Injective Affine Linear Map for (Z/2Z)^3 to (Z/2Z)^3 Sending (1,1,1) to (0,0,0)

[Firepanda]

What is the number of bit permutations of the set {0,1}ⁿ and the number of circular right shifts of the set {0,1}ⁿ.

I think the number of bit permuations is 2ⁿ, so is there 4 bit permutations here? Namely (0,0), (0,1), (1,0) and (1,1).

And the right shift is just sending each element one space right isn't it?

i.e. (0,0) -> (0,0) , (0,1) -> (1,0) etc

So what does he mean by the number of shifts? 4?

Find an injective affine linear map (Z/2Z)³ -> (Z/2Z)³ that sends (1,1,1) to (0,0,0).

In my notes all I have is that

f(v) = Av + b

Here I'm trying to find A, with given v = (1,1,1) and f(v) = (0,0,0)

I also have that for this to be a bijection

=> for (Z/mZ)ⁿ -> (Z/mZ)^l then n=l and gcd(det(A), m) = 1

So since this is needed for a bijection am I ok in saying this can be the criteria for it to be injective?

So if I find an A where (det(A), 2) = 1 , => det(A) can equal 1

i.e. A =

0 1 0
0 0 1
1 0 0

and so Av = (1,1,1) and now I can take b = (1,1,1) so that Av+b = (2,2,2) = (0,0,0)?

Is this all correct? Thanks.

 

[Firepanda]

If something doesn't make sense just say and I'll explain what I meant.

 

[Firepanda]

I don't have long until I have to hand this in, would be nice for some input (any at all)!

 

[Firepanda]

last bump i spose