Determinants

A determinant is a number; it is obtained by performing an operation on the elements of a matrix, but after all of that multiplication, addition and subtraction is done the end result is just a number, a constant.

In your case you are forming each determinant by selecting from one large fixed set of numbers, so you are using the same set of numbers throughout your calculation.

So using your notation the large matrix has elements

A_1,1, A_1,2, ... A_1,n+1,

A_2,1, A_2,2,... A_2,n+1

:

A_n+1,1,A_n+1,2,...A_n+1,n+1

Let's call this matrix M1 and from this matrix we form the smaller matrices M2, M3, M4 and M5. From these matrices we form the determinants |M1|, M2|, etc. So your statement is thus:

(|M2|*|M3|-|M4|*|M5|) / |M1| = K

Well, each term here is a determinant so these are all just numbers; i.e., constants. Simple math operations on any set of constants yields another constant!

Even if one of the terms in M1 is an irrational number such as pi (3.14159....), the net result of the math operations can be reduced to: (k1*pi)/(k2*pi) = k3, where k1*pi = (|M2|*|M3|-|M4|*|M5|), k2*pi = |M1| and k1, k2, and k3 are all rational constants.

I know this is not a rigorous proof since it assumes the elements of M1 are numbers. But the math needed for a rigorous proof only involves simple operations of multiplication, etc. So if you wanted to spend the time, you could expand the matrices and do the sums yourself. In doing so, you could prove the more rigorous case where the elements of M1 are not all constants; i.e., you could prove the general case where some or all of the elements of M1 are functions.

Dear usbport,

Thanks for your interest.

Here the constant I referred to is a common factor in the expansion of the expression on the left hand side of the equation and is a product of some of the individual elements of the matrix such as 'Amn' 'm' and 'n' taking different values. The elements with the same indices on both sides have the same value i.e. A45 occuring on the L.H.S is the same as the one on the R.H.S. of the equation.

The L.H.S. when expanded will show up as sum of the products of (n+n=2n) elements. The R.H.S. when expanded will show up as products of n+1 elements. Thus the constant should be a product of [2n-(n+1)=(n-1)] elements.

Cheers.