Uni Maths Questions

[Deleted User]

How do I do this question?

Find all solutions in mod 2 of the following linear system:
x1  +  x3         + x5 = 1
x1  +  x3 + x4 + x5  = 0
                 x4 + x5  = 0

Express your answer as a finite list.

So I've put the system into an augmented matrix and reduced it to RREF:
1 0 1 0 1 | 1
0 0 0 1 0 | 1
0 0 0 0 1 | 1

How do I find the solutions from here? Thanks

[Mao]

The full RREF in is

Code: [Select]

1 0 1 0 0 0
0 0 0 1 0 1
0 0 0 0 1 1

x1=t, x2=s, x3=t, x4=1, x5=1,   

[M-D]

how can i express the following in Cartesian form a+ib:



the answer in the solutions is

thanks

[mark_alec]

Since this is university maths, I assume you have come across the identity:
.

[Deleted User]

That RREF isn't right. It should be

Code: [Select]

1 0 1 0 0 0
0 0 0 1 0 -1
0 0 0 0 1 1

The answer is x1=t, x3=-t, x4=-1, x5=1,

I thought in mod 2 there are only two numbers 0 and 1?

[M-D]

Since this is university maths, I assume you have come across the identity:
.

could you show me the working out please. thanks

[mark_alec]

Use that identity and the usual index law:

[Alwin]

Since this is university maths, I assume you have come across the identity:
.

could you show me the working out please. thanks

Hmm, M-D I'm not sure if you're doing Uni Maths, or UMEP. Perhaps clarifying would help

Anyways, from Complex Numbers in the UMEP course:

To Prove:


Let:


Hence:








This is a constant function, since gradient = 0

Also,


Thus:






As Required

[Jeggz]

could you show me the working out please. thanks


Hmm, M-D I'm not sure if you're doing Uni Maths, or UMEP. Perhaps clarifying would help

There's a difference? 

[Alwin]

There's a difference? 

Hmm, not really. Just that our course, MAST10013 UMEP Mathematics for High Achieving Students is actually only a 1 semester subject at Melbourn University haha.

Just wasn't sure what Maths M-D does in uni, like pure maths and statistics-based maths would vary by a LOT

EDIT: Deleted 2nd post, just added it here:

how can i express the following in Cartesian form a+ib:



the answer in the solutions is

thanks

AHAHAHAHAHAHA SORRY M-D I did't actually answer your question.

So, you have the identity:
  See my other post, Re: Uni Maths Questions, if you still want the formal proof.

Continuing, so you now have:








TADAAAA

[M-D]

thanks Alwin. how to do get the theta and phi symbols in Latex?

[Lasercookie]

thanks Alwin. how to do get the theta and phi symbols in Latex?

\phi \Phi
\theta \Theta

The case of the first letter alternates between upper or lowercase. You'd be using the lowercase ones in this case.

If you click 'quote post' on a post with latex, you can look through the latex code they used that way too. (Or right click -> open image in new tab and read the latex code from the URL).

[Mao]

I thought in mod 2 there are only two numbers 0 and 1?

Oops, I didn't even read the question. I'll update the answer.

[Mao]

So, you have the identity:
  See my other post, Re: Uni Maths Questions, if you still want the formal proof.

Continuing, so you now have:








TADAAAA

At this level, it's probably more intuitive to stay away from :

[Deleted User]

Thanks Mao!

Could someone help me with this question as well?

Without having to find the linear transformation matrix, how do I calculate the dimensions of the image and kernel for the linear transformation:

R: R^3->R^3 by reflection in the plane x+y+z=1 ?

So I know that dim(Ker) = nullity of R and dim(Im) = rank of R but I don't know whether these facts are of any use in this question?