Please help me solve this question. If $ >0$ means to be positive definite, then obtain matrix A. (Obtain matrix A in terms of other matrices) $$\left[\begin{matrix} A & B' \\ B & A^{-1} \end{matrix}\right]>0$$
I tried to use Schur complement but failed
You cannot solve for $A$ uniquely using just this condition. If we assume that the positive definite matrices must be symmetric, then the condition that $\begin{pmatrix} A & B' \\ B & A^{-1} \end{pmatrix} > 0$ is exactly equivalent to the condition that $A$ is positive definite, and that the Schur complement with respect to the bottom right block is positive definite, i.e. $$A-B'AB > 0.$$ If, for example, $B = \frac{1}{2} I$, then this in fact holds for any positive definite $A$.
