$$A*B^{T}*A^{-1}=\begin{pmatrix} 8_{} & 0_{} & -1_{}\\ 11_{} & -6_{} & -3_{}\\ -7_{} & 11_{} & 4_{} \end{pmatrix}$$
Find matrix B
After I transfer it to: $$B^{T}=A^{-1}\begin{pmatrix} 8_{} & 0_{} & -1_{}\\ 11_{} & -6_{} & -3_{}\\ -7_{} & 11_{} & 4_{} \end{pmatrix}A$$
Is it valid to transpose both sides of the equation, so I can get (possibly) the value of B?
Is this general direction for solving the problem, legit?
Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^{-1}$
D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors