$A,B$ are matrices $3x3$ so that $B^2A=-2B^3$ and $B^3+AB^2=3I$ express $A^{-1}$ and $B^{-1}$ using $B$

Question

I have the follow question :

$A,B$ are matrices $3x3$ so that $B^2A=-2B^3$ and $B^3+AB^2=3I$ express $A^{-1}$ and $B^{-1}$ using $B$

I tried to "play" with the equations but I always get stuck with $A^{-1}$ or $B^{-1}$ in the wrong side.

Any ideas?

Thank you!

Answer 1

Using the comment of @DavidC.Ullrich we consider $B^3+AB^2=3I$ or that $(B+A)B^2=3I$. If $B$ is not invertible, then it does not have full rank and any product of $B$ would not have full rank, but since $I$ has full rank, this is impossible. Therefore, $B$ is invertible.

Now, taking the first equation, $B^2A=-2B^3$, we have $2B^3+B^2A=0$ so that $B^2(2B+A)=0$. Since $B$ is invertible, we know that $2B+A=0$ or that $A=-2B$. Therefore, $A$ is also invertible.

Now, observe that $B^3+AB^2=B^3-2B^3=-B^3$. Therefore, $-B^3=3I$. Multiplying through by $B^{-1}$ gives $-B^2=3B^{-1}$. Since $A=-2B$, the result for $A$ follows similarly.

Answer 2

$(B^2+AB)B=3I$, so $B$ is invertible, and in fact $B^{-1}=(B^2+AB)/3$. Since $B$ is invertble, $B^2A=-2B^3$ shows that $A=-2B$. Put that back into the expression for $B^{-1}$ and you have $B^{-1}$ in terms of $B$. And then you get $A^{-1}$ as well, because...

Answer 3

Since $B^3+AB^2=3I$ it follows $(\frac{1}{3}B^2+\frac{1}{3}AB)B=I$ hence $B$ is invertible and $$B^{-1}=\frac{1}{3}B^2+\frac{1}{3}AB$$ Then, $A=-2B$, also $-B^3=3I$, $B^{-1}=-\frac{1}{3}B^2$ and $A^{-1}=\frac{1}{6}B^2$.