change of basis conflicted definitions the 2 Definitions I'm talking about are
$P_{B->B'}$ is the change of basis matrix from B to $B'$ such as $B=(x_1,x_2,x_3)$ and $ B'=(x_1',x_2',x_3')$
the two Definitions that I'm talking about are:
$$ \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=P_{B->B'}\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix} 
$$
And
$$
P_{B->B'}=Mat_{B',B}(Id_E)
$$
Let's take an example:
$$ 
\begin{equation}
    \begin{cases}
     x_1'=x_1+3x_2+4x_3\\x_2'=x_2+2x_3\\x_3'=x_3
    \end{cases}\,.
\end{equation}  
$$
$$\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}=\begin{bmatrix}1&3&4\\0&1&2\\0&0&1\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$
so
$$
 P_{B->B'}={\begin{bmatrix}1&3&4\\0&1&2\\0&0&1\end{bmatrix}}^{-1}={\begin{bmatrix}1&-3&2\\0&1&-2\\0&0&1\end{bmatrix}}
$$
using the  second definition :
$$ Id_{E}(x_1')=(1,3,4) \hspace{0.25cm} in \hspace{0.25cm} B\\
Id_{E}(x_2')=(0,1,2) \hspace{0.25cm} in \hspace{0.25cm} B\\
Id_{E}(x_3')=(0,0,1) \hspace{0.25cm}in \hspace{0.25cm}B\\
$$so$$ P_{B->B'}=Mat_{B',B}(Id_E)={\begin{bmatrix}1&0&0\\3&1&0\\4&2&1\end{bmatrix}}
$$
Equations source:Wikipedia(in frensh)
 A: Maybe this helps:
As written in the wikipage, let $E$ be a $K$-vector space of dimension $d$. Let us for  simplicity assume that $K$ is either $\mathbb{R}$ or $\mathbb{C}$. Now, the first thing to understand is, that $E$ must not be $\mathbb{R}^d$ nor $\mathbb{C}^d$, that is, the elements in $E$ could be anything as long as the axioms of a vector space are fulfilled. So, for instance $E$ could be the space of polynomials with degree smaller or equal to $d$ over the real numbers. However, one awesome result of linear algebra is that this vector space is still isomorphic to $\mathbb{R}^d$. In other words, you can uniquely identify every polynomial in that vector space with a $d$-tuple of real numbers. However, this identification depends on your choice of bases! But it this identification that makes it possible to still identify every linear map on a finite dimensional vector space with a matrix. Notice that this is not trivial.
Now, to the concrete problem. I think the confusion comes from the fact that you wrote down an equality that isn't defined. But you want the following:
$$
\begin{pmatrix}x_1' & x_2' & x_3' 
\end{pmatrix}
=
P_{B'\to B}
\begin{pmatrix}
x_1 & x_2 & x_3
\end{pmatrix}.
$$
Notice, that like this the dimension fit. Suppose that $E=\mathbb{R}^3$ then on
the left there is written a $3\times 3 $ matrix. And, if and only if $P_{B'\to B}$
is a $3\times 3$ matrix then so it the right side a $3 \times 3$ matrix. In your case, that didn't
work, because $\begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}^T$ is a $9$ tuple.
Now, the matrix $Mat_{B,B'}(Id_E)$ is constructed by taking that columns to be
the bases elements of $B'$ written in terms of the bases of $B$. So, that in
your case
$$
Mat_{B,B'}(Id_E) = 
\begin{pmatrix}
1 & 0 & 0
\\
3 & 1 & 0
\\
4 & 2 & 1
\end{pmatrix}.
$$
You see that then indeed
$$
P_{B' \to B} = Mat_{B,B'}(Id_E)
\quad
\text{or equivalently}
\quad
P_{B\to B'} = Mat_{B',B}(Id_E).
$$
