Prove that $\ AB$, $\ A^{-1}$ and the adjunct of $\ A$ have also this property

Question

Let $\ A, B$ be two real matrices of dimension n so that they are invertible and the inverse is equal to the transposition. Prove that $\ AB$, $\ A^{-1}$ and the adjunct of $\ A$ have also this property and find the matrices defined in the natural set, of dimension n, that have this property and are also symmetric and prove their trace differs from two. I am not sure on how to prove $\ A^{*}$ has the property, but I have a clue about proving $\ AB$, $\ A^{-1}$ has the property, but I am not sure. Any help?

Answer 1

If you want to check that the transpose of some matrix is the inverse of that matrix, you should multiply it with its transpose and see that you get the identity. For instance, I will show that $AB(AB)^T=I$. First, $(AB)^T=B^TA^T$, so we insert that: $$ AB(AB)^T=ABB^TA^T $$ Next, we are told that $A^T=A^{-1}$ and $B^T=B^{-1}$, so we insert that and we get $$ ABB^TA^T=ABB^{-1}A^{-1}=AIA^{-1}=I $$ and we're (almost) done. Technically we have to show that $(AB)^TAB=I$ as well, but the proof of that is completely analogous. The other proofs use similar techniques as well: multiply with the transpose, use that $A^{-1}=A^T$, and calculate.