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If $A_{N \times N} (x)$ is complex $N \times N$ positive definite matrix whose components $a_{ij}(x)$ are $C^\infty$ functions, then can we say that the positive definite matrix $A^{-1}$ (the inverse of $A$) is also consists of $C^\infty$ functions?

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Use the formula $A^{-1}=\frac 1{\det A}(A^*)^t$, where $A^*_{ij}=(-1)^{i+j}\Delta_{ij},$ and $\Delta_{ij}$ is the determinant of the matrix $A$ without the row $i$ and column $j$. The map $x\mapsto \delta A(x)$ is $C^{\infty}$ as a finite sum of product of such functions. As it never vanishes, $x\mapsto \frac 1{\det A(x)}$ is smooth too. As $x\mapsto \Delta_{i,j}(x)$ is smooth, we are done.

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Thank you very much. –  Ann Oct 21 '12 at 22:07
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