# Calculation of a Determinant?

• Dec 13th 2011, 09:20 AM
Tesla2011
Calculation of a Determinant?
So our professor at university gave us this assignment for home, we need to calculate the determinant of the following square(N*N) matrix

http://img210.imageshack.us/img210/556/matrixuq.png

I have already found that |AN|=a|AN-1| - |AN-2| , but from that point I can't go any further, heard that I need to use difference equations, but unfortunately I am not familiar with them. Thank you in advance, I would be really grateful for any help given
• Dec 13th 2011, 11:12 AM
Johngalt13
Re: Can anyone help me with the Calculation of a Determinant???
You're on the right track trying to find a recursive formula. I would say solve the first few determinants and see if a pattern begins to form, it usually will.

Also, take a look at this, it should help.

Recurrence relation - Wikipedia, the free encyclopedia
• Dec 13th 2011, 11:16 AM
Tesla2011
Re: Can anyone help me with the Calculation of a Determinant???
Thanks for your answer, I already solved the first few determinants but I am unable to find the recursive formula, the link you provided was very helpful I will look into it. Thanks again
• Dec 13th 2011, 11:24 AM
Johngalt13
Re: Can anyone help me with the Calculation of a Determinant???
No problem, I'm actually just in linear now so I'm no expert but if I finish my take home test and you haven't solved I'll look into it more.
• Dec 13th 2011, 12:05 PM
Opalg
Re: Can anyone help me with the Calculation of a Determinant???
Just a few hints about solving recurrence relations, which is actually not too difficult and is explained very well in the Wikipedia article that Johngalt13 linked to. To solve the equation $|A_N| - a|A_{N-1}|+|A_{N-2}| = 0$, you first need to solve the auxiliary equation $\lambda^2-a\lambda+1=0$. Its roots $\lambda_1$ and $\lambda_2$ are $\tfrac12(a\pm r)$, where $r=\sqrt{a^2-4}$. The solution to the difference equation is $|A_N| = C\lambda_1^N + D\lambda_2^N$, where C and D are constants that you have to determine by ensuring that the formula works for N = 1 and 2.

The solution works well if a>2. If a<2 then the roots of the auxiliary equation are complex numbers, but the formula for $|A_N|$ still works.

If a=2 then r=0 and you will find that the formula breaks down completely. But in fact the formula for $|A_N|$ becomes much simpler when a=2. You can guess what it is by evaluating $|A_N|$ directly from the determinant for small values of N, and then verifying your guess using a proof by induction.