MHB [ASK] Determinant of a Matrix with Polynomial Elements

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To simplify the determinant of a matrix with polynomial elements, it is recommended to use row and column operations. The process begins by subtracting one row from another to simplify the elements, as demonstrated with specific calculations. Continuing this method leads to a series of transformations that ultimately simplify the matrix significantly. Following these steps will yield a final result of the constant value -8. This approach effectively reduces the complexity of the determinant calculation.
Monoxdifly
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Help me if what I have done so far can be simplified further.
 

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Monoxdifly said:
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 $-8$.
 
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Okay, thanks Opalg!
 
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