which is the direct and inverse transformation matrix? for a linear transformation :
$$x'=ax+by$$
$$y'=cx+dy$$
is this the direct transformation matrix or the inverse transformation matrix?
$$T = \begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix} \tag{1}\label{1}$$
if its direct, I'm having confusion with contravariant components of a vector... a curvilinear transformation can be considered linear in the vicinity of a point and hence (if (1) is true) the direct transformation matrix would be :
$$T = \begin{bmatrix}
\frac{\partial {x_1}'}{\partial x_1} &\frac{\partial {x_1}'}{\partial x_2} \\
\frac{\partial {x_2}'}{\partial x_1}&\frac{\partial {x_2}'}{\partial x_2}\\
\end{bmatrix} \tag{2}$$
3Blue1Brown series on linear algebra taught me that $\begin{bmatrix}a \\ c \\ \end{bmatrix}$ in (1) and
$\begin{bmatrix}\frac{\partial {x_1}'}{\partial x_1} \\ \frac{\partial {x_2}'}{\partial x_1} \\ \end{bmatrix}$ in (2) represent the covariant basis vectors of new (primed) coordinate system.
So the covariant basis vectors are transforming following the direct matrix and that means the contravariant components of a vector would transform following the inverse transformation matrix:
$$g = \begin{bmatrix}
\frac{\partial {x_1}}{\partial {x_1}'} &\frac{\partial {x_1}}{\partial {x_2}'} \\
\frac{\partial {x_2}}{\partial {x_1}'}&\frac{\partial {x_2}}{\partial {x_2}'}\\
\end{bmatrix} \tag{3}$$
Thus, (from (3)), in Index notation, contravariant components of a vector transforms like :
$$A'^{i}= \frac{\partial {x_j}}{\partial x_i}A^{j} $$
this is contrary to the definition of contravariance :
$$A'^{i}= \frac{\partial {x_i}'}{\partial x_j}A^{j} $$
please clarify where I'm wrong
 A: Your coordinated transformation can be written as
$$\left(
\begin{array}{c}
x'\\y'
\end{array}
\right)
=
\left(
\begin{array}{cc}
a&b\\c&d
\end{array}
\right)
\left(
\begin{array}{c}
x\\y
\end{array}
\right)$$
which has derivatives
$$\frac{\partial x'}{\partial x}=a\quad ,\quad \frac{\partial x'}{\partial y}=b,
$$
$$\frac{\partial y'}{\partial x}=c\quad ,\quad \frac{\partial y'}{\partial y}=d.
$$
If you use the canonical base
$e_1=\left(\begin{array}{c}1\\0\end{array}\right)$ and
$e_2=\left(\begin{array}{c}0\\1\end{array}\right)$ you will get the components of new base by transforming the old via
$$\left(
\begin{array}{cc}
a&b\\c&d
\end{array}
\right)
\left(
\begin{array}{c}
1\\0
\end{array}
\right)=\left(\begin{array}{c}a\\c\end{array}\right)
\quad ,\quad
\left(
\begin{array}{cc}
a&b\\c&d
\end{array}
\right)
\left(
\begin{array}{c}
0\\1
\end{array}
\right)=\left(\begin{array}{c}b\\d\end{array}\right).$$
You could write the same as
$$e'_1=ae_1+ce_2,$$
$$e'_2=be_1+de_2.$$
Now, solving for $e_1, e_2$ will give
$$e_1=\frac{d}{ad-bc}e'_1+\frac{-c}{ad-bc}e'_2,$$
$$e_2=\frac{-d}{ad-bc}e'_1+\frac{a}{ad-bc}e'_2.$$
Observe that this correspond to the inverse coordinated transformation
$$\left(
\begin{array}{c}
x\\y
\end{array}
\right)
=
\left(
\begin{array}{cc}
\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}
\end{array}
\right)
\left(
\begin{array}{c}
x'\\y'
\end{array}
\right),$$
or in other words
$$x=\frac{d}{ad-bc}x'+\frac{-b}{ad-bc}y',$$
$$y=\frac{-c}{ad-bc}x'+\frac{a}{ad-bc}y'.$$
So
$$
\frac{\partial x}{\partial x'}=\frac{d}{ad-bc}\quad ,\quad \frac{\partial x}{\partial y'}=\frac{-b}{ad-bc}
$$
$$
\frac{\partial y}{\partial x'}=\frac{-c}{ad-bc}\quad ,\quad \frac{\partial y}{\partial y'}=\frac{a}{ad-bc}
$$
Finally, if you has a vector $A^1e_1+A^2e_2$ written in old coordinates then you will get its new components by subbing the solve above to get
\begin{eqnarray*}
A^1e_1+A^2e_2
&=&A^1\left(\frac{d}{ad-bc}e'_1+\frac{-c}{ad-bc}e'_2\right)+
A^2\left(\frac{-b}{ad-bc}e'_1+\frac{a}{ad-bc}e'_2\right),\\
&=&\left(A^1\frac{d}{ad-bc}+A^2\frac{-b}{ad-bc}\right)e'_1+
\left(A^1\frac{-c}{ad-bc}+A^2\frac{a}{ad-bc}\right)e'_2.
