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Thread: find [MATH]\alpha , \beta[/MATH]

  1. #1
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    Exclamation find [MATH]\alpha , \beta[/MATH]

    If A=$\displaystyle \begin{pmatrix}
    0 & 1 \\
    -1 & 0
    \end{pmatrix} $, find $\displaystyle \alpha$,$\displaystyle \beta$ so that $\displaystyle (\alpha l + \beta A)^2 = A
    $
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  2. #2
    Super Member Gamma's Avatar
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    I think that is supposed to be the Identity matrix?

    So you need
    $\displaystyle \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2$

    Well multiply it out and see what you get. What I got was:
    $\displaystyle \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta$
    and
    $\displaystyle 2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2$

    But then use 1 and we see we need $\displaystyle \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$
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  3. #3
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    I got this ?

    Quote Originally Posted by Gamma View Post
    I think that is supposed to be the Identity matrix?

    So you need
    $\displaystyle \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}\alpha & \beta \\-\beta & \alpha\end{pmatrix}^2$

    Well multiply it out and see what you get. What I got was:
    $\displaystyle \alpha^2=\beta^2 \Rightarrow \alpha = +/- \beta$
    and
    $\displaystyle 2\alpha \beta = 1 \Rightarrow \alpha \beta = 1/2$

    But then use 1 and we see we need $\displaystyle \alpha (+/- \alpha)=1/2 \Rightarrow \alpha = \frac{1}{\sqrt{2}}=\beta$

    $\displaystyle ( \alpha I + \beta A)^2$ = $\displaystyle \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}$


    Is this right and what should i do next ????
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  4. #4
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    Quote Originally Posted by zorro View Post
    $\displaystyle ( \alpha I + \beta A)^2$ = $\displaystyle \begin{bmatrix} \alpha^2 + \beta^2 & 0 \\ 0 & \alpha^2 - \beta^2 \end{bmatrix}$


    Is this right and what should i do next ????
    According to the time stamp, gamma gave you the answer 8 months ago. Why are you asking now?
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