There are two matrices $$A = \begin{bmatrix}3&2\\1&4\end{bmatrix}$$ and $$B = \begin{bmatrix}2&1&-4\\0&-2&2\end{bmatrix}$$ I have to solve for matrix $X$ in this equation $$6X + B^T = XA$$ I tried getting rid of $A$ through multiplying both sides by the inverse of $A$, but I can't get any further than that. What is it that I'm missing?


Note that$$6X+B^T=XA\iff B^T=X(A-6\operatorname{Id}_2).$$So, take$$X=B^T(A-6\operatorname{Id}_2)^{-1}=\begin{bmatrix}-1 & -1 \\ 0 & 1 \\ \frac{3}{2} & \frac{1}{2}\end{bmatrix}.$$

  • $\begingroup$ I didn't consider taking out X as a common factor, thank you! $\endgroup$
    – Jonathan
    Mar 24 at 7:26
  • $\begingroup$ I'm glad I could help. $\endgroup$ Mar 24 at 7:33

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