# Determinant proof/evaluation ?

• Aug 13th 2010, 05:26 AM
Doktor_Faustus
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 ?!?!
• Aug 16th 2010, 10:55 PM
ajskim
Hmm, I'm not sure if that is sufficient, essentially you are calculating the determinants. Is there a way you can cite the general row adding proof? Or a version of that proof you can emulate?
• Aug 17th 2010, 01:23 AM
Ackbeet
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:

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

You can double-click this determinant to get the LaTeX code that generated it.
• Aug 17th 2010, 02:13 AM
Vlasev
Are you trying to prove it for n by n matrices or just for the specific example? If you need to prove it for general n by n matrices, there is no way you can calculate the determinant, because you don't know the entries. However, you can expand it along row i. Then you can recognize that the obtained minor is the same for both matrices A and B and thus you can break up your determinant into two.

Suppose you have an n by n matrix M. Let M(i,j) denote the minor obtained by removing row i and column j. Suppose the matrices A and B differ only in the entries of a certain row i. Then you surely have that A(i,j) = B(i,j) = C(i,j) since we have removed all the entries that make the determinants different.

Now

$\displaystyle \det(C)$

$\displaystyle = \sum_{j=1}^n (-1)^{i+j}C_{i,j}\det C(i,j) =$

$\displaystyle = \sum_{j=1}^n (-1)^{i+j}(A_{i,j}+B_{i,j})\det C(i,j)$

$\displaystyle = \sum_{j=1}^n (-1)^{i+j} A_{i,j}\det C(i,j) + \sum_{j=1}^n (-1)^{i+j} B_{i,j}\det C(i,j)$

$\displaystyle = \sum_{j=1}^n (-1)^{i+j} A_{i,j}\det A(i,j) + \sum_{j=1}^n (-1)^{i+j} B_{i,j}\det B(i,j)$

$\displaystyle = \det(A) + \det(B)$

I'm afraid that this still counts as a calculation of sorts, but its not an EXPLICIT one as we are not calculating the whole determinant, just a small portion of it.
• Aug 17th 2010, 03:50 AM
Doktor_Faustus
Hi,

Yes it's just for the specific example I quoted. The questions requires that you prove the identity.
• Aug 17th 2010, 04:17 AM
Doktor_Faustus
Quote:

Originally Posted by ajskim
Hmm, I'm not sure if that is sufficient, essentially you are calculating the determinants. Is there a way you can cite the general row adding proof? Or a version of that proof you can emulate?

If you have Howard Anton's "Elementary Linear Algebra " refer to the chapter on determinants - page 93 in the fifth edition. Anton shows that for a 2x2 matrix det (C) = det (A) + det (B) when the matrices A and B differ only in a single row.

Here it is:

Consider two 2x2 matrices that differ only in a single row.

A = a11 a12
a21 a22

B= a11 a12
b21 b22

det (A) + det (B) = (a11a22 - a12a21) + (a11b22 - a12b21 )

= a11 ( a22 + b22 ) - a12 ( a21 + b21 )

= det a11 a12
a21+b21 a22+b22

Sorry about the crappy formatting of the determinants. I haven't got used to using Latext yet.

Cheers.
• Aug 23rd 2010, 04:37 AM
halbard
Another way, tailored to your original question, would be to write your determinants using the scalar triple product,

$\left|\begin{matrix} a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{matrix}\ right|=\mathbf a.(\mathbf b\times\mathbf c)$ where $\mathbf a=\left(\begin{matrix}a_1\\a_2\\a_3\end{matrix}\ri ght)$, $\mathbf b=\left(\begin{matrix}b_1\\b_2\\b_3\end{matrix}\ri ght)$ and $\mathbf c=\left(\begin{matrix}c_1\\c_2\\c_3\end{matrix}\ri ght)$.

The identity you require can be obtained by considering the formula

$(\mathbf a+\mathbf d).(\mathbf b\times\mathbf c)=\mathbf a.(\mathbf b\times\mathbf c)+\mathbf d.(\mathbf b\times\mathbf c)$
• Aug 23rd 2010, 05:39 AM
ModusPonens
I think the intended answer is that it is because the determinant is a multilinear function of its vectors.