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Math Help - determinant

  1. #1
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    determinant

    prove that

    |a b ax+by |
    |b c bx+cy |
    |ax+by bx+cy 0 |

    =(b^2-ac)(ax^2+2bxy+cy^2)
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  2. #2
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    Quote Originally Posted by minaxijgupta
    prove that

    |a b ax+by |
    |b c bx+cy |
    |ax+by bx+cy 0 |

    =(b^2-ac)(ax^2+2bxy+cy^2)
    Compute this determinant by using the 3rd collossus.
    Thus, we have,
    (-1)^{1+3}(ax+by)\left| \begin{array}{cc} b&c\\ ax+by& bx+cy \end{array} \right|  +(-1)^{2+3}(bx+cy)\left| \begin{array}{cc} a&b\\ ax+by&bx+cy \end{array} \right| +(-1)^{3+3}0 \left| \begin{array}{cc} a&b\\b&c \end{array} \right|
    Thus,
    (ax+by)(b^2x+bcy-acx-bcy) -(bx+cy)(abx+acy-abx-b^2y)
    Thus,
    (ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)
    Now, just open this expression up and also open up the expression you are trying to prove to see that they match!
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    ...
    Thus,
    (ax+by)(b^2x+bcy-acx-bcy) -(bx+cy)(abx+acy-abx-b^2y)
    Thus,
    (ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)
    Now, just open this expression up and also open up the expression you are trying to prove to see that they match!
    Hello,

    with this result (which I never would have found) I suggest factorization:

    (ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)

    (ax+by)\cdot x\cdot (b^2-ac)-(bx+cy)\cdot (-y)\cdot (b^2-ac)

    (b^2-ac)\cdot \left[(ax+by)\cdot x+(bx+cy)\cdot (+y) \right]

    And now, minaxijgupta, the next steps are for you.

    Best wishes

    EB
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  4. #4
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    Hello, minaxijgupta!

    I think I've come up with a good approach . . .


    Prove that: . \begin{vmatrix}a & b &  ax+by \\ b &  c & bx+cy \\ ax+by & bx+cy &  0\end{vmatrix}\;=\;(b^2-ac)(ax^2+2bxy+cy^2)

    Perform some row operations to simplify the determinant . . .


    \begin{array}{ccc}x\cdot R_1 \\ y\cdot R_2 \\. \end{array}\;\begin{vmatrix}ax & bx & ax^2+bxy \\ by & cy & bxy + cy^2 \\ ax+by & bx + cy & 0\end{vmatrix}


    \begin{array}{ccc}R_1+R_2\\.\\.\end{array}\;\begin  {vmatrix}ax+by & bc + cy & ax^2 + 2bxy + cy^2 \\ by & cy & bxy + cy^2 \\ ax + by & bx + cy & 0\end{vmatrix}


    \begin{array}{ccc}R_1-R_3\\R_2\div xy\\ .\end{array}\;\begin{vmatrix}0 & 0 & ax^2+2bxy+cy^2\\ \frac{b}{x} & \frac{c}{x} & \frac{bx + cy}{x} \\ ax + by & bx + cy & 0\end{vmatrix}


    Now we "expand by minors" (using the top row):

    (ax^2 + 2bxy + cy^2)\,\begin{vmatrix}\frac{b}{x} & \frac{c}{x} \\ ax + by & bx + cy\end{vmatrix}

    . . =\;(ax^2+2bxy+cy^2)\,\left[\frac{b}{x}(bx+cy) - \frac{c}{x}(ax + by)\right]

    . . = \;(ax^2 + 2bxy + cy^2)\,\left(b^2 + \frac{bcy}{x} - ac - \frac{bcy}{x}\right)

    . . = \;(ax^2 + 2bxy + cy^2)\,(b^2 - ac) . . . ta-DAA!

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