parisnaomi
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'The rotation represented by the 2 X 2 matrix B is such that B^3=I and B=/=I

Write down a possible matrix B and describe the transformation fully.'

Could anyone help explain how to do this? The previous question asked for when A^2=I and I answered that fine, but I cant seem to use the same method for this one.

'I' is the identity matrix, just to clarify.
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mnot
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So start by building the foundations:
lets call Matrx B = [a b, c d]
B^3 = I = [a b, c d] . [a b, c d] . [a b, c d] = [1 0, 0 1]

hence we can generate expressions for a b c d. We also know: [a b, c d] =/= [1 0, 0 1]

this should get you started...
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parisnaomi
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(Original post by mnot)
So start by building the foundations:
lets call Matrx B = [a b, c d]
B^3 = I = [a b, c d] . [a b, c d] . [a b, c d] = [1 0, 0 1]

hence we can generate expressions for a b c d. We also know: [a b, c d] =/= [1 0, 0 1]

this should get you started...
thank you, i'll try this out. i already kind of did this but i let B equal the general transformation matrix for any angle and that didn't get me anywhere haha
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RDKGames
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(Original post by parisnaomi)
'The rotation represented by the 2 X 2 matrix B is such that B^3=I and B=/=I

Write down a possible matrix B and describe the transformation fully.'

Could anyone help explain how to do this? The previous question asked for when A^2=I and I answered that fine, but I cant seem to use the same method for this one.

'I' is the identity matrix, just to clarify.
B is a rotation about the origin by some angle, and if you apply it 3 times then thats the same as rotating 3 times.

The fact that this is I means if you rotate 3 times about the origin you end up where you started...

So how much is each rotation by?? Construct its matrix!
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parisnaomi
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(Original post by RDKGames)
B is a rotation about the origin by some angle, and if you apply it 3 times then thats the same as rotating 3 times.

The fact that this is I means if you rotate 3 times about the origin you end up where you started...

So how much is each rotation by?? Construct its matrix!
ahhhh ok! this makes so much sense. i was honestly thinking of something like this but couldn't quite grasp the right idea. and now im kicking myself for not realising it myself hahaha. thank you
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RDKGames
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(Original post by parisnaomi)
ahhhh ok! this makes so much sense. i was honestly thinking of something like this but couldn't quite grasp the right idea. and now im kicking myself for not realising it myself hahaha. thank you
Another, but less common approach, is to treat B^3 = I like a cubic equation b^3 = 1 for you to solve.

You can rewrite this as B^3 - I = 0 (where 0 is now not a regular scalar 0 but just a 2x2 matrix of zeroes) and factorise the LHS just as you would normally;

(B-I)(B^2 + B + I) = 0

One solution is of course B=I but we are explicitly told that B \neq I in the question, which is always good to refer back to.

So we have that B^2 + B + I = 0.

If you pretend B isn't a matrix, and just a scalar b, and similarly I is just 1, then you would solve b^2 + b + 1 = 0 as b = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i.

This is where knowledge of complex numbers is applicable, because we can treat our normal 2D plane simply as an Argand diagram. Notice how multiplying by i corresponds to a rotation about the origin in the Argand diagram, it precisely means anticlockwise by angle \frac{\pi}{2}.

So, based on the previous part, you could claim (and even verify) that

B = \left( -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i \right) I

It is not immediatly clear what this means but this complex number can be rewritten as

-\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i = \cos \left( \dfrac{2\pi}{3} \right) \pm i \sin \left( \dfrac{2\pi}{3} \right)

hence we have that

B = \left[ \cos \left( \dfrac{2\pi}{3} \right) \pm i \sin \left( \dfrac{2\pi}{3} \right) \right] I = \cos \left( \dfrac{2\pi}{3} \right) I\pm i \sin \left( \dfrac{2\pi}{3} \right) I

or if we rewrite it in matrix form;

B = \begin{pmatrix} \cos\left( \frac{2\pi}{3} \right) & 0 \\ 0 & \cos (\frac{2\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{2\pi}{3} \right)\cdot i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}


As mentioned previously, i corresponds to rotating about the origin by pi/2 radians anticlockwise. Applying it to the identity matrix means we simply rotate the unit vectors \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 1 \end{pmatrix} by this description to obtain \begin{pmatrix} 0 \\ 1 \end{pmatrix} and \begin{pmatrix} -1 \\ 0 \end{pmatrix}, respectively.

Hence i\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} in this sense.


