# Confusing Matrix Question

• Mar 26th 2010, 08:25 AM
craig
Confusing Matrix Question
Let an $\displaystyle n \times n$ determinant be:

$\displaystyle C_n = \left| \begin{array}{cccc} \frac{1}{a_1 + b_1} & \frac{1}{a_1 + b_2} & ... & \frac{1}{a_1 + b_n} \\ \frac{1}{a_2 + b_1} & \frac{1}{a_2 + b_2} & ... & \frac{1}{a_2 + b_n} \\ ... & ... & ... & ...\\ \frac{1}{a_n + b_1} & \frac{1}{a_n + b_2} & ... & \frac{1}{a_n + b_n} \\ \end{array} \right|$

In the question we are also given that:

$\displaystyle C_n = \frac{\Pi_{j>k}(a_j-a_k)(b_j-b_k)}{\Pi_{j,k}(a_j+b_k)}$

Where $\displaystyle \Pi_{j,k}$ symbolises a product over all possible $\displaystyle j$ and $\displaystyle k$ and $\displaystyle \Pi_{j>k}$ a product subject to $\displaystyle j > k$.

Prove the above equation for both $\displaystyle n =2$ and $\displaystyle n=3$.

Have got no idea where to start with this, I'm not even sure if I understand the equation itself?

If someone could help me with proving the $\displaystyle n =2$ case I'm sure I could figure out the $\displaystyle n =3$ part.

• Mar 26th 2010, 11:56 AM
Black
For n = 2, we have

$\displaystyle C_2=\left|\begin{array}{cc}\frac{1}{a_1+b_1}&\frac {1}{a_1+b_2} \\ \frac{1}{a_2+b_1}&\frac{1}{a_2+b_2}\end{array}\rig ht|$

$\displaystyle =\frac{1}{(a_1+b_1)(a_2+b_2)}-\frac{1}{(a_1+b_2)(a_2+b_1)}$

$\displaystyle =\frac{(a_1+b_2)(a_2+b_1)-(a_1+b_1)(a_2+b_2)}{(a_1+b_1)(a_2+b_2)(a_1+b_2)(a_ 2+b_1)}$

$\displaystyle =\frac{a_2b_2-a_2b_1-a_1b_2+a_1b_1}{(a_1+b_1)(a_2+b_2)(a_1+b_2)(a_2+b_1 )}$

$\displaystyle =\frac{(a_2-a_1)(b_2-b_1)}{(a_1+b_1)(a_2+b_2)(a_1+b_2)(a_2+b_1)}$
• Mar 26th 2010, 02:48 PM
craig
Ahh I see thankyou! I'll attempt the 3x3 in the morning and post up my (hopefully correct) work ;)

Thanks again
• Mar 28th 2010, 05:15 AM
craig
Finally got round to calculating the 3x3 determinant, and - with a little help from Maple - got:

$\displaystyle \frac{(a_1 - a_2)(a_1 - a_3)(a_2 - a_3)(b_1 - b_2)(b_1 - b_3)(b_2 - b_3)}{(a_1 + b_1)(a_2 + b_2)(a_3 + b_3)(a_2 + b_3)(a_3 + b_2)(a_1 + b_2)(a_2 + b_1)(a_3 + b_1)(a_1 + b_3)}$

This looks ok doesn't it?

Thanks again for the help, wouldn't have been able to start without it.
• Mar 28th 2010, 05:33 AM
Black
Yep, it equals to

$\displaystyle \frac{(a_2-a_1)(b_2-b_1)(a_3-a_1)(b_3-b_1)(a_3-a_2)(b_3-b_2)}{(a_1+b_1)(a_1+b_2)(a_1+b_3)(a_2+b_1)(a_2+b_2 )(a_2+b_3)(a_3+b_1)(a_3+b_2)(a_3+b_3)}$,

which coincides with the formula for $\displaystyle C_n$.
• Mar 28th 2010, 05:34 AM
craig
Thanks again