Abstract Algebra: Let Q be the group of 2x2 matrices with

[Sydney]

Let Q = {alpha gamma} alpha, gamma are complex numbers)
{____ ______}
{-gamma alpha}
___
Ex. 2x2 matirces with complex entries (alpha=a+bi gamma=c+di alpha=a-bi
______ _____
gamma=c-di) alpha is the conjugate of alpha. Prove Q is a division ring. Use ordinary matrix addition and multiplication. For inverse use 1/k (____ )
(alpha -gamma)
(_____ )
(gamma alpha)
_____ ______
k= a^2 + b^2 + c^2 + d^2 = alpha alpha + gamma gamma

 

[pka]

I simply cannot read that question!
At the top of this page is a tab "Forum Help". On are some suggestions on how one types mathematics for the web. Please make the question readable.
At this level, you should learn to use LaTeX.

 

[Sydney]

Q is a 2x2 matrix with alpha and gamma. alpha = a+bi and gamma = c+di the conjugate of alpha is a-bi and the conjugate of gamma is c-di. Prove that Q is a division ring using ordinary matrix addition and multiplication. HINT for the inverse use 1/k being k=a^2 + b^2 +C^2 + d^2 which equals (alpha)(conjugate alpha) + (gamma)(conjugte gamma) I hope this helps, in plain english I believe this is what the problem is saying. I had a problem downloading LaTex on this computer. It is not mine I am borrowing from a friend. Sorry!!

 

[pka]

Working with this problem it would be easier to write the matrices as:
\(\displaystyle \L \left[ {\begin{array}{rr}
z & w \\ - {\overline w } & {\overline z } \\ \end{array}} \right]\).

Now clearly these form a ring with unity. All that has to done is to show that each such non-zero matrix has an inverse.

 

[Sydney]

Such that....? That is my problem? Is it by using the inverse that was giving:
_ __
You take Z-1(Z)(Z)=(W)(W) Z-1
__ __
Z E=(W)(W) Z-1
__ __
W-1( Z)E=(W)(W)Z-1W-1
__ __
W-1(Z)E=(W)Z-1E

E=E

 

[galactus]

You don't have to 'download' LaTex. You can type the code yourself.

\(\displaystyle \L\\{\gamma}~{\alpha}\)

 

[pka]

Recall that \(\displaystyle \L z\bar z = |z|^2\) and then determinate \(\displaystyle \L k = \left| {\begin{array}{rr}
z & w \\ { - \bar w} & {\bar z} \\ \end{array}} \right| = z\bar z + w\bar w\).
So the inverse is \(\displaystyle \L \left( {\begin{array}{rr}
z & w \\ { - \bar w} & {\bar z} \\ \end{array}} \right)^{ - 1} = \frac{1}{k}\left( {\begin{array}{rr}
{\bar z} & {- w} \\ { \bar w} & z \\ \end{array}} \right)\)

 

[Sydney]

Thank you for your help. I can see it now.