Let $A$ be a non-singular matrix such that $3ABA^{-1}+A=2A^{-1}BA$ then determine whether $A+B$ and $ABA^{-1}-A^{-1}BA$ are singular or non-singular

Question

Let $A$ be a non-singular matrix such that $$3ABA^{-1}+A=2A^{-1}BA$$ then determine whether $A+B$ and $ABA^{-1}-A^{-1}BA$ are singular or non-singular.

My Attempt:

If I put $A=I$ then both $A+B$ and $ABA^{-1}-A^{-1}BA$ are singular. But I am not able to generate a proper way to prove

Answer 1

\begin{align} A+B&= A(I+A^{-1}B)=(I+BA^{-1})A\\ (A+B)(2A)&=A(I+A^{-1}B)(2A)\\ (A+B)(2A)&=A(2A+2A^{-1}BA)\\ &=A(2A+3ABA^{-1}+A)\\ &=A^2(3I+3BA^{-1})\\ &= 3A^2(I+BA^{-1})\\ 2(I+BA^{-1})A^{2}&=3A^{2}(I+BA^{-1})\\ 2^n\det(I+BA^{-1})&=3^n\det(I+BA^{-1}) \end{align}

Hence $I+BA^{-1}$ is not invertible which implies that $A+B$ is not invertible.

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\begin{align} ABA^{-1}-A^{-1}BA &= \frac13 \left(3ABA^{-1}-3A^{-1}BA \right)\\ &=\frac13(-A-A^{-1}BA)\\ &= -\frac13(I+A^{-1}B)A \end{align}

Since $I+A^{-1}B$ is singular, $ABA^{-1}-A^{-1}BA$ is singular as well.

Answer 2

Instead of trying to evaluate it for a particular case of $A$, try solving the equation for $A+B$ and $ABA^{-1}-A^{-1}BA$.