$AB*\text{adjoint}(BA)=I$

Question

$AB*\text{adj}(BA)=I$

Prove:

$1$. $|AB|=1$

$2$. $AB=BA$

As for $2$. what I have menage is $AB*AB^{-1}=AB^{-1}*AB=AB*\text{adj}$(BA)=I$ \rightarrow BA=AB$

How do I solve $1$. and is $2$. is valid?

Answer 1

From your assumptions it follows that $A$, $B$, and hence also $\text{adj}(AB)$ and $\text{adj}(BA)$ are invertible. Just apply determinant to the equation $AB\cdot\text{adj}(BA)=I$, and use the fact $M\text{adj}M=\det(M)I$.

Moreover, the assumption implies $$ (\text{adj}(BA))^{-1} = AB. $$ On the other hand using $M\text{adj}M=\det(M)I$ we find $$ (\text{adj}(BA))^{-1} = (\det(BA))^{-1} BA $$ This implies $$ \det(BA)AB = BA. $$ Applying determinant to this equation yields $\det(BA)=1$, which gives and $AB=BA$.

Answer 2

For $1$, try calculating the determinant of the left side of the hypothesis. If $A$ and $B$ are both $n \times n$, you should find that $\det(LHS) = \det(AB)^n$, from which the result follows.

Following this, $1 \implies 2$, as applying $M\text{adj}(M) = \det(M)I$ to the hypothesis then implies that $\text{adj}(AB) = \text{adj}(BA)$.