Determinant proof/evaluation ?

Hi guys,

I'm trying to answer the following:

Prove the following identity without evaluating the determinants:

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a + b c+d e+f a c e b d f

p q r = p q r + p q r

u v w u v w u v w

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It comes from the theorem that if A, B and C are n x n matrices who differ in only one row, the ith row, and assume the ith row of C can be obtained by adding corresponding entries in the ith rows of A and B then,

det (C) = det (A) + det (B)

I think I proved it by using co - factor expansion of the determinants on the right hand side of the identity above. I ended up with the following co-factor expansion:

a + b ( qw - rv ) + c + d ( ru - pw ) + e + f ( pv - qu )

And this equals the determinant on the left hand side with a + b being an entry and (qw- rv ) being a minor of that entry etc.

Thus the proof is complete. Or is it ? My query is this: have I just evaluated the determinants without actually proving anything ? My understanding is that the result I showed above will hold for any integer, and therefore the identity is proved for all integers.

Any help would be appreciated.

Many thanks.

PS anyone know how I can properly represent determinants on a post ?!?!

Just the Formatting Aspect of Your Question

Here's how you can write a determinant: use LaTeX. Use the Advanced button, and click the TeX button. Here's a determinant:

$\displaystyle \det(M)=\left|\begin{matrix}1 &2 &3\\ 4 &5 &6\\ 7 &8 &9\end{matrix}\right|=1\left|\begin{matrix}5 &6\\ 8 &9\end{matrix}\right|-2\left|\begin{matrix}4 &6\\ 7 &9\end{matrix}\right|+3\left|\begin{matrix}4 &5\\ 7 &8\end{matrix}\right|=...$

You can double-click this determinant to get the LaTeX code that generated it.