Physical/geometrical interpretation of the determinant of a matrix
do all Lorentz transformations have determinant equal to 1? Because they preserve the space-time interval?
Yes, they do, but preservation of the interval is not the right way to think about the "why."
We have two facts:
Lorentz transformations have Jacobian determinant 1.
Lorentz transformations preserve the spacetime interval.
There is no close or simple relationship between these facts. In particular, 1 does not imply 2. For example, the Galilean transformations have Jacobian determinant 1, but they do not preserve the spacetime interval. Similarly, we could define a rotation in the $x-t$ plane, and it would have Jacobian determinant 1 but not preserve the spacetime interval.
It is true that 2 implies 1, but this is simply because 2 is sufficient to completely characterize the Lorentz transformations and therefore give all their properties indirectly.
For a general proof of the unit Jacobian that applies to both Galilean transformations and Lorentz transformations, see the answer to this question: Motivation for preservation of spacetime volume by Lorentz transformation?
Yes, the determinant of the matrix expressing the components of a Lorentz transformation is dictated by the need to keep the spacetime interval invariant.
In the following I will use natural units such that $c=1$, use the "East coast" sign convention, and use the Einstein summation convention. Primed indices indicate vector or tensor components as expressed in the primed coordinate system.
In any coordinate system considered by special relativity, the spacetime interval between two events can be written
$$s^2=\ -\ (\Delta t)^2\ +\ (\Delta x)^2\ +\ (\Delta y)^2\ +\ (\Delta z)^2\ \ .$$
This means that if we define
$$A=\left\lbrack \begin{matrix}\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{matrix}\right\rbrack$$
we can write the spacetime interval as
$$s^2=A^{\mu} g_{\mu\nu} A^{\nu}\ \ ,$$
where $g$ is the Minkowski metric, whose components are:
$$g_{\mu\nu}=\left\lbrack \begin{matrix}-1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{matrix}\right\rbrack \ \ .$$
The Minkowski metric is what defines the geometry of the spacetime used in special relativity, Minkowski space. The Minkowski metric defines what intervals are, in the same way that the Euclidean metric defines what lengths are in Euclidean space. A key point about the Minkowski metric is that its components are the same in any coordinate system considered by special relativity. The invariance of intervals is a direct, easy consequence of the invariance of the components of the Minkowski metric.
To convert a vector from one coordinate system to another, you use a Lorentz transformation once:
$$x^{\mu '}={\Lambda^{\mu '}}_{\nu}x^{\nu}\ \ .$$
But the Minkowski metric is a rank two tensor. To convert a rank two tensor from one coordinate system to another, you use a Lorentz transformation twice, like
$$g_{\mu' \nu'}={\Lambda^{\sigma}}_{\mu '}g_{\sigma\rho}{\Lambda^{\rho}}_{\nu '}\ \ ,$$
which you could write in matrix notation as
$$g'\ =\ {\Lambda}^{T} g \Lambda\ \ .$$
But since the components of the Minkowski metric are the same in any coordinate system, we can also write
$$g\ =\ {\Lambda}^{T} g \Lambda\ \ .$$
Taking the determinant of both sides of that equation gives
$$\det(g)=\det(\Lambda)\det(g)\det(\Lambda)\ \ .$$
The only way for that equation to hold is if
$$[\det(\Lambda)]^2=1\ \ .$$
There are two possible ways that that equation can hold. The most commonly considered possibility is that
$$\det(\Lambda)\ =\ 1\ \ ,$$
in which case the transformation is called a "proper" Lorentz transformation. The other possibility is that
$$\det(\Lambda)\ =\ -1\ \ ,$$
in which case the transformation is called an "improper" Lorentz transformation. Improper Lorentz transformations are less frequently useful than proper Lorentz transformations.