How to prove that $B$ is invertible and to find $\det A$

Question

I try to solve a question but can't find a way.

Given $A,B$ are $4\times4$ matrices and $$ A-A^2B-2I = 0 \quad\text{and}\quad 3BA^2+A^2-3A = 0 $$ I need to prove that $A,B$ are invertible and to find $\det A$.

I succeeded to find that $A$ is invertible using $A(I-AB) = 2I$.

But I can't find why $B$ is invertible no matter how I do algebraic manipulations.

Please help, even a hint will help, because I have no clue.

Answer 1

Observe that $3BA^2+A^2-3A=0$ implies $3BA+A-3I=0$, from which we get $$ (3B+I)A=3I $$ which, by using the fact that for an invertible matrix $M$ we have $MM^{-1}=M^{-1}M=I$, we achieve $$ A(3B+I)=3AB+A=3I \implies AB=I-\frac{1}{3}A. $$ Now, replace for $AB$ in $A-A^2B-2I=0$ and enjoy the rest of the journey for yourself!

Answer 2

The first equation gives $$AB=I-2A^{-1}$$ Multiply the second equation by $A$ on the left side and by $A^{-2}$ on the right side to get $$AB=I-{1\over 3}A $$ Now you can solve for $A$ to get $A^2=6I,$ which implies that $\pm \sqrt{6}$ are the only possible eigenvalues of $A. $ The equation $B={1\over 6}A-{1\over 2}I$ implies that $B$ is invertible, as $0$ cannot be its eigenvalue.

The condition $A^2=6I$ does not determine $\det A, $ but $(\det A)^2.$