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Math Help - confusion involving method to compute determinant

  1. #1
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    confusion involving method to compute determinant

    1) Do all square matrices have a unique determinant?
    2) If the answer to the above question is yes, then what part of the following procedure am I over looking?

    I am trying to use "computation via reduction to a triangular form" to find the determinant of \left[\begin{matrix}<br />
4 & -3 & 5 \\<br />
5 & 2 & 0 \\<br />
2 & 0 & 4<br />
\end{matrix} \right]<br />

    by adding -2.5 times the first row to the second gives: \left[\begin{matrix}<br />
4 & -3 & 5 \\<br />
0 & -5.75 & -6.25 \\<br />
0 & 1.5 & 1.5<br />
\end{matrix} \right]<br />

    Here's my confusion. To eliminate the first 1.5 I can either add 1/2 the first row to the third or -5.75/1.5 the second row to the third. Depending on which route I take the last element in the third row will be differnt, and when I multiply the numbers along the diagonal I will get 60 or 72 for the determinant.
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  2. #2
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    Quote Originally Posted by superdude View Post
    1) Do all square matrices have a unique determinant?
    2) If the answer to the above question is yes, then what part of the following procedure am I over looking?

    I am trying to use "computation via reduction to a triangular form" to find the determinant of \left[\begin{matrix}<br />
4 & -3 & 5 \\<br />
5 & 2 & 0 \\<br />
2 & 0 & 4<br />
\end{matrix} \right]<br />

    by adding -2.5 times the first row to the second gives: \left[\begin{matrix}<br />
4 & -3 & 5 \\<br />
0 & -5.75 & -6.25 \\<br />
0 & 1.5 & 1.5<br />
\end{matrix} \right]<br />
    Well, you also subtracted 1/2 of the first line from the third line!

    Here's my confusion. To eliminate the first 1.5 I can either add 1/2 the first row to the third or -5.75/1.5 the second row to the third. Depending on which route I take the last element in the third row will be differnt, and when I multiply the numbers along the diagonal I will get 60 or 72 for the determinant.
    If you "add 1/2 the first row to the third" you get
    \left[\begin{matrix}4 & -3 & 5 \\0 & -5.75 & -6.25 \\2 & 0 & 4<br />
\end{matrix} \right]
    Which is NOT triangular because you have added (1/2)(4)= 2 to the 0 that was in the first column of the third line.

    That is why you should always work with the line that has only "0"s to the left of the pivot.
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