Integration.Matrix.Determinant.Inverse.Trace.

Question

Given $$ I_n=\int_0^1\frac{x^n}{x^{2012}-1}{\rm d}x\text{ and }J_n=\int_0^1\frac{x^n}{x^{2013}+1}{\rm d}x\quad\forall n>2012, n\in\mathbb N$$ If the matrix $$\rm A=[a_{ij}]_{3\times3}\text{ where }{\rm a_{ij}}=\begin{cases}I_{(2012+i)}-I_i&i=j\\0&i\ne j\end{cases}$$

Now $$I_{(2012+i)}-I_i= \int_0^1\frac{x^{2012+i}}{x^{2012}-1}{\rm d}x-\int_0^1\frac{x^i}{x^{2012}-1}{\rm d}x=\int_0^1x^i{\rm d}x=\frac1{i+1}$$

and the matrix $$\rm B=[b_{ij}]_{3\times3}\text{ where }{\rm b_{ij}}=\begin{cases}J_{(2016+j)}+J_{j+3}&i=j\\0&i\ne j\end{cases}$$

Now $$J_{(2016+j)}+J_{j+3}= \int_0^1\frac{x^{2016+j}}{x^{2013}+1}{\rm d}x-\int_0^1\frac{x^{j+3}}{x^{2013}+1}{\rm d}x=\int_0^1x^{j+3}{\rm d}x=\frac1{j+4}$$

then the value of $\rm tr(A^{-1})+det(B^{-1})$ is?

$${\rm A}= \begin{pmatrix} \frac12 & 0 & 0 \\ 0 & \frac13 & 0 \\ 0 & 0 &\frac14 \\ \end{pmatrix}, {\rm B}= \begin{pmatrix} \frac15 & 0 & 0 \\ 0 & \frac16 & 0 \\ 0 & 0 &\frac17 \\ \end{pmatrix}. $$

Now $\rm det(B^{-1})=(det(B))^{-1}$ can be calculated easily, but what about $\rm tr(A^{-1})$. Is there any easy way or I have to do find inverse using minors and cofactors and determinant.

Answer 1

As Jack said:

in general, the trace is the sum of the diagonal elements as well as the sum of the eigenvalues, and the eigenvalues of the inverse matrix are just the inverse of the eigenvalues of the original matrix.

So it must work adding the reciprocals of the elements.