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Thread: matrices problem

  1. #1
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    matrices problem

    Question:
    Given that $\displaystyle P=\begin{pmatrix} 3 &4 \\ -4 &3 \end{pmatrix}$ and $\displaystyle A = \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix} $, find the martix M where $\displaystyle \mathbf{M=P^{-1}AP} $. Hence or otherwise find $\displaystyle \mathbf{M^{5}} $


    Solution:


    $\displaystyle \mathbf{M=P^{-1}AP}$
    $\displaystyle M=\frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix} \begin{pmatrix} 3 & 4\\ -4 & 3 \end{pmatrix}$
    $\displaystyle M= \frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 6+0 &8+0 \\ 0+(-4) &0+3 \end{pmatrix}$
    $\displaystyle M=\frac{1}{25} \begin{pmatrix} 3&-4 \\ 4& 3 \end{pmatrix} \begin{pmatrix} 6 &8 \\ -4&3 \end{pmatrix} $
    $\displaystyle M= \frac{1}{25} \begin{pmatrix} 18+16&24-12 \\ 24-12& 32+9 \end{pmatrix} $
    $\displaystyle \frac{1}{25} \begin{pmatrix} 34&12 \\ 12& 41 \end{pmatrix}$


    My problem now is how do i find


    $\displaystyle \mathbf{M^{5}} $
    Am i suppose to raise each number in the matrix to the power 5 ( eg; $\displaystyle 34^5$) or raise the matrix M to the power 5


    the answer given by the book is
    $\displaystyle \frac{1}{25} \begin{pmatrix} 304&372 \\ 372& 521 \end{pmatrix} $
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  2. #2
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    Re: matrices problem

    The idea is that $M= P^{-1}AP$ where $A$ is diagonal.

    $M^5 = P^{-1}AP \cdot P^{-1}AP \cdot \cdot \cdot P^{-1}AP = P^{-1}A^5 P$

    $(A^5)_{i,i} = (A_{i,i})^5$
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  3. #3
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    Re: matrices problem

    Quote Originally Posted by bigmansouf View Post
    Question:
    Given that $\displaystyle P=\begin{pmatrix} 3 &4 \\ -4 &3 \end{pmatrix}$ and $\displaystyle A = \begin{pmatrix} 2 &0 \\ 0 &1 \end{pmatrix} $, find the martix M where $\displaystyle \mathbf{M=P^{-1}AP} $. Hence or otherwise find $\displaystyle \mathbf{M^{5}} $
    Here is the trick:
    $\displaystyle \underbrace {{P^{ - 1}}AP \cdots {P^{ - 1}}AP}_{5 \times } = {P^{ - 1}}{A^5}P$

    But $\displaystyle {\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}}\right )^5 ={\left( {\begin{array}{*{20}{c}}32&0\\0&1\end{array}} \right)}}$
    Last edited by Plato; Apr 11th 2019 at 01:20 PM.
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  4. #4
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    Re: matrices problem

    With matrices, $\displaystyle M^5$ means raise the matrix M to the fifth ie MxMxMxMxM.

    For diagonal matrices (all non-diagonal elements are 0) such as A you can calculate $\displaystyle A^5$ by simply raising the diagonal elements to the fifth. I suggest you find $\displaystyle A^2$ by multiplying $\displaystyle A*A$ and it'll make sense why this is so for diagonal matrices.


    As Romsek said
    $\displaystyle M^5 = (P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP) *(P^{-1}AP)
    = P^{-1}*(AP *P^{-1})*(AP *P^{-1})*(AP *P^{-1})*(AP *P^{-1})*AP$ using the associativity of matrix multiplication
    $\displaystyle = P^{-1}*A*A*A*A*A*P$
    $\displaystyle = P^{-1}*A^5* P$


    Now since $\displaystyle A^5$ is easy to calculate (since A is diagonal), it is easier to calculate $\displaystyle = P^{-1}*A^5* P$ than $\displaystyle M^5$.
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