the Desmos calculator to see if I could graph the triangle: https://www.desmos.com/calculator/yuqwtbnzes

Answer 5

Ugh these matrix transformations aren't as familiar as they should be... Is this linear algebra?

Answer 6

Intuitively, I can see through f4, (1,0) becomes (-1/2, sqrt3/2)
:) it's not linear algebra, it's abstract algebra.

Answer 7

\[f_4=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\]

Answer 8

\[\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\left[\begin{matrix}1\\0\end{matrix}\right]=\left[\begin{matrix}-1/2\\\sqrt3/2\end{matrix}\right]\]

Answer 9

which gives me a = -1/2 and c = sqrt3/2
but I don't know how to find b, d

Answer 10

It looks like \(b\) and \(d\) can be any number (i.e. infinite number of solutions).

Answer 11

Or are you supposed to get unique solutions for all constants?

Answer 12

Yes, I have to have a unique one.

Answer 13

Is there anything preventing you from fixing \(b\) and \(d\)? We know the equation holds for any \(b,d\), so why not let them be 0 for simplicity?

Answer 14

It's a symmetry function, so that, it must apply for both 1 and 2. I mean 1 to 2 and 2 to 1. that means f_4 applies to (1,0) we get (-1/2, sqrt3/2)
f_4 applies to (-1/2,sqrt3/2) we get (1,0) back.
That's why b, d cannot be arbitrary values.

Answer 15

I see, so basically, you need to satisfy
\[f_n\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}-1/2\\\sqrt3/2\end{bmatrix}~~\iff~~\begin{bmatrix}1\\0\end{bmatrix}={f_n}^{-1}\begin{bmatrix}-1/2\\\sqrt3/2\end{bmatrix}\]
which requires \(f\) to be nonsingular, correct?

Answer 16

yes,

Answer 17

For example: the reflection about y =x will take (1,0) to (0,1) and vice versa
and the matrix is \(\left[\begin{matrix}0&1\\1&0\end{matrix}\right]\)
so that if applying to (1,0), we get (0,1)
and IT applies to (0,1) we get (1,0)

Answer 18

but our line is y = sqrt3 x, ha!!

Answer 19

http://planetmath.org/derivationof2dreflectionmatrix

Scroll down to "Reflections across a line of given slope." In this case, \(m=\sqrt3\). I hope that's what you're looking for.

Answer 20

Thanks a ton. I will

Answer 21

@SithsAndGiggles I got it. :)

Answer 22

Glad to have helped!