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Math Help - Use Cramer's Rule to find x_2 when Ax=...

  1. #1
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    Use Cramer's Rule to find x_2 when Ax=...

    I think I was handed a trick question.

    For the matrix \left(\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\1 & -1 & -1\end{array}\right), use Cramer's Rule to find x_2 when Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right).

    My current answer is x_2=\frac{16}{3}. Is this correct, or do I have to determine what x is by inverting A?
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  2. #2
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    Quote Originally Posted by Runty View Post
    I think I was handed a trick question.

    For the matrix \left(\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\1 & -1 & -1\end{array}\right), use Cramer's Rule to find x_2 when Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right).

    My current answer is x_2=\frac{16}{3}. Is this correct, or do I have to determine what x is by inverting A?
    You are correct as you are only asked for x_2 .

    Solving for x means solving for (x_1,x_2,x_3)

    as x = \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right)

    This requires inverting A
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  3. #3
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    Quote Originally Posted by pickslides View Post
    You are correct as you are only asked for x_2 .

    Solving for x means solving for (x_1,x_2,x_3)

    as x = \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right)

    This requires inverting A
    Alright, doing that, my new answer is x_2=-\frac{65}{9}, with x=\left(\begin{array}{c}7/3 \\ 16/3 \\ -4 \end{array}\right). Is that correct, or have I made a mistake somewhere?

    Remember, the question says to use Cramer's Rule to find x_2.
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  4. #4
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    Have a look at Online Matrix Calculator
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  5. #5
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    The problem told you to use "Cramer's rule":

    x_2= \frac{\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & -2 & 3 \\ 1 & 1 & -1\end{array}\right|}{\left|\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\ 1 & -1 & 3\end{array}\right|}.

    I have no idea why you would consider that a "trick question".
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    The problem told you to use "Cramer's rule":

    x_2= \frac{\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & -2 & 3 \\ 1 & 1 & -1\end{array}\right|}{\left|\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\ 1 & -1 & 3\end{array}\right|}.

    I have no idea why you would consider that a "trick question".
    I say it because I believe that by stating Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right), it could mean that you need to do this: x=A^{-1}\left(\begin{array}{c}1 \\-2\\1\end{array}\right), before you apply Cramer's Rule. This would come out to x=\left(\begin{array}{c}7/3 \\ 16/3 \\ -4 \end{array}\right), and this x would be used in Cramer's Rule, not Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right).

    Of course, I could be wrong. That's why I thought this might be a trick question.
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  7. #7
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    Nevermind, I just realized the mistake I made. It was a notation mistake. The correct answer is x_2=\frac{16}{3}.
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