\end{eqnarray*}
These are
$$A'^1=A^1\frac{d}{ad-bc}+A^2\frac{-b}{ad-bc}\quad {\rm and}\quad A'^2=A^1\frac{-c}{ad-bc}+A^2\frac{a}{ad-bc}.$$
That in terms of derivatives are
$$A'^1=A^1\frac{\partial x}{\partial x'}+A^2\frac{\partial x}{\partial y'}\quad {\rm and}\quad A'^2=A^1\frac{\partial y}{\partial x'}+A^2\frac{\partial y}{\partial y'},$$
which coincide with the law you seek to grasp.
A: Direct transformation:
$$
dx_1' = \frac{\partial x_1'}{\partial x_1}dx_1 + \frac{\partial x_1'}{\partial x_2}dx_2 \\
dx_2' = \frac{\partial x_2'}{\partial x_1}dx_1 + \frac{\partial x_2'}{\partial x_2}dx_2
$$
$$
\begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} = \Large
\begin{bmatrix} \frac{\partial x_1'}{\partial x_1} & \frac{\partial x_1'}{\partial x_2} \\
                \frac{\partial x_2'}{\partial x_1} & \frac{\partial x_2'}{\partial x_2}
\end{bmatrix} \normalsize
\begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix}
$$
Inverse transformation:
$$
dx_1 = \frac{\partial x_1}{\partial x_1'}dx_1' + \frac{\partial x_1}{\partial x_2'}dx_2' \\
dx_2 = \frac{\partial x_2}{\partial x_1'}dx_1' + \frac{\partial x_2}{\partial x_2'}dx_2'
$$
$$
\begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} = \Large
\begin{bmatrix} \frac{\partial x_1}{\partial x_1'} & \frac{\partial x_1}{\partial x_2'} \\
                \frac{\partial x_2}{\partial x_1'} & \frac{\partial x_2}{\partial x_2'}
\end{bmatrix} \normalsize
\begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix}
$$
With the direct transformation, calculating the inverse results in:
$$
\begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} = \Large
\begin{bmatrix} \frac{\partial x_2'}{\partial x_2} & -\frac{\partial x_1'}{\partial x_2} \\
                -\frac{\partial x_2'}{\partial x_1} & \frac{\partial x_1'}{\partial x_1}
\end{bmatrix} / \normalsize \det
\begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} \\ \mbox{with} \qquad
\det = \frac{\partial x_1'}{\partial x_1}\frac{\partial x_2'}{\partial x_2}
     - \frac{\partial x_2'}{\partial x_1}\frac{\partial x_1'}{\partial x_2}
$$
With the inverse transformation, calculating the inverse results in:
$$
\begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} = \Large
\begin{bmatrix} \frac{\partial x_2}{\partial x_2'} & -\frac{\partial x_1}{\partial x_2'} \\
                -\frac{\partial x_2}{\partial x_1'} & \frac{\partial x_1}{\partial x_1'}
\end{bmatrix} / \normalsize \det'
\begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} \\ \mbox{with} \qquad
\det' = \frac{\partial x_1}{\partial x_1'}\frac{\partial x_2}{\partial x_2'}
      - \frac{\partial x_2}{\partial x_1'}\frac{\partial x_1}{\partial x_2'}
$$
Hope you can take it from here.
