Please help

- jango_IN_DTOWN

Please help

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- jango_IN_DTOWN

##### 1 Attachment

- jango_IN_DTOWN

I am having trouble with the notations

- ganeshie8

Hey!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- jango_IN_DTOWN

hiii @ganeshie8

- ganeshie8

are you familiar with special linear group ?

- jango_IN_DTOWN

yes

- ganeshie8

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

- jango_IN_DTOWN

and the determinant value will be 1

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

ok

- ganeshie8

try below two matrices :
\(\begin{bmatrix} 1&2&1\\0&1&3\\0&0&1 \end{bmatrix}\)
\(\begin{bmatrix} 1&1&2\\3&0&1\\1&0&0 \end{bmatrix}\)
do they belong to the given set in part i ?

- jango_IN_DTOWN

|dw:1444570260285:dw|

- jango_IN_DTOWN

yes the determinant value is 1

- ganeshie8

more importantly the element at first row first column is 1

- jango_IN_DTOWN

i guess closure property is not satisfied

- ganeshie8

Exactly!
\(A=\begin{bmatrix} 1&2&1\\0&1&3\\0&0&1 \end{bmatrix}\)
\(B=\begin{bmatrix} 1&1&2\\3&0&1\\1&0&0 \end{bmatrix}\)
\(BA\) belongs to the given set, but \(AB\) does not.

- jango_IN_DTOWN

thanks we are done with the first one..

- ganeshie8

for part ii, you need to think of (1, 0) as column vector, not a row vector as it was shown in the problem

- jango_IN_DTOWN

yes but i dont know the notation what does it mean

- ganeshie8

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

- ganeshie8

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

- ganeshie8

\[\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

- ganeshie8

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

- jango_IN_DTOWN

oh now it make sense.

- jango_IN_DTOWN

|dw:1444571091740:dw|

- ganeshie8

Good, as you can see the first column of the matrix has to be fixed at \(\begin{array}{} 1\\ 0\end{array}\) and the second column could be anything :
\[\begin{bmatrix}1&b\\0& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]
above relation works for all \(a,b\in\mathbb{R}\)

- jango_IN_DTOWN

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

- ganeshie8

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

- ganeshie8

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

- ganeshie8

My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set

- ganeshie8

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

- jango_IN_DTOWN

yes and matrix multiplication is associative and closure will be satisfied since the product will give us members of R in the last column

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

yes

- ganeshie8

what can you say about its inverse

- jango_IN_DTOWN

its singular matrix and hence not invertible

- jango_IN_DTOWN

so it will not form a group?

- jango_IN_DTOWN

|dw:1444572164470:dw|

- ganeshie8

Yes, every element in a group must have an inverse. Since few matrices in the set don't have inverses, the given set doesn't form a group under multiplication

- jango_IN_DTOWN

so infinitely many elements dont have inverse...

- jango_IN_DTOWN

thanks

- jango_IN_DTOWN

and the last one we have orthogonal matrices?

- ganeshie8

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

- jango_IN_DTOWN

wow

- ganeshie8

By definition, \(A\) is called orthogonal matrix iff \(AA^T = I\).
so clearly, the inverse of \(A\) is \(A^T\)

- ganeshie8

so all the elements in the set trivially have inverses

- jango_IN_DTOWN

yes

- ganeshie8

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

- jango_IN_DTOWN

so it will form a group

- ganeshie8

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

- jango_IN_DTOWN

|dw:1444572740427:dw|

- ganeshie8

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

- ganeshie8

ofcourse associativity is always there because matrix multiplication is associative

- jango_IN_DTOWN

|dw:1444572917065:dw|

- jango_IN_DTOWN

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

- jango_IN_DTOWN

oh b and b will be always 1or -1

- jango_IN_DTOWN

since the determinant is always +1 or -1

- jango_IN_DTOWN

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

- ganeshie8

good observation, yeah we only have two matrices in the group I think
http://www.wolframalpha.com/input/?i=solve+%7B%7B1%2Ca%7D%2C%7B0%2Cb%7D%7D*%7B%7B1%2C0%7D%2C%7Ba%2Cb%7D%7D+%3D+%7B%7B1%2C0%7D%2C%7B0%2C1%7D%7D

- ganeshie8

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

- jango_IN_DTOWN

Yes sure I will.

Looking for something else?

Not the answer you are looking for? Search for more explanations.