Active and passive transformations in Linear Algebra

I am trying to understand what each transformation means and what their differences are but many books that don’t state which transformation they are referring to make it a bit confusing to understand which is which.
Also, I want to particularly know how do matrices change when we deal with each transformation, so if anybody could help it would be really appreciated.

Note: I also want to know how is a matrix represented when we change a basis through such transformations

Solutions Collecting From Web of "Active and passive transformations in Linear Algebra"

In an active transformation, given a basis, we start from a vector and we find a new vector in the same basis.

In a passive transformation we have a vector expressed in a basis and we express it in a new basis.

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The figure illustrate the action of a matrix $A$ as an active transformation and of $A^{-1}$ as the corresponding passive transformation.

Here we have:
$$
A=\begin{bmatrix}
1&2\\
-2&4
\end{bmatrix}
\qquad A^{-1}=\frac{1}{8}
\begin{bmatrix}
4&-2\\
2&1
\end{bmatrix}
$$

The matrix $A$ acts on a vector $\mathbf{x}$ that in the standard basis $S$ (represented in black) has components $\mathbf{x}=[3,2]_S^T$, and, as active transformation, gives the vector $\mathbf{x’}=A\mathbf{x}=[7,2]_S^T$.

Note that in the new basis $B$ that has as basis vectors the columns of $A$ (represented in blue) this vector has components $\mathbf{x’}=[3,2]_B^T$.

The inverse matrix $A^{-1}$ represents the passive transformation that gives the components of the vector $\mathbf{x}$ in the new basis $B$:

$$
A^{-1}\mathbf{x}=
\frac{1}{8}
\begin{bmatrix}
4&-2\\
2&1
\end{bmatrix}
\begin{bmatrix}
3\\
2
\end{bmatrix}=
\begin{bmatrix}
1\\
1
\end{bmatrix}
$$