How to conclude that matrices $A$ and $B$ are congruent?

Question

Let $A$ and $B$ be real $n×n$ matrices. Which of the following statements is false ?

1. If $A^{-1}$ and $B^{-1}$ are congruent then so are $A$ and $B$.

2. If $A^t$ and $B^t$ are congruent then so are $A$ and $B$. Where $A^t$ denotes transpose of matrix $A$

3. If both $A$ and $B$ are congruent to the identity matrix then they are congruent.

4. If both $A$ and $B$ have the same rank then they are congruent.

My Attempt:

I know that two square matrices $A$ and $B$ over a field are congruent if $\exists$ an invertible matrix $P$ over same field such that $P^tAP = B$

Option $1$, Since $A^{-1}$ and $B^{-1}$ are congruent so there exist a non singular matrix $P$ such that $P^tA^{-1}P = B^{-1}$ Taking inverse on both sides,we get $(P^tA^{-1}P)^{-1} = (B^{-1})^{-1} \implies P^{-1}A(P^t)^{-1} = B$

let $P^{-1} = Q^t$ where $Q^t$ is non singular , then we get $Q^tAQ = B$ . Hence $A$ and $B$ are congruent.

Option $2$, Given that $A^t$ and $B^t$ are congruent so there non singular matrix $P$ such that $P^tA^tP = B^t$ Taking transpose on both sides, we get $P^tAP = B$ so $A$ and $B$ are congruent.

Option 3, given that $A$ and $B$ are congruent to $ I =I_n$ then there exist two invertible matrices $P$ and $Q$ such that $P^tAP = I$ ....(i) and $Q^tBQ = I$ ....(ii). From (i) and (ii) we get $P^tAP = Q^tBQ \implies (P^t)^{-1}P^tAPP^{-1} = (P^t)^{-1}Q^tBQP^{-1} \implies A = (P^t)^{-1}Q^tBQP^{-1} \implies A = (P^{-1})^tQ^tBQP^{-1} \implies A = (QP^{-1})^tB(QP^{-1}) \implies A = C^tBC$ , where $C = QP^{-1}$ is invertible as product of two invertible is invertible. Hence $A$ and $B$ are congruent. I think option 4 is false. Please help me how to discard it.

Answer 1

Hint :

If $A$ and $B$ are congruent, then their determinant have the same sign. Try to construct a counterexample to option $4.$, using (and proving !) this statement.