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This is my first question, so I’m sorry if I made mistakes.

$\left(\begin{array}{ccccc}1&1&1&\cdots&1\\2&2^2&2^3&\cdots&2^n\\3&3^2&3^3&\cdots&3^n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n&n^2&n^3&\cdots&n^n\end{array}\right)$

In order to solve the above inverse matrix, I tried to solve the below liner simultaneous equations in n-th unknowns.

$kx_1+k^2x_2+\cdots+k^nx_n=1+2^{n-1}+\cdots+k^{n-1}$
$k=1,2,\cdots,n$

Inductively, I know the solution of the liner simultaneous equations.

$x_k=\frac{B_{n-k}}{n-k} {n-1 \choose n-k-1} (1≦k<n-1)$
$x_{n-1}=\frac{1}{2}$
$x_n=\frac{1}{n}$

But even though I solved the solutions of the equations, I have no ideas how to use it.

I thought Vandermonde’s déterminant may be effective, but I have no ideas how to use it.

Gymnast
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  • @DietrichBurde: this matrix is Vandermonde, but with very specific coefficients so that another resolution path might exist. I wouldn't consider the question as a duplicate. –  Oct 23 '17 at 10:11
  • &DietrichBurde: This matrix is a little different from Vandermonde’s determinant. That’s why I can’t apply the theorem to this matrix... – Gymnast Oct 23 '17 at 10:13
  • @Gymnast: in what way is it different ? –  Oct 23 '17 at 10:14
  • You can apply it, see Vandermonde matrix. As Yves says, such a matrix is even more general than yours. But perhaps the more general case is nice to know. – Dietrich Burde Oct 23 '17 at 10:14
  • After all, no, this is not a Vandermonde matrix. A column of all ones is missing. –  Oct 23 '17 at 10:17
  • @Yves Daoust I meant Vandermonde matrix’s coefficients begins with 1 but this matrix with k – Gymnast Oct 23 '17 at 10:20

1 Answers1

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If that can help, a few inverses:

$$\frac12\begin{pmatrix}4&-1\\-2&1\end{pmatrix}$$ $$\frac16\begin{pmatrix}18& -9& 2\\-15& 12& -3\\3& -3& 1\end{pmatrix}$$

$$\frac1{24}\begin{pmatrix}96& -72& 32& -6\\-104& 114& -56& 11\\36& -48& 28& -6\\-4& 6& -4& 1\end{pmatrix}$$

$$\frac1{120}\begin{pmatrix}600& -600& 400& -150& 24\\-770& 1070& -780& 305& -50\\355& -590& 490& -205& 35\\-70& 130& -120& 55& -10\\5& -10& 10& -5& 1\end{pmatrix}$$

$$\frac1{720}\begin{pmatrix}4320& -5400& 4800& -2700& 864& -120\\-6264& 10530& -10160& 5940& -1944& 274\\3480& -6915& 7440& -4605& 1560& -225\\-930& 2055& -2420& 1605& -570& 85\\120& -285& 360& -255& 96& -15\\-6& 15& -20& 15& -6& 1\end{pmatrix}$$

No easily discernible pattern except for the binomial coefficients in the last row.