# Negative determinant as an area

In the week 4 determinant as an area

so the positive determinant matrix is [[3,1],[1,2]], what he means by " It’s the same as the other matrix, except in the opposite order."?

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

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 someone can help, thx

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!

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@Joseph_Kim1, it looks like the LaTeX notation in your reply is broken. There are a lot of “Math Processing Error” messages.

try refresh, it works for me

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@TMosh thanks for letting me know.

It seems to work on my end, so I’m not sure what might be causing the issue. Maybe refreshing like @Bio_J suggested, or clearing the browser cache might fix it.

Refresh worked. Which is weird because some of the LaTex had displayed correctly, but the last three didn’t.

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