I have a matrix which is composed by these elements:

$$\mathcal{K}_{11} = \frac{3\alpha^{4/3}(4\mathcal{V} - \xi +6\alpha \sum_{k = 2}^n\lambda_k \tau_k^{3/2})}{4(2\mathcal{V}+\xi)^2(\mathcal{V}+\alpha\sum_{k=2}\lambda_k\tau_k^{3/2})^{1/3}}$$

$$\mathcal{K_{ij}} = \frac{9\alpha^2\lambda_i\lambda_j\sqrt{\tau_i}\sqrt{\tau_j}}{2(2\mathcal{V}+\xi)^2}$$

$$\mathcal{K}_{1j} = -\frac{9\alpha^{5/3}\lambda_i\sqrt{\tau_j}(\mathcal{V} + \alpha \sum_{k = 2}^n \lambda_k \tau_k^{3/2})^{1/3}}{2(2\mathcal{V} + \xi)^2}$$

$$\mathcal{K}_{ii} = \frac{3\alpha\lambda_i(2\mathcal{V} + \xi + 6\alpha\lambda_i\tau_i^{3/2})}{4(2\mathcal{V} + \xi)^2\sqrt{\tau_i}}$$

How can I give Mathematica those elements in order to invert the matrix? Is it possible?

The paper I'm studying in just gave me the inverse elements, but I don't know how to calculate them. I want to do that in order to check if what they wrote is correct and to learn something really new and important.

Thank you!

  • 1
    $\begingroup$ Actually, I don't know where to start. You can use Table for constructing the matrix and Inverse or LinearSolve for its inversion. Their usage is described in the documentation. I would suggest to start with very small and simple examples and with numerical data. $\endgroup$ – Henrik Schumacher Feb 1 '19 at 11:56
  • $\begingroup$ If you could code the above relations into MMa, that would help us to help you. Also, link to the paper you cite. $\endgroup$ – MikeY Feb 1 '19 at 15:23

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