Matrix Question Help!!

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That 36 Boy
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#1
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Can anyone please explain how to go about this c) part? Question is down below. Thank you!!
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That 36 Boy
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RDKGames
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(Original post by That 36 Boy)
...
Note that M^4 is just -I. This means that M^8 = M^4 \cdot M^4 = (-I)(-I) = I^2 = I.

Hence (M^8)^k = I^k \implies M^{8k} = I for any integer k, so you can choose k so that you get M raised to a very large power being equal to just I. The question asks for M^{2006}, so you might think about getting k so that 8k = 2006 ... but unfortunately k is not an integer if we solve this equation... but is there a number *close* to 2006 which would give you an appropriate k to choose?
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That 36 Boy
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Ummm how about 2008 which when divided by 8 is 251
(Original post by RDKGames)
Note that M^4 is just -I. This means that M^8 = M^4 \cdot M^4 = (-I)(-I) = I^2 = I.

Hence (M^8)^k = I^k \implies M^{8k} = I for any integer k, so you can choose k so that you get M raised to a very large power being equal to just I. The question asks for M^{2006}, so you might think about getting k so that 8k = 2006 ... but unfortunately k is not an integer if we solve this equation... but is there a number *close* to 2006 which would give you an appropriate k to choose?
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RDKGames
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(Original post by That 36 Boy)
Ummm how about 2008 which when divided by 8 is 251
Sounds good, but since you're not confident on this question yet, I suggest you pick k = 250 as to then obtain the power of 2000 [since 8*250 = 2000]

So then, M^{2000} = I.

But you're interested in M^{2006}. What can you do to the LHS of our equation in order to go from M^{2000} to M^{2006} ?? Do the same to the RHS. What do you get?
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That 36 Boy
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Multiply by M^6 on both sides? Then i get M^2006 = M^6 which is M^4 * M^2 calculated previously?
(Original post by RDKGames)
Sounds good, but since you're not confident on this question yet, I suggest you pick k = 250 as to then obtain the power of 2000 [since 8*250 = 2000]

So then, M^{2000} = I.

But you're interested in M^{2006}. What can you do to the LHS of our equation in order to go from M^{2000} to M^{2006} ?? Do the same to the RHS. What do you get?
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RDKGames
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(Original post by That 36 Boy)
Multiply by M^6 on both sides? Then i get M^2006 = M^6 which is M^4 * M^2 calculated previously?
That's right, and using the fact that M^4 = -I you shouldn't need to multiply any matrices out.
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That 36 Boy
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Is there any way to know that M^4 = -I without calculating M^4?
(Original post by RDKGames)
That's right, and using the fact that M^4 = -I you shouldn't need to multiply any matrices out.
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RDKGames
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(Original post by That 36 Boy)
Is there any way to know that M^4 = -I without calculating M^4?
It's not so obvious without calculating it. Of course, in this question you're asked to calculate it in the previous part, so you know what it is. The rest is just to realise that is looks very similar to the identity matrix... indeed it's just the -ve of it.

There is a way without any calculation, but it hinges on your understanding of matrix transformations visually. For instance, this matrix M represents a rotation clockwise about the origin by 45 degrees. M^4 means you apply this 4 times, so it's a 45 x 4 = 180 degree rotation. Where do the unit vectors \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 1 \end{pmatrix} go after this transformation? Obviously to \begin{pmatrix} -1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ -1 \end{pmatrix} respectively... so the matrix for M^4 is \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.
Last edited by RDKGames; 2 years ago
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That 36 Boy
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This makes so much sense now. But doesn’t M represent 45° rotation clockwise about the origin since It's in the 4th quadrant?
(Original post by RDKGames)
It's not so obvious without calculating it. Of course, in this question you're asked to calculate it in the previous part, so you know what it is. The rest is just to realise that is looks very similar to the identity matrix... indeed it's just the -ve of it.

There is a way without any calculation, but it hinges on your understanding of matrix transformations visually. For instance, this matrix M represents a rotation anticlockwise about the origin by 45 degrees. M^4 means you apply this 4 times, so it's a 45 x 4 = 180 degree rotation. Where do the unit vectors \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 1 \end{pmatrix} go after this transformation? Obviously to \begin{pmatrix} -1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ -1 \end{pmatrix} respectively... so the matrix for M^4 is \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.
Last edited by That 36 Boy; 2 years ago
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RDKGames
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(Original post by That 36 Boy)
This makes so much sense now. But doesn’t M represent 45° rotation clockwise about the origin since It's in the 4th quadrant?
Yeah, that's what I meant it's 4 lots of it at the end of the day so it doesn't matter whether we go clockwise or anticlockwise for M^4
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That 36 Boy
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#12
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Thanks a lot for your help.Out of curiosity what if i wanted M^2007? I tried it and it doesn't seem to get me anywhere cause 2007 is an odd number?
(Original post by RDKGames)
Yeah, that's what I meant it's 4 lots of it at the end of the day so it doesn't matter whether we go clockwise or anticlockwise for M^4
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RDKGames
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(Original post by That 36 Boy)
Thanks a lot for your help.Out of curiosity what if i wanted M^2007? I tried it and it doesn't seem to get me anywhere cause 2007 is an odd number?
Just multiply M^{2000} = I through by M^7 = M^{4} \cdot M^{2} \cdot M.


Alternatively, if you're OK with taking inverse matrices, then clearly we have established by that M^{2008} = I, so multiplying through by M^{-1} will also yield the result.
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That 36 Boy
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#14
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#14
Ah i see. That's just perfect. Thanks for your priceless help mate 😊✌
(Original post by RDKGames)
Just multiply M^{2000} = I through by M^7 = M^{4} \cdot M^{2} \cdot M.


Alternatively, if you're OK with taking inverse matrices, then clearly we have established by that M^{2008} = I, so multiplying through by M^{-1} will also yield the result.
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