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Let $A$ be an $n \times n$ invertible matrix with \begin{align} \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{array}\right)^{-1}= \left(\begin{array}{ccc} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end{array}\right) \end{align} Is there an explicit formula for $b_{ij}$ in terms of the elements of $A$ and the determinant of $A$?

Edit: Here is a link to the different possible methods to invert a matrix http://en.m.wikipedia.org/wiki/Invertible_matrix#Methods_of_matrix_inversion.

Originally I had in mind using the blockwise inversion method but I think all of the methods require using some submatrix of $A$ and their determinants and / or inverses. I don't think it's possible to write something explicit in terms of $\{a_{ij}\}_{i=1,j=1}^{n,n}$

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Hint: $A^{-1}=\frac{1}{|A|}\cdot \text{Adj}(A)$

$\text{Adj}(A)$ is the adjunct matrix of A.

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  • $\begingroup$ Also called the adjugate matrix or the classical adjoint. $\endgroup$ Commented Oct 27, 2014 at 1:22
  • $\begingroup$ Thanks. It still seems like this would not be completely explicit but rather in terms of determinants of submatrices? e.g. $b_{11}=\frac{1}{|A|} \left|\begin{array}{ccc}a_{22} &\cdots & a_{2n} \\ \vdots & \ddots & \vdots \\ a_{n2} & \cdots & b_{nn} \end{array}\right|$. $\endgroup$
    – user103828
    Commented Oct 27, 2014 at 12:27
  • $\begingroup$ At the moment I don´t see a possibility to write it without the determinants of the submatrices: $ A^{-1}=\frac{1}{|A|}\cdot\begin{pmatrix} |A_{11}| & -|A_{12}| & \ldots & (-1) ^{1+n}|A_{1n}|\\ \vdots & \vdots & \vdots & \vdots \\ (-1) ^{n+1}|A_{n1}| & (-1) ^{n+2}|A_{22}| & \ldots & (-1) ^{n+n}|A_{nn}| \end{pmatrix}^T$ $\endgroup$ Commented Oct 27, 2014 at 13:19

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