Let L = $\frac{f^{2n-2}(0)}{\left(2n-2\right)!}$ and Let M = $\sum_{n=1}^{102}(-1)^{n-1}(L)$. Evaluate M.

Question

Let A = $\left[a_{ij}\right]_{(2n-1)(2n-1)}$ (n ∈ N) denote a square matrix of order [ 2n - 1 ] and \begin{equation} [a_{ij}] = \left\{ \begin{array}{lr} \frac{1}{2}(2n^2+i+j+2-4n), i = j\\ \frac{1}{2}(2n^2+i-j+2+4n),i≠j \end{array} \right\} \end{equation}

Let f(x) = | xI - A |. Let L = $\frac{f^{2n-2}(0)}{\left(2n-2\right)!}$ and Let M = $\sum_{n=1}^{102}(-1)^{n-1}(L)$.

Here, $f^k(o)$ represents the $k^{th}$ derivative of y = f(x) at x = 0,

[I represents an Identity matrix and |D| represents the determinant value of any matrix 'D']

Evaluate M.

This question is from highschool test and during the test, I started with taking the case n = 2 and create the matrix which in my case was $[a_{ij}]_{(3 \,x \, 3)}$,

$$A = \left[\begin{array}{rrr} 2 & 0.5 & 0 \\ 1.5 & 3 & 0.5 \\ 2 & 1.5 & 4 \end{array}\right] $$

which makes 'I' naturally a $3\,x\,3$ matrix,

$$f(x) = \left[\begin{array}{rrr} x-2 & -0.5 & 0 \\ -1.5 & x-3 & 0.5 \\ -2 & -1.5 & x-4 \end{array}\right] $$ Then, on taking n = 2 in L, we get L = $\frac{f^2(0)}{2!}$, before which we need f'(x) so to get that obviously we expand the matrix f(x) giving us;

| f(x) | = $x^3-9x^2-26x-23 $ after which, L = -9 and similarly M.

NOTE-I am interested to learn a method which doesn't involve characteristic equation. The question however has two elegant answers below using characteristic equation.

Answer 1

It may be worth reading more about the characteristic polynomial (its expressions and interesting math around it) to get a sense of direction since that is what your f(x) is coincidentally describing.

As for giving you a starting point, this may be helpful: using Jacobi's formula, you can evaluate $$ \begin{align} f^1(x)&=det|xI-A|\cdot tr\left((xI-A)^{-1} \frac{d (xI-A)}{dx}\right) \\ &= f(x)\cdot tr\left((xI-A)^{-1} I\right) \\ &= f(x)\cdot tr\left(\frac{1}{det|xI-A|}adj(xI-A) \right) \\ &= f(x)\cdot tr\left(\frac{1}{f(x)}adj(xI-A) \right) \\ &= tr\left(adj(xI-A) \right) \\ \end{align} $$ and you could continue from there.

PS: What high school test is this a part of (just curious)?

Answer 2

First of all, $A,f,L$ should be indexed by $n.$

$$\begin{align}-L_n&=\operatorname{tr}(A_n)\\&=\sum_{j=1}^{2n-1}(n^2+j+1-2n)\\&=(2n-1)(n-1)^2+\sum_{j=1}^{2n-1}j\\&=(2n-1)(n-1)^2+n(2n-1)\\&=(2n-1)(n^2-n+1)\\&=2n^3-3n^2+3n-1\\&=n^3+(n-1)^3. \end{align}$$

$$\begin{align}M&=\sum_{n=1}^{102}(-1)^n(n^3+(n-1)^3)\\&=102^3, \end{align}$$ by telescoping.