Hi @Bio_J,

Does he mean the e1= [1,0] was transfer to [3,1] and now transfer to [1,2]?

Yes, it means that the transformations of the basis vectors would be switched:

- For the positive determinant matrix, e1 = [1, 0] will go to [3, 1] while e2 = [0, 1] will go to [2, 1].
- For the negative determinant matrix, e1 will go to [2, 1] and e2 to [3, 1] instead.

Also I don’t get what it means by " **take the vectors in counterclockwise order and positive if we take them in clockwise order** "

If we look at the example image you provided, we have two basis vectors on the left graph, and if we take them in clockwise order, starting from [0, 1], they are [0, 1] and [1, 0].

When applying the positive determinant matrix to the first vector in clockwise order, [0, 1], you would get:

\left(\begin{array}{cc}
3 & 1\\
1 & 2
\end{array}\right)
\left(\begin{array}{cc}
0\\
1
\end{array}\right)
=
\left(\begin{array}{cc}
1\\
2
\end{array}\right)

And then applying the positive determinant matrix to the second vector in clockwise order, you get:

\left(\begin{array}{cc}
3 & 1\\
1 & 2
\end{array}\right)
\left(\begin{array}{cc}
1\\
0
\end{array}\right)
=
\left(\begin{array}{cc}
3\\
1
\end{array}\right)

The second vector [3, 1] is clockwise from the first vector [1, 2] after the transformation.

We can also see what happens for the negative determinant matrix.

Again, going in clockwise order of the vectors, the first vector would become:

\left(\begin{array}{cc}
1 & 3\\
2 & 1
\end{array}\right)
\left(\begin{array}{cc}
0\\
1
\end{array}\right)
=
\left(\begin{array}{cc}
3\\
1
\end{array}\right)

And then applying the negative determinant matrix to the second point in clockwise order, you would get:

\left(\begin{array}{cc}
1 & 3\\
2 & 1
\end{array}\right)
\left(\begin{array}{cc}
1\\
0
\end{array}\right)
=
\left(\begin{array}{cc}
1\\
2
\end{array}\right)

In this case, the second vector [1, 2] is counterclockwise from the first vector [3, 1].

Hope this helps!