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Thread: Multiplication

  1. #1
    Junior Member lanierms's Avatar
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    Multiplication

    $\displaystyle a, b, c, d$ are real numbers.
    When they satisfy the equations $\displaystyle a^2+b^2=1$, $\displaystyle c^2+d^2=1$, $\displaystyle ac+bd=0$, find the value of $\displaystyle ab+cd$.

    I got as far as getting $\displaystyle a^2+c^2=1$, but I can't seem to go any further.

    Please help.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by lanierms View Post
    $\displaystyle a, b, c, d$ are real numbers.
    When they satisfy the equations $\displaystyle a^2+b^2=1$, $\displaystyle c^2+d^2=1$, $\displaystyle ac+bd=0$, find the value of $\displaystyle ab+cd$.

    I got as far as getting $\displaystyle a^2+c^2=1$, but I can't seem to go any further.

    Please help.
    In terms of matrices, these equations can be written $\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix }a&c\\b&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. But that implies that those two matrices are inverses of each other, so that $\displaystyle \begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix }a&b\\c&d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$. Hence $\displaystyle ab+cd=0.$
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  3. #3
    Junior Member lanierms's Avatar
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    Aren't there any other ways than using matrix? Because I didn't learn matrix yet and I have no idea what you are saying.
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  4. #4
    Senior Member BAdhi's Avatar
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    $\displaystyle a=-\displaytype{\frac{bd}{c}}$

    $\displaystyle a^2+b^2=1$

    $\displaystyle \displaytype{\frac{b^2d^2}{c^2}}+b^2=1$

    $\displaystyle b^2(d^2+c^2)=c^2$

    since $\displaystyle d^2+c^2=1$

    $\displaystyle b^2=c^2$

    I think you can use this result
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  5. #5
    Junior Member lanierms's Avatar
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    Thank you so much BAdhi. I got this:

    $\displaystyle b^2=c^2$

    $\displaystyle b=c$ or $\displaystyle b=-c$

    i) When $\displaystyle b=c$,

    $\displaystyle ac+bd=ab+bd=ab+cd=0$

    ii) When $\displaystyle b=-c$,

    $\displaystyle ac+bd=-ab+bd=-ab-cd=-(ab+cd)=0$

    Is there anything wrong with it?
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  6. #6
    Senior Member BAdhi's Avatar
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    just the way i thought
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