Quadrivectors in relativity

This is what I understood about 4-vectors in relativity.

We define the contravariant and covariant vectors like this : $$ A^\mu=\begin{bmatrix} A^0 \\ A^1 \\ A^2 \\ A^3 \end{bmatrix}$$

$$ A_\mu=\begin{bmatrix} A_0 \\ A_1 \\ A_2 \\ A_3 \end{bmatrix}$$

The relationship between them will be :

$$ A^\mu=\eta^{\mu \nu}A_\nu $$

In +--- convention it will lead to :

$$ A^\mu=\begin{bmatrix} A_0 \\ -A_1 \\ -A_2 \\ -A_3 \end{bmatrix}$$

Great.

But it doesn't give me information on the "absolute" sign of 4-vectors. For example if I take the 4-position.

I have an even at time $t$ at space coordinates $(x,y,z)$.

Will I have $$X^\mu=\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$$ Or

$$X_\mu=\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$$ I think it is the first answer because $A^\mu$ should transform the same way that the "real" coordinates $(t,x,y,z)$ transform, but I am not totally sure ?

Thank you.