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Math Help - Find the determinant D if D equals...

  1. #1
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    Find the determinant D if D equals...

    Hey, hopefully this is the correct place to post! I was looking for help with this textbook problem...

    Find det D if...
    Find the determinant D if D equals...-matrix.png

    Where the back of the book gives the answer as (-1)Nd1d2...dn, N = n(n-1)/2.

    This problem was actually asked before (Determinant Question), and I read the answer that was given and it made some sense, but still left me with questions. For example, can you even have a diagonal matrix from a 5x4 matrix? No matter what won't there always be a "gap" since there would always be a column of zeroes? Also in the book the number of "switches" is given as n(n-1)/2 which using the logic of swapping rows to create a diagonal doesn't make sense (like if n = 5 that would be 10 swaps!)
    Before consulting the web I thought the answer would have to do with cofactors and using the formula Find the determinant D if D equals...-formula.png where the -1 and M multiply to be the cofactor of element a. As you can probably tell I'm pretty lost so any help would be great!
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  2. #2
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    Re: Find the determinant D if D equals...

    Quote Originally Posted by abcdoll View Post
    Hey, hopefully this is the correct place to post! I was looking for help with this textbook problem...

    Find det D if...
    Click image for larger version. 

Name:	matrix.png 
Views:	1 
Size:	2.5 KB 
ID:	30807

    Where the back of the book gives the answer as (-1)Nd1d2...dn, N = n(n-1)/2.

    This problem was actually asked before (Determinant Question), and I read the answer that was given and it made some sense, but still left me with questions. For example, can you even have a diagonal matrix from a 5x4 matrix?
    Why are you asking about a "5x4 matrix"? This question is clearly about an "nxn matrix" so definitely about square matrices.

    The first thing I would do is try some simple examples: with n= 2, this would be \left|\begin{array}{cc}0 & a_1 \\ a_2 & 0 \end{array}\right| which, with one swap of rows gives -\left|\begin{array}{cc}a_2 & 0 \\ 0 & a_1 \end{array}\right|= -a_1a_2
    With n= 2, there was 2(1)/2= 1 swap required.

    With n= 3, we have \left|\begin{array}{ccc}0 & 0 & a_1 \\0 & a_2 & 0\\ a_3 & 0 & 0 \end{array}\right|.
    Swapping the first and third rows immediately gives -\left|\begin{array}{ccc}a_3 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_1\end{array}\right|= -a_1a_2a_3
    But we could also swap the first and second rows, to get -\left|\begin{array}{ccc}0 & a_2 & 0 \\ 0 & 0 & a_1 \\ a_3 & 0 & 0\end{array}\right| then swap second and third rows, +\left|\begin{array}{ccc}0 & a_2 & 0 \\ a_3 & 0 & 0 \\ 0 & 0 & a_1 \end{array}\right| and, finally, swap first and third rows to get -\left|\begin{array}{ccc}a_3 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_1 \end{array}\right|= -a_1a_2a_3 again: that is 3(2)/2= 3 swaps. In any case the point is that that those have the same parity so the same sign.

    No matter what won't there always be a "gap" since there would always be a column of zeroes? Also in the book the number of "switches" is given as n(n-1)/2 which using the logic of swapping rows to create a diagonal doesn't make sense (like if n = 5 that would be 10 swaps!)
    Before consulting the web I thought the answer would have to do with cofactors and using the formula Click image for larger version. 

Name:	formula.png 
Views:	1 
Size:	2.3 KB 
ID:	30808 where the -1 and M multiply to be the cofactor of element a. As you can probably tell I'm pretty lost so any help would be great!
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