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Math Help - Solve determinant

  1. #1
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    Solve determinant

    <br />
\left(<br />
\begin<br />
{array}<br />
{ccccc}<br />
1&n&n&...&n\\<br />
n&2&n&...&n\\<br />
n&n&3&...&n\\<br />
...&...&...&...&...\\<br />
n&n&n&...&n<br />
\end<br />
{array}<br />
\right)<br />

    The solution is  (-1)^{(n-1)}*n!

    Would someone be so kind and explain me how to do it? Thanks in advance!
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  2. #2
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    Subtract the last row from every other row, this leaves us

    \left(\begin{array}{ccccc}1-n&0&0&...&0\\0&2-n&0&...&0\\0&0&3-n&...&0\\...&...&...&...&...\\n&n&n&...&n\end{arra  y}\right)
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  3. #3
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    Quote Originally Posted by SimonM View Post
    Subtract the last row from every other row, this leaves us

    \left(\begin{array}{ccccc}1-n&0&0&...&0\\0&2-n&0&...&0\\0&0&3-n&...&0\\...&...&...&...&...\\n&n&n&...&n\end{arra  y}\right)
    then subtract the last column from every other column and you're done!
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  4. #4
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    re

    thanks guys. But what will the result be in your cases?
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  5. #5
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    Quote Originally Posted by babanner View Post
    thanks guys. But what will the result be in your cases?
    Have you done what was suggested? What do you get?
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  6. #6
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    re

    well, i have posted in the first post what the final result should be. However in the above posts the final result seems to be different. In your cases i think the result will be:  (n-1)*(n-2)* ... *n
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  7. #7
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    Re

    Thanks guys. I figured it out

    I didn't notice that  n*(n-1)*(n-2)*(n-3) ... *1 is the same as  1*2*3*4*...*(n-1)*n which both equal  n!
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