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    This is my working for 7(i).

    A^2=\begin{pmatrix} 1 & 6 \\0 & 1\end{pmatrix}

    A^3=\begin{pmatrix} 1 & 9 \\0 & 1\end{pmatrix}

    so it appears a formula for A^n could be

    A^n=\begin{pmatrix} 1 & 3n \\0 & 1\end{pmatrix}

    To prove this by induction let p(n)=A^n=\begin{pmatrix} 1 & 3n \\0 & 1\end{pmatrix}

    Now p(1) is obviously true by definition.

    Assume p(k) is true.

    p(k+1)=\begin{pmatrix} 1 & 3k+3 \\0 & 1\end{pmatrix}

    Can I conclude that the general matrix A^n is proved by induction from this? What is the inductive assumption?
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    If I could use latex, I would explain. But I can't so sorry.
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    (Original post by TheEd)
    To prove this by induction let p(n)=A^n=\begin{pmatrix} 1 & 3n \\0 & 1\end{pmatrix}
    A pedantic point. p(n) is a sentence, not your actual matrix itself. So p(n) is the statement "A^n = blah".

    (Original post by TheEd)
    [p(n):] A^n=\begin{pmatrix} 1 & 3n \\0 & 1\end{pmatrix}

    [...]

    Assume p(k) is true.

    p(k+1)=\begin{pmatrix} 1 & 3k+3 \\0 & 1\end{pmatrix}
    Firstly, I assume you mean A^(k+1) here (see above note). Your inductive hypothesis is that p(k) is true, i.e. that A^k=\begin{pmatrix} 1 & 3k \\0 & 1\end{pmatrix}. You have to prove from here that A^(k+1) is what you said it is, by actually multiplying A^k by A.
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    to find A^(k+1), multiply A^k by A
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    ..don't remember any of that in the film before! Badum dum!
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    (Original post by generalebriety)
    A pedantic point. p(n) is a sentence, not your actual matrix itself. So p(n) is the statement "A^n = blah".


    Firstly, I assume you mean A^(k+1) here (see above note). Your inductive hypothesis is that p(k) is true, i.e. that A^k=\begin{pmatrix} 1 & 3k \\0 & 1\end{pmatrix}. You have to prove from here that A^(k+1) is what you said it is, by actually multiplying A^k by A.
    Thanks
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    The method is as follows:

    Show that P(1) is true.
    Assume that P(k) is true where k is an element of the set of positive integers.
    Show that P(k+1) is true by multiplying A^k by A.
    P(1) is true, P(k) is true and implies that P(k+1) is true, so P(n) is true for n∈Z^+.

    Sorry for my LaTeX deficiency.
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    (Original post by CocoPop)
    The method is as follows:

    Show that P(1) is true.
    Assume that P(k) is true where k is an element of the set of positive integers.
    Show that P(k+1) is true by multiplying A^k by A.
    P(1) is true, P(k) is true and implies that P(k+1) is true, so P(n) is true for n∈Z^+.

    Sorry for my LaTeX deficiency.
    For the second matrix the general matrix is:

    A^n=\begin{pmatrix} cos(n\theta) & sin(n\theta) \\-sin(n\theta) & cos(n\theta)\end{pmatrix} .

    Proving it by induction you get:

    A^{k+1}=A^kA=\begin{pmatrix} {cos(k \theta)cos(\theta)-sin(k\theta)sin(\theta)} & {cos(k\theta)sin(\theta)+sin(k\t  heta)cos(\theta)} \\ {-sin(k\theta)cos(\theta)-cos(k\theta)sin(\theta)} & {cos(k\theta)cos(\theta)-sin(k\theta)sin(\theta)} \end{pmatrix} .

    But I don't see what this proves?
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    Addition formulae for sine and cosine.
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    (Original post by TheEd)
    Proving it by induction
    :confused:

    You don't quite seem to understand the concept. You haven't proved anything by multiplying A^k by A; you now need to show that A^(k+1) is equal to what you predicted it would be before (i.e. that p(k+1) is true). Do this by using trig addition formulas.
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    (Original post by generalebriety)
    :confused:

    You don't quite seem to understand the concept. You haven't proved anything by multiplying A^k by A; you now need to show that A^(k+1) is equal to what you predicted it would be before (i.e. that p(k+1) is true). Do this by using trig addition formulas.
    Ah I get it now Thanks for the clarification!
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    Sorry for not making that clear before. A good way of showing your method is by referring to the "left hand side" (LHS) and "right hand side" (RHS) of the equation.

    So you could say that:

    LHS = A^(k+1), here you can show what your predicted outcome is according to the statement for A^n.

    Then on the Right hand side you could say that A^(k+1)=A^k * A

    So:

    RHS = A^k * A

    You can then show that LHS = RHS... Don't know if this is the method that other people use, but I find it helpful.
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    (Original post by CocoPop)
    Sorry for not making that clear before. A good way of showing your method is by referring to the "left hand side" (LHS) and "right hand side" (RHS) of the equation.

    So you could say that:

    LHS = A^(k+1), here you can show what your predicted outcome is according to the statement for A^n.

    Then on the Right hand side you could say that A^(k+1)=A^k * A

    So:

    RHS = A^k * A

    You can then show that LHS = RHS... Don't know if this is the method that other people use, but I find it helpful.
    Notationally it's a bit bad. It's quite obvious that A^(k+1) = A^k * A regardless of how you think the answer should come out looking. But whatever works for you, I guess...
 
 
 
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