# Thread: Find Determinant of the Matrix

1. ## Find Determinant of the Matrix

Hi! My HW task is to find determinant of given matrix. I tried to do elementary row operations, but did not get anything. How to get rid of $\beta$?
Answer should be 0 for p>2.

$\begin{bmatrix} a^{n}-\beta & a^{n+1}-\beta & ... & a^{n+p-1}-\beta\\ a^{n+p}-\beta & a^{n+p+1}-\beta & ...& a^{n+2p-1}-\beta\\ .& .& .& \\ .& .& .& \\ a^{n+p(p-1)}-\beta& a^{n+p(p-1)+1}-\beta& ...&a^{n+p^2-1}-\beta \end{bmatrix}$

2. ## Re: Find Determinant of the Matrix

I don't think it's necessary to get rid of B.

I would try to apply the definition of determinant directly and use the principle of induction to show the determinant is 0 for p>2.
The key, I think, is using the inductive step when asked to calculate the determinant of minors for matricies larger the 2x2.

3. ## Re: Find Determinant of the Matrix

After looking at the problem more, I think the problem can be solved using row reduction, and not induction like I originally suspected. I would consider looking at the case for p=3 first and observe the powers of a. Row reduce. I'll leave it to you to generalize. Process should be similar for greater than 3

Original Matrix for p = 3

a^n - B & a^n+1 - B & a^n+2 - B
a^n+3 - B & a^n+4 - B & a^n+5 -B
a^n+6 - B & a^n+7 - B & a^n+8 - B

Row 2 minus Row 1 and Row 3 minus Row 2
a^n - B & a^n+1 - B & a^n+2 - B
a^n * (a^3 - 1) & a^n+1 * (a^3-1) & a^n+2 * (a^3-1)
a^(n+3)* (a^3 - 1) & a^n+4 * (a^3 -1) & a^n+5 (a^3 - 1)

Now note that Row 3 is a^3 * Row 2. Hence the last row can be reduced to 0, hence proving the determinant is 0.