So, we just finish up and note that

B = \begin{pmatrix} \cos\left( \frac{2\pi}{3} \right) & 0 \\ 0 & \cos (\frac{2\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{2\pi}{3} \right) \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \cos\left( \frac{2\pi}{3} \right) & \mp \sin \left( \frac{2\pi}{3} \right) \\ \pm \sin \left( \frac{2\pi}{3} \right) & \cos (\frac{2\pi}{3}) \end{pmatrix}

which is precisely the matrix for B as required, and clearly B is a rotation by pi/3 radians.

The \pm and \mp signs only indicate the direction in which we are rotating, so technically there are two possible answers for B (notice that B^2 + B+ 1 = 0 is a quadratic so by Fundamental Theorem of Algebra it has at exactly 2 roots)

We can either rotate 2pi/3 anticlockwise, or clockwise. This is why the question is phrased in the way 'write down a possible matrix B' implying there are more than one but they are all equally as valid.



Anyway, I do not suggest you take this approach for this question. I merely mention it because it's interesting to see a link between the complex plane and 2D transformations, and treating a matrix equation as a regular polynomial is helpful at times because you can still use factorisation just as you would normally.
For instance, you can use this if you wish to determine two possible matrices A such that A^2 + A - 6I = 0.

This approach comes with a caveat in that you should be careful when there is more than one matrix in an equation, simply due to the fact that matrices most of the time do not commute; ie AB \neq BA.
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mnot
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(Original post by RDKGames)
Another, but less common approach, is to treat B^3 = I like a cubic equation b^3 = 1 for you to solve.

You can rewrite this as B^3 - I = 0 (where 0 is now not a regular scalar 0 but just a 2x2 matrix of zeroes) and factorise the LHS just as you would normally;

(B-I)(B^2 + B + I) = 0

One solution is of course B=I but we are explicitly told that B \neq I in the question, which is always good to refer back to.

So we have that B^2 + B + I = 0.

If you pretend B isn't a matrix, and just a scalar b, and similarly I is just 1, then you would solve b^2 + b + 1 = 0 as b = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i.

This is where knowledge of complex numbers is applicable, because we can treat our normal 2D plane simply as an Argand diagram. Notice how multiplying by i corresponds to a rotation about the origin in the Argand diagram, it precisely means anticlockwise by angle \frac{\pi}{2}.

So, based on the previous part, you could claim (and even verify) that

B = \left( -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i \right) I

It is not immediatly clear what this means but this complex number can be rewritten as

-\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i = \cos \left( \dfrac{\pi}{3} \right) \pm i \sin \left( \dfrac{\pi}{3} \right)

hence we have that

B = \left[ \cos \left( \dfrac{\pi}{3} \right) \pm i \sin \left( \dfrac{\pi}{3} \right) \right] I = \cos \left( \dfrac{\pi}{3} \right) I\pm i \sin \left( \dfrac{\pi}{3} \right) I

or if we rewrite it in matrix form;

B = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & 0 \\ 0 & \cos (\frac{\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{\pi}{3} \right)\cdot i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}


As mentioned previously, i corresponds to rotating about the origin by pi/2 radians anticlockwise. Applying it to the identity matrix means we simply rotate the unit vectors \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 1 \end{pmatrix} by this description to obtain \begin{pmatrix} 0 \\ 1 \end{pmatrix} and \begin{pmatrix} -1 \\ 0 \end{pmatrix}, respectively.

Hence i\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} in this sense.


So, we just finish up and note that

B = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & 0 \\ 0 & \cos (\frac{\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{\pi}{3} \right) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & \pm \sin \left( \frac{\pi}{3} \right) \\ \mp \sin \left( \frac{\pi}{3} \right) & \cos (\frac{\pi}{3}) \end{pmatrix}

which is precisely the matrix for B as required, and clearly B is a rotation by pi/3 radians.

The \pm and \mp signs only indicate the direction in which we are rotating, so technically there are two possible answers for B (notice that B^2 + B+ 1 = 0 is a quadratic so by Fundamental Theorem of Algebra it has at exactly 2 roots)

We can either rotate pi/3 anticlockwise, or clockwise. This is why the question is phrases in the way 'write down a possible matrix B' implying there are more than one but they are all equally as valid.



Anyway, I do not suggest you take this approach for this question. I merely mention it because it's interesting to see a link between the complex plane and 2D transformations, and treating a matrix equation as a regular polynomial is helpful at times because you can still use factorisation just as you would normally.
For instance, you can use this if you wish to determine two possible matrices A such that A^2 + A - 6 = 0.

