# Thread: Find the determinant D if D equals...

1. ## 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...

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 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!

2. ## Re: Find the determinant D if D equals...

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

Find det D if...

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?
The first thing I would do is try some simple examples: with n= 2, this would be $\displaystyle \left|\begin{array}{cc}0 & a_1 \\ a_2 & 0 \end{array}\right|$ which, with one swap of rows gives $\displaystyle -\left|\begin{array}{cc}a_2 & 0 \\ 0 & a_1 \end{array}\right|= -a_1a_2$
With n= 3, we have $\displaystyle \left|\begin{array}{ccc}0 & 0 & a_1 \\0 & a_2 & 0\\ a_3 & 0 & 0 \end{array}\right|$.
Swapping the first and third rows immediately gives $\displaystyle -\left|\begin{array}{ccc}a_3 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_1\end{array}\right|= -a_1a_2a_3$
But we could also swap the first and second rows, to get $\displaystyle -\left|\begin{array}{ccc}0 & a_2 & 0 \\ 0 & 0 & a_1 \\ a_3 & 0 & 0\end{array}\right|$ then swap second and third rows, $\displaystyle +\left|\begin{array}{ccc}0 & a_2 & 0 \\ a_3 & 0 & 0 \\ 0 & 0 & a_1 \end{array}\right|$ and, finally, swap first and third rows to get $\displaystyle -\left|\begin{array}{ccc}a_3 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_1 \end{array}\right|= -a_1a_2a_3$ 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.