Matrix question

Part a I found fine thanks to earlier help but part b is confusing.

I tried to use determinant to guide me but to no avail.

How do I do this? It's only worth one mark so I feel like I'm missing something easy.

Reply 1

tory88

Original post by Ravster

If you multiply a 2x2 matrix by the 2x2 identity matrix you end up with the same matrix.

As A^n is just doing the transformation n times, what it's really asking you is how many times do you need to carry out the transformation to get back to where you started.

Reply 2

Kvothe the Arcane

Original post by Ravster

If you have some coordinates

\begin{pmatrix}x \\ y \end{pmatrix}

and you multiply it by the identity matrix I, you will get the same thing:

\begin{pmatrix}x \\ y \end{pmatrix}

.

As A is the rotation about the origin by certain angle, how many times will it have rotate by that angle to get to its original position

k360^{\circ}

Edit: Zacken has pointed out that you would have to get to multiple of

360^{\circ}

which would be the LCM of the angle you found in part(a) and 360.

(edited 7 years ago)

Reply 3

Zacken

Original post by Ravster

Think about what the angle of rotation A represents and think about how many multiples of that rotation you need to get to a multiple of 360 degrees, which will be the identity matrix.

Remember that if A represents a rotation of

\alpha

then

A^2

is a rotation of

\alpha

followed by a rotation of

\alpha

for a total rotation of

2\alpha

. In general here

A^n

is a rotation of

n\alpha

degrees. So you need to find a

n

s,t

n\alpha

is a multiple of 360.

Reply 4

Zacken

Original post by shivtek

Part was 135 I think. Or 45. I can't remember. I is the identity as you know so what it's asking is how many times does one apply A to get back to the start. So you do 360/(135 or 45). I got 8/3 I think

Original post by Kvothe the arcane

If you have some coordinates

\begin{pmatrix}x \\ y \end{pmatrix}

and you multiply it by the identity matrix I, you will get the same thing:

\begin{pmatrix}x \\ y \end{pmatrix}

A

represents a transformation and it's saying that

\mathrm{\underbrace{A \times A \times \times A \times .... \times A}_{n \ times}=X}

As A is the rotation about the origin by certain angle, how many times will it have rotate by that angle to get to its original position

360^{\circ}

I'm going to have to butt in here and say that you want a natural

n

; so

n=\frac{8}{3}

is incorrect. You don't want to rotate to get to

360^{\circ}

but rather, you want to rotate to get to a natural multiple of

360^{\circ}

. Which is an important point here.

Reply 5

Kvothe the Arcane

Original post by Zacken

...

That makes sense. You can't have a fraction of a multiplication. It would have made sense to evaluate a.
Thanks for pointing it out.

(edited 7 years ago)

Reply 6

shivtek

Original post by Zacken

I'm going to have to butt in here and say that you want a natural

n

; so

n=\frac{8}{3}

is incorrect. You don't want to rotate to get to

360^{\circ}

but rather, you want to rotate to get to a natural multiple of

360^{\circ}

. Which is an important point here.

Away good point I didn't read "positive integer"
So really, raising it to the power of 8 would work, as it would be 0 rotation modulo 360. Thanks for pointing that out!

Reply 7

Zacken

Original post by shivtek

Away good point I didn't read "positive integer"
So really, raising it to the power of 8 would work, as it would be 0 rotation modulo 360. Thanks for pointing that out!