jango_IN_DTOWN
  • jango_IN_DTOWN
Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
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jango_IN_DTOWN
  • jango_IN_DTOWN
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jango_IN_DTOWN
  • jango_IN_DTOWN
I am having trouble with the notations
ganeshie8
  • ganeshie8
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jango_IN_DTOWN
  • jango_IN_DTOWN
hiii @ganeshie8
ganeshie8
  • ganeshie8
are you familiar with special linear group ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yes
ganeshie8
  • ganeshie8
\(SL_3(\mathbb{Z})\) is the special linear group of 3x3 matrices with "integer" entries
jango_IN_DTOWN
  • jango_IN_DTOWN
and the determinant value will be 1
ganeshie8
  • ganeshie8
yes, also we're considering only the matrices whose element at first row first column is 1
ganeshie8
  • ganeshie8
as a start, consider any two such matrices, just to get a feel..
jango_IN_DTOWN
  • jango_IN_DTOWN
ok
ganeshie8
  • 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
  • jango_IN_DTOWN
|dw:1444570260285:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
yes the determinant value is 1
ganeshie8
  • ganeshie8
more importantly the element at first row first column is 1
jango_IN_DTOWN
  • jango_IN_DTOWN
i guess closure property is not satisfied
ganeshie8
  • 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
  • jango_IN_DTOWN
thanks we are done with the first one..
ganeshie8
  • 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
  • jango_IN_DTOWN
yes but i dont know the notation what does it mean
ganeshie8
  • ganeshie8
\(\mathbb{R}^{2*2}\) means, all 2x2 matrices with real entries
ganeshie8
  • ganeshie8
\[A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]
ganeshie8
  • ganeshie8
\[\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]
ganeshie8
  • ganeshie8
to get a feel, try finding few matrices that satisfy above relation
jango_IN_DTOWN
  • jango_IN_DTOWN
oh now it make sense.
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1444571091740:dw|
ganeshie8
  • 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
  • jango_IN_DTOWN
yes... the second column doesn't contribute anything
ganeshie8
  • ganeshie8
so, now that we know what things belong to the set, we can test if those elements form a group
ganeshie8
  • ganeshie8
before even starting to test anything, whats ur gut feeling, do they form a group ?
ganeshie8
  • ganeshie8
My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set
ganeshie8
  • ganeshie8
so, lets try and prove that the set is indeed a group under multiplication
jango_IN_DTOWN
  • 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
  • ganeshie8
right, what about inverses ? does every element in the set has an inverse ?
ganeshie8
  • ganeshie8
consider below matrix : \[\begin{bmatrix}1&0\\0& 0\end{bmatrix} \] does it belong to the set ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yes
ganeshie8
  • ganeshie8
what can you say about its inverse
jango_IN_DTOWN
  • jango_IN_DTOWN
its singular matrix and hence not invertible
jango_IN_DTOWN
  • jango_IN_DTOWN
so it will not form a group?
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1444572164470:dw|
ganeshie8
  • 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
  • jango_IN_DTOWN
so infinitely many elements dont have inverse...
jango_IN_DTOWN
  • jango_IN_DTOWN
thanks
jango_IN_DTOWN
  • jango_IN_DTOWN
and the last one we have orthogonal matrices?
ganeshie8
  • ganeshie8
Yes, 2x2 orthogonal matrices quick review http://mathworld.wolfram.com/OrthogonalGroup.html
jango_IN_DTOWN
  • jango_IN_DTOWN
wow
ganeshie8
  • ganeshie8
By definition, \(A\) is called orthogonal matrix iff \(AA^T = I\). so clearly, the inverse of \(A\) is \(A^T\)
ganeshie8
  • ganeshie8
so all the elements in the set trivially have inverses
jango_IN_DTOWN
  • jango_IN_DTOWN
yes
ganeshie8
  • ganeshie8
we have seen earlier in part ii that closure is satisfied, so...
jango_IN_DTOWN
  • jango_IN_DTOWN
so it will form a group
ganeshie8
  • ganeshie8
I feel so, let me think a bit more..
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1444572740427:dw|
ganeshie8
  • ganeshie8
we have identity element in the set we have closure, and we have inverses so we must have a group !
ganeshie8
  • ganeshie8
ofcourse associativity is always there because matrix multiplication is associative
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1444572917065:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
closure will fail if we can contruct bd not equal to +1 or -1
jango_IN_DTOWN
  • jango_IN_DTOWN
oh b and b will be always 1or -1
jango_IN_DTOWN
  • jango_IN_DTOWN
since the determinant is always +1 or -1
jango_IN_DTOWN
  • jango_IN_DTOWN
thanks you helped me a lot, I really had problems with the notations.
ganeshie8
  • 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
  • ganeshie8
np, do tag me in all your future group theory questions, i wanted to review these too...
jango_IN_DTOWN
  • jango_IN_DTOWN
Yes sure I will.

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