# Integration.Matrix.Determinant.Inverse.Trace.

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.

• What a bad title for this question! – Siminore Apr 14 '15 at 14:50
• math.stackexchange.com/questions/391128/… – tired Apr 14 '15 at 15:15
• in your case $Tr(A^{-1})=\sum_i(1/a_{ii})$ – tired Apr 14 '15 at 15:17
• @tired i was looking for something like that. – RE60K Apr 14 '15 at 15:27
• @ADG: 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. – Jack D'Aurizio Apr 14 '15 at 15:48