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Math Help - Multiplication

  1. #1
    Junior Member lanierms's Avatar
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    Multiplication

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

    I got as far as getting 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
    a, b, c, d are real numbers.
    When they satisfy the equations a^2+b^2=1, c^2+d^2=1, ac+bd=0, find the value of ab+cd.

    I got as far as getting 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 \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 \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 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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    a=-\displaytype{\frac{bd}{c}}

    a^2+b^2=1

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

    b^2(d^2+c^2)=c^2

    since d^2+c^2=1

    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:

    b^2=c^2

    b=c or b=-c

    i) When b=c,

    ac+bd=ab+bd=ab+cd=0

    ii) When b=-c,

    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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