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Year 13 Maths Help Thread

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    Area of a circle sector = 30
    Arc Length = 10
    Find theta

    someone pls help I've been trying for a while and can't figure it out, also it's in radians.
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    (Original post by medhelp)
    Area of a circle sector = 30
    Arc Length = 10
    Find theta

    someone pls help I've been trying for a while and can't figure it out, also it's in radians.
    Arc length is r\theta and sector area is \frac{r^2{\theta}}{2}.
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    (Original post by Palette)
    Arc length is r\theta and sector area is \frac{r^2{\theta}}{2}.
    Yeah I know the formulae... I just dk how to do it without knowing r or theta
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    (Original post by medhelp)
    Yeah I know the formulae... I just dk how to do it without knowing r or theta
    Use a substitution for r from the arc length. Get r in terms of \theta.
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    (Original post by RDKGames)
    Use a substitution for r from the arc length. Get r in terms of \theta.
    **** yes I remember now!
    I'm trying to go over some AS stuff lol, thanks dude
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    Does anyone here know who stickied this thread?
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    (Original post by Palette)
    Does anyone here know who stickied this thread?
    Regular users have no way of knowing I'm not sure if the people with permission to do so can even see whodunnit, if my memory serves my correctly.
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    (Original post by SeanFM)
    Regular users have no way of knowing I'm not sure if the people with permission to do so can even see whodunnit, if my memory serves my correctly.
    I suggested these to be stickied either on this thread or the Y12 one a few weeks ago. Deffo don't have any thing to do with that :teehee:
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    I'm stuck on a weird question.
    Show that the following relationship between the non-zero matrices
     \mathbf{AB} =\mathbf{AC} + \mathbf{I} , where  \mathbf{I} is the identity matrix, only holds when  \mathbf{A} is non-singular.
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    (Original post by Ano9901whichone)
    I'm stuck on a weird question.
    Show that the following relationship between the non-zero matrices
     \mathbf{AB} =\mathbf{AC} + \mathbf{I} , where  \mathbf{I} is the identity matrix, only holds when  \mathbf{A} is non-singular.
    idk what singular means but doing the following:
    consider A as singular and find a contradiction, then find an example where A works where it is non-singular
    suffices
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    (Original post by Ano9901whichone)
    I'm stuck on a weird question.
    Show that the following relationship between the non-zero matrices
     \mathbf{AB} =\mathbf{AC} + \mathbf{I} , where  \mathbf{I} is the identity matrix, only holds when  \mathbf{A} is non-singular.
    if singular means 'has only one element' the question is essentially

    if a natural number greater than 1 divides n, show it does not divide kn+1
    which you can do via modular arithmetic
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    (Original post by MathMoFarah)
    if singular means 'has only one element' the question is essentially

    if a natural number greater than 1 divides n, show it does not divide kn+1
    which you can do via modular arithmetic
    Singular means it has no inverse.
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    (Original post by MathMoFarah)
    if singular means 'has only one element' the question is essentially

    if a natural number greater than 1 divides n, show it does not divide kn+1
    which you can do via modular arithmetic
    (Singular matrices have a determinant of zero i.e. They are NOT invertible)
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    (Original post by Ano9901whichone)
    Singular means it has no inverse.
    (Original post by Xsk)
    (Singular matrices have a determinant of zero i.e. They are invertible)
    which is it?
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    (Original post by MathMoFarah)
    which is it?
    Lmao meant to say NOT invertible, sorry
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    I never did any matrices so idk but it sounds like the fact that I = AA^-1 is important (Shows that it has to have an inverse for I to be a factor and the sum to work type-thinking)
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    (Original post by Ano9901whichone)
    I'm stuck on a weird question.
    Show that the following relationship between the non-zero matrices
     \mathbf{AB} =\mathbf{AC} + \mathbf{I} , where  \mathbf{I} is the identity matrix, only holds when  \mathbf{A} is non-singular.
    Hmm... What if we say AB - AC = I, then A(B - C) = I as matrices are distributive...
    let (B - C) = D for clarity, then AD = I.

    Det(AD) = Det(A) x Det(D)
    If A is singular, then Det(A) is 0, therefore Det(AD) = 0.
    However, the determinant of the Identity is 1.
    This contradicts, therefore the original statement doesn't hold if A is singular. Go on some spiel to prove it then holds for non-singular matrices.

    Or that's how I've taken the question anyhow. It's been a year since I did matrices in fp1, so sorry if there is some flawed logic there lmao
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    (Original post by Xsk)
    Hmm... What if we say AB - AC = I, then A(B - C) = I as matrices are distributive...
    let (B - C) = D for clarity, then AD = I.

    Det(AD) = Det(A) x Det(D)
    If A is singular, then Det(A) is 0, therefore Det(AD) = 0.
    However, the determinant of the Identity is 1.
    This contradicts, therefore the original statement doesn't hold if A is singular. Go on some spiel to prove it then holds for non-singular matrices.

    Or that's how I've taken the question anyhow. It's been a year since I did matrices in fp1, so sorry if there is some flawed logic there lmao
    Yeah seems right like that. I suppose that B-C must be inverse of A, so it A is singular it can't happen.
    Thanks.
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    (Original post by Ano9901whichone)
    Yeah seems right like that. I suppose that B-C must be inverse of A, so it A is singular it can't happen.
    Thanks.
    Oh. Yeah that seems a lot less convoluted and realllyyy simple. Ahahahhaa well I tried, well done
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    (Original post by Ano9901whichone)
    I'm stuck on a weird question.
    Show that the following relationship between the non-zero matrices
     \mathbf{AB} =\mathbf{AC} + \mathbf{I} , where  \mathbf{I} is the identity matrix, only holds when  \mathbf{A} is non-singular.
    \mathbf{AB}-\mathbf{AC}=\mathbf{I}

    \Rightarrow \mathbf{A}(\math{B-C})=\mathbf{I} therefore \math{A} must have an inverse which is equal to \mathbf{B-C} hence it is non-singular.

    Just my thoughts at 1am lol

    EDIT: nvm I guess I was beaten to the answer
 
 
 
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