Welcome to our community

Be a part of something great, join today!

[ASK] Determinant of a Matrix with Polynomial Elements

Monoxdifly

Well-known member
Aug 6, 2015
267
Help me if what I have done so far can be simplified further.
 

Attachments

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,656
Leeds, UK
Help me if what I have done so far can be simplified further.
You would do better to evaluate this determinant using row and column operations. Start by subtracting row 2 from row 3, using calculations like $(n+2)^2 - (n+1)^2 = n^2+4n+4 - (n^2+2n+1) = 2n+3$: $$ \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} = \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix}.$$ Then continue like this: $$\begin{aligned} \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} &= \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix} \\ \\ \text{(Subtract row 1 from row 2)}\qquad &= \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ 2n+1 & 2n+3 & 2n+5 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix} \\ \\ \text{(Subtract col 2 from col 3)}\qquad &= \begin{vmatrix}n^2 & (n+1)^2 & 2n+3 \\ 2n+1 & 2n+3 & 2\\ 2n+3 & 2n+5 & 2 \end{vmatrix} \\ \\ \text{(Subtract col 1 from col 2)}\qquad &= \begin{vmatrix}n^2 & 2n+1 & 2n+3 \\ 2n+1 & 2& 2\\ 2n+3 & 2 & 2 \end{vmatrix} .\end{aligned}$$ Now subtract col 2 from col 3. Proceed in this way and you should end with a very simple answer, namely the constant $-8$.
 
Last edited:

Monoxdifly

Well-known member
Aug 6, 2015
267
Okay, thanks Opalg!