At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

I am having trouble with the notations

Hey!

hiii @ganeshie8

are you familiar with special linear group ?

yes

\(SL_3(\mathbb{Z})\) is the special linear group of 3x3 matrices with "integer" entries

and the determinant value will be 1

yes, also we're considering only the matrices whose element at first row first column is 1

as a start, consider any two such matrices, just to get a feel..

|dw:1444570260285:dw|

yes the determinant value is 1

more importantly the element at first row first column is 1

i guess closure property is not satisfied

thanks we are done with the first one..

yes but i dont know the notation what does it mean

\(\mathbb{R}^{2*2}\) means, all 2x2 matrices with real entries

\[A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

to get a feel, try finding few matrices that satisfy above relation

oh now it make sense.

|dw:1444571091740:dw|

yes... the second column doesn't contribute anything

so, now that we know what things belong to the set, we can test if those elements form a group

before even starting to test anything, whats ur gut feeling, do they form a group ?

so, lets try and prove that the set is indeed a group under multiplication

right, what about inverses ?
does every element in the set has an inverse ?

consider below matrix :
\[\begin{bmatrix}1&0\\0& 0\end{bmatrix} \]
does it belong to the set ?

yes

what can you say about its inverse

its singular matrix and hence not invertible

so it will not form a group?

|dw:1444572164470:dw|

so infinitely many elements dont have inverse...

thanks

and the last one we have orthogonal matrices?

Yes, 2x2 orthogonal matrices
quick review
http://mathworld.wolfram.com/OrthogonalGroup.html

wow

so all the elements in the set trivially have inverses

yes

we have seen earlier in part ii that closure is satisfied, so...

so it will form a group

I feel so, let me think a bit more..

|dw:1444572740427:dw|

we have identity element in the set
we have closure, and we have inverses
so we must have a group !

ofcourse associativity is always there because matrix multiplication is associative

|dw:1444572917065:dw|

closure will fail if we can contruct bd not equal to +1 or -1

oh b and b will be always 1or -1

since the determinant is always +1 or -1

thanks you helped me a lot, I really had problems with the notations.

np, do tag me in all your future group theory questions, i wanted to review these too...

Yes sure I will.