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FP1 matrix

2016-05-14.png for part C, isnt it QT' = T ? but it is wrong and i cant get the answer, can someone help me?
Reply 1
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Zacken
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Original post by alesha98
2016-05-14.png for part C, isnt it QT' = T ? but it is wrong and i cant get the answer, can someone help me?


Well T is a triangle and not a matrix.

I agree that QT' = T. And we know that to go from T to T' we need to use P and to go from T' to T we need to use P^(-1).

So PT = T^(-1).

So QT' = T is QPT = T. This gets you Q = P^(-1).
(edited 7 years ago)
Reply 2
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alesha98
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Original post by Zacken
Well T is a transformation and not a matrix.

I agree that QT' = T. And we know that to go from T to T' we need to use P and to go from T' to T we need to use P^(-1).

So QT' = T is QP^(-1) = P. This gets you Q = PP.


So from what you are saying, should it be QT'=T, but PT = T'. So if i sub PT = T' in to QT' = T, i will get QPT = T, then QP=I, Q = P-1?
Reply 3
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Zacken
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Original post by alesha98
So from what you are saying, should it be QT'=T, but PT = T'. So if i sub PT = T' in to QT' = T, i will get QPT = T, then QP=I, Q = P-1?


Yes, sorry. Slip up on my part, that's correct.
Reply 4
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alesha98
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Original post by Zacken
Yes, sorry. Slip up on my part, that's correct.


Thankyou , i got another question, can you help me as well please?
Reply 5
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Zacken
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Original post by alesha98
Thankyou , i got another question, can you help me as well please?


Sure, post it.
Reply 6
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alesha98
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Original post by Zacken
Sure, post it.

2016-05-14 (1).png I totally struggled to do 9(d), How to undo arg?
Reply 7
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Zacken
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Original post by alesha98
2016-05-14 (1).png I totally struggled to do 9(d), How to undo arg?


There's no need to undo the arg. You just need to put in the value of w to get: arg⁡((λ+10)+4i)=π4\displaystyle \arg((\lambda + 10) +4i) = \frac{\pi}{4}.

We know that if arg⁡z=π4\arg z = \frac{\pi}{4} then it must mean that the imaginary part of z has to equal the real part of z. Because, well the half-line is y = x. The diagonal line passing through the centre. That is Im(z) =Re(z).

So we need λ+10=4\lambda + 10 = 4.
Reply 8
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alesha98
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4
Original post by Zacken
There's no need to undo the arg. You just need to put in the value of w to get: arg⁡((λ+10)+4i)=π4\displaystyle \arg((\lambda + 10) +4i) = \frac{\pi}{4}.

We know that if arg⁡z=π4\arg z = \frac{\pi}{4} then it must mean that the imaginary part of z has to equal the real part of z. Because, well the half-line is y = x. The diagonal line passing through the centre. That is Im(z) =Re(z).

So we need λ+10=4\lambda + 10 = 4.

Oh ... It is a tricky question for me, but thanks for your explanation, it makes sense to me now :smile:
Reply 9
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Zacken
22
Original post by alesha98
Oh ... It is a tricky question for me, but thanks for your explanation, it makes sense to me now :smile:


No problem. :-)

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