Row operations is the key here

Algebra Level 5

Let $f(x) = \dfrac d{dx} \left(x + x^2 + \cdots + x^{677} \right)$ .

And define $e_n$ as the $n^\text{th}$ symmetric sum of the roots of $f(x)$ .

Evaluate the determinant $\begin{vmatrix}{e_1} && {e_2} && {\ldots} && {e_{26}} \\ {e_{27}} && {\ddots} && {\phantom0} && {\vdots} \\ {\vdots} && {\phantom0} && {\ddots} && {\vdots} \\ {e_{651}} && {\ldots} && {\ldots} && {e_{676}}\end{vmatrix}$ .

Bonus: Generalize this.

I ripoff from this problem .

The answer is 0.

1 solution

Mark Hennings
Nov 21, 2017

Since $f(x) \; = \; \sum_{j=0}^{676}(j+1)x^j$ we deduce that $e_j \; = \; (-1)^j \tfrac{677-j}{677} \hspace{2cm} 1 \le j \le 676$ We are interested in the matrix $M$ with coefficients $M_{u,v} \; = \; e_{26u+v-26} \hspace{2cm} 1 \le u,v \le 26$ and therefore $M_{u+1,v} - M_{u,v} \; = \; e_{26u+v} - e_{26u+v-26} \; = \; (-1)^v\left[\tfrac{677 - 26u - v}{677} - \tfrac{651 - 26u - v}{677}\right] \;= \; (-1)^v \tfrac{26}{677} \hspace{2cm} 1 \le u \le 25\,,\,1 \le v \le 26$ and hence $M_{u,v} + M_{u+2,v} \; =\; M_{u+1,v} \hspace{2cm} 1 \le u \le 24\,,\,1 \le v \le 26$ Thus any three successive rows of $M$ are linearly dependent, and hence the determinant of $M$ is $\boxed{0}$ .

I'm impressed that you don't have to draw out the matrix (with some of the coefficients) to realize that there are rows that are linearly dependent. You must be in a whole new level than me.

Thanks for the solution again. You're the best!!

Pi Han Goh - 3 years, 6 months ago

Row operations is the key here

The answer is 0.

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