This approach comes with a caveat in that you should be careful when there is more than one matrix in an equation, simply due to the fact that matrices most of the time do not commute; ie AB \neq BA.
:yy: nicely done.

I love reading these posts its is nice to see different ways people approach problems like this and 'elegantly worked' solutions. I dont really ever do line by line mathematics anymore but its great to see.

But i see this a lot on TSR, people keep writing i instead of j :lol: (why cant everyone adopt the imo superior notation).
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RDKGames
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(Original post by mnot)
:yy: nicely done.

I love reading these posts its is nice to see different ways people approach problems like this and 'elegantly worked' solutions. I dont really ever do line by line mathematics anymore but its great to see.

But i see this a lot on TSR, people keep writing i instead of j :lol: (why cant everyone adopt the imo superior notation).
This is proper mathematics, we do not accept the notation \mathbf{j} = \sqrt{-1} around these parts, keep it in the engineering department :hitler: :fuhrer:
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parisnaomi
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(Original post by RDKGames)
Another, but less common approach, is to treat B^3 = I like a cubic equation b^3 = 1 for you to solve.

You can rewrite this as B^3 - I = 0 (where 0 is now not a regular scalar 0 but just a 2x2 matrix of zeroes) and factorise the LHS just as you would normally;

(B-I)(B^2 + B + I) = 0

One solution is of course B=I but we are explicitly told that B \neq I in the question, which is always good to refer back to.

So we have that B^2 + B + I = 0.

If you pretend B isn't a matrix, and just a scalar b, and similarly I is just 1, then you would solve b^2 + b + 1 = 0 as b = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i.

This is where knowledge of complex numbers is applicable, because we can treat our normal 2D plane simply as an Argand diagram. Notice how multiplying by i corresponds to a rotation about the origin in the Argand diagram, it precisely means anticlockwise by angle \frac{\pi}{2}.

So, based on the previous part, you could claim (and even verify) that

B = \left( -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i \right) I

It is not immediatly clear what this means but this complex number can be rewritten as

-\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i = \cos \left( \dfrac{\pi}{3} \right) \pm i \sin \left( \dfrac{\pi}{3} \right)

hence we have that

B = \left[ \cos \left( \dfrac{\pi}{3} \right) \pm i \sin \left( \dfrac{\pi}{3} \right) \right] I = \cos \left( \dfrac{\pi}{3} \right) I\pm i \sin \left( \dfrac{\pi}{3} \right) I

or if we rewrite it in matrix form;

B = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & 0 \\ 0 & \cos (\frac{\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{\pi}{3} \right)\cdot i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}


As mentioned previously, i corresponds to rotating about the origin by pi/2 radians anticlockwise. Applying it to the identity matrix means we simply rotate the unit vectors \begin{pmatrix} 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 1 \end{pmatrix} by this description to obtain \begin{pmatrix} 0 \\ 1 \end{pmatrix} and \begin{pmatrix} -1 \\ 0 \end{pmatrix}, respectively.

Hence i\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} in this sense.


So, we just finish up and note that

B = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & 0 \\ 0 & \cos (\frac{\pi}{3}) \end{pmatrix} \pm \sin \left( \dfrac{\pi}{3} \right) \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \cos\left( \frac{\pi}{3} \right) & \mp \sin \left( \frac{\pi}{3} \right) \\ \pm \sin \left( \frac{\pi}{3} \right) & \cos (\frac{\pi}{3}) \end{pmatrix}

which is precisely the matrix for B as required, and clearly B is a rotation by pi/3 radians.

The \pm and \mp signs only indicate the direction in which we are rotating, so technically there are two possible answers for B (notice that B^2 + B+ 1 = 0 is a quadratic so by Fundamental Theorem of Algebra it has at exactly 2 roots)

We can either rotate pi/3 anticlockwise, or clockwise. This is why the question is phrased in the way 'write down a possible matrix B' implying there are more than one but they are all equally as valid.



Anyway, I do not suggest you take this approach for this question. I merely mention it because it's interesting to see a link between the complex plane and 2D transformations, and treating a matrix equation as a regular polynomial is helpful at times because you can still use factorisation just as you would normally.
For instance, you can use this if you wish to determine two possible matrices A such that A^2 + A - 6I = 0.

This approach comes with a caveat in that you should be careful when there is more than one matrix in an equation, simply due to the fact that matrices most of the time do not commute; ie AB \neq BA.
thanks a lot! that was interesting to read
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