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Math Help - One more determinant related question.

  1. #1
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    One more determinant related question.

    Alright, here is another question that I need some help with, I think I have an idea of how to do it but it never hurts to make sure, though

    The question states:
    solve the following equation.

    <br />
\begin{vmatrix}<br />
0 & 0 & -1\\<br />
3 & x & 0\\<br />
2 & 0 & 3<br />
\end{vmatrix}=x^2<br />

    My idea is to either, expand by minors, or, to evaluate it by using the main diagonals(?) method.
    Which way would be quickest and easiest to do?
    Or am I headed in the wrong direction entirely?
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  2. #2
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    Quote Originally Posted by quikwerk View Post
    Alright, here is another question that I need some help with, I think I have an idea of how to do it but it never hurts to make sure, though

    The question states:
    solve the following equation.

    <br />
\begin{vmatrix}<br />
0 & 0 & -1\\<br />
3 & x & 0\\<br />
2 & 0 & 3<br />
\end{vmatrix}=x^2<br />


    My idea is to either, expand by minors, or, to evaluate it by using the main diagonals(?) method.
    Which way would be quickest and easiest to do?
    Or am I headed in the wrong direction entirely?
    \begin{vmatrix}<br />
 0 & 0 & -1\\<br />
 3 & x_1 & 0\\<br />
 2 & 0 & 3<br />
 \end{vmatrix}=2x_1=x^2\rightarrow x_1=\frac{x^2}{2}
    Last edited by dwsmith; May 14th 2010 at 04:12 PM.
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  3. #3
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    Hello, quikwerk!

    Solve: . \begin{vmatrix}0 & 0 & \text{-}1\\ 3 & x & 0\\ 2 & 0 & 3 \end{vmatrix}\;=\;x^2

    My idea is to either, expand by minors, or, to evaluate it by using the main diagonals method.
    Which way would be quickest and easiest to do?
    I'd expand by minors.

    I can already see all of this: . \begin{vmatrix}0&0&\text{-}1 \\ 3&x&0 \\ 2&0&3\end{vmatrix} \;=\;x^2

    . . 0\begin{vmatrix}x&0\\0&3\end{vmatrix} \:-\: 0\begin{vmatrix}3&0\\2&3\end{vmatrix} \:-\: 1\begin{vmatrix}3&x\\2&0\end{vmatrix} \;\;=\;\;x^2 \qquad\Rightarrow\qquad 0 \;-\; 0 \;-\;1(0 - 2x) \;\;=\;\;x^2


    Hence: . 2x \:=\:x^2 \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0

    Therefore: . x \;=\;0,\:2

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  4. #4
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    Take the determinant of the LHS and solve, should be pretty easy.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, quikwerk!

    I'd expand by minors.

    I can already see all of this: . \begin{vmatrix}0&0&\text{-}1 \\ 3&x&0 \\ 2&0&3\end{vmatrix} \;=\;x^2

    . . 0\begin{vmatrix}x&0\\0&3\end{vmatrix} \:-\: 0\begin{vmatrix}3&0\\2&3\end{vmatrix} \:-\: 1\begin{vmatrix}3&x\\2&0\end{vmatrix} \;\;=\;\;x^2 \qquad\Rightarrow\qquad 0 \;-\; 0 \;-\;1(0 - 2x) \;\;=\;\;x^2

    Hence: . 2x \:=\:x^2 \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0

    Therefore: . x \;=\;0,\:2


    Don't we want the determinant to be
    x^2?
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