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    Can someone check my answer to this question please? There's no mark scheme to the question.

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    (Original post by Glavien)
    Can someone check my answer to this question please? There's no mark scheme to the question.

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    Seems good to me, you can always check your answer by see if 2+i squared equals 3 + 4i
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    Any recommendations on what to do after doing all the edexcel papers?
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    (Original post by AmarPatel98)
    Any recommendations on what to do after doing all the edexcel papers?
    Chill out.
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    (Original post by alfmeister)
    Seems good to me, you can always check your answer by see if 2+i squared equals 3 + 4i
    Thanks!



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    Is part b of this question correct?

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    (Original post by Glavien)
    Is part b of this question correct?

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    Yes it seems correct however on the second to last line of working why do you have -λ
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    I have no clue what they're doing in this question, can someone please explain?

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    (Original post by fullmetal heart)
    I have no clue what they're doing in this question, can someone please explain?

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    which part don't you understand?
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    (Original post by fullmetal heart)
    I have no clue what they're doing in this question, can someone please explain?

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    for part a

    they are multiplying by the inverse of B on each side. So the inverse of B times the identity matrix equuals the inverse of B. Anything multiplied by the identity matrix is the same . And B*B^-1 = I so you get AB= B^-1 then you multiply agin by B^-1 to get A*B*B^-1 =B^-1B^-1
    which equals A=B^-1B^-1

    Hope that helps
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    (Original post by maruchan)
    which part don't you understand?
    How they know what to do next
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    (Original post by maruchan)
    for part a

    they are multiplying by the inverse of B on each side. So the inverse of B times the identity matrix equuals the inverse of B. Anything multiplied by the identity matrix is the same . And B*B^-1 = I so you get AB= B^-1 then you multiply agin by B^-1 to get A*B*B^-1 =B^-1B^-1
    which equals A=B^-1B^-1

    Hope that helps
    Thanks
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    (Original post by fullmetal heart)
    Thanks
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    In proof by induction for divisibility proofs, you know how you might typically do
    f(k+1) - f(k), find this, and then make f(k+1) the subject

    i.e. f(k+1) - f(k) = ABC
    so f(k+1) = ABC + f(k)

    Can you do f(k+1) + f(k) ???

    i.e. f(k+1) + f(k) = ABC
    so f(k+1) = ABC - f(k)
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    What's some of the hardest things they could ask?
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    (Original post by AmarPatel98)
    In proof by induction for divisibility proofs, you know how you might typically do
    f(k+1) - f(k), find this, and then make f(k+1) the subject

    i.e. f(k+1) - f(k) = ABC
    so f(k+1) = ABC + f(k)

    Can you do f(k+1) + f(k) ???

    i.e. f(k+1) + f(k) = ABC
    so f(k+1) = ABC - f(k)
    Do you think you can? I think you can

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    (Original post by AmarPatel98)
    In proof by induction for divisibility proofs, you know how you might typically do
    f(k+1) - f(k), find this, and then make f(k+1) the subject

    i.e. f(k+1) - f(k) = ABC
    so f(k+1) = ABC + f(k)

    Can you do f(k+1) + f(k) ???

    i.e. f(k+1) + f(k) = ABC
    so f(k+1) = ABC - f(k)
    Do you understand it?

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    (Original post by AmarPatel98)
    In proof by induction for divisibility proofs, you know how you might typically do
    f(k+1) - f(k), find this, and then make f(k+1) the subject

    i.e. f(k+1) - f(k) = ABC
    so f(k+1) = ABC + f(k)

    Can you do f(k+1) + f(k) ???

    i.e. f(k+1) + f(k) = ABC
    so f(k+1) = ABC - f(k)
    Yeah, of course.
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    Name:  IMG-20160517-WA0044.jpg
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Size:  99.3 KBEDIT: SHOULD READ 'Divisible by 18' not 8

    can someone check this pls ... im not sure if this would get full marks. It's from the paper linked below - June 14 IAL F1 - and the method i did it by isn't in the markscheme. Thanks!

    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
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    For proof by induction, what does all the notation (e.g. nEZ+) actually mean?
 
 
 
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