Solving Matrix Equation $(A^T)^{-1}X+(\frac12B)^{-1}=(B^{-1}A^T)^{-1}$

Question

$$(A^T)^{-1}X+(\tfrac12B)^{-1}=(B^{-1}A^T)^{-1}\implies X=B-2A^TB^{-1}$$ I've asked yesterday about one similar equation and I always get stuck, if someone could help me! :)

Answer 1

Let $A,B\in M_{nxn}$ two matrices with the same dimension, There some properties that I used \begin{eqnarray*} (B^{-1}A^{T})^{-1}=(A^{T})^{-1}(B^{-1})^{-1}; \hspace{0.3cm} (\frac{1}{2}B)^{-1}=2B^{-1}; \hspace{0.3cm} A^{T}(A^{T})^{-1}=I \end{eqnarray*} So: \begin{eqnarray*} (A^{T})^{-1}X+(\frac{1}{2}B)^{-1}&=&(B^{-1}A^{T})^{-1}\\ (A^{T})^{-1}X&=&(B^{-1}A^{T})^{-1}-(\frac{1}{2}B)^{-1}\\ X&=&(A^{T})((B^{-1}A^{T})^{-1}-(\frac{1}{2}B)^{-1}))\\ X&=&A^{T}(B^{-1}A^{T})^{-1}-A^{T}(\frac{1}{2}B)^{-1}))\\ \end{eqnarray*} Then $X=B-2A^{T}B^{-1}$

Answer 2

$$(A^T)^{-1} X+(1/2B)^{-1} = (B^{-1} A^T)^{-1} $$Hi welcome to MSE

as a hint $$(A^T)^{-1} =(A^{-1})^T$$and $$(B^{-1} A^T)^{-1}=(A^{T})^{-1}(B^{-1})^{-1}$$also $$(1/2B)^{-1}=2B^{-1}$$