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### Course: Linear algebra>Unit 2

Lesson 6: More determinant depth

# Determinant and area of a parallelogram

Realizing that the determinant of a 2x2 matrix is equal to the area of the parallelogram defined by the column vectors of the matrix. Created by Sal Khan.

## Want to join the conversation?

• This is kind of off topic but can we also use cross product to calculate the area of the parallelogram? like v1 cross v2? If so, they would be different method to achieve the same answer?
• Yes, you can. The cross product is used to do this is the last few videos in the calculus playlist. A good way to see why is to consider the alternate way of calculating the cross product.
• Does this work for any kind of area or only for parallelograms?
• Half of the parallelogram is the triangle created by v1 and v2 so you can find the area of a triangle as being the absolute value of half of the determinant.
• Does this extend to higher dimensional vectors?
• I think it at least applies to 3 dimensions:
http://en.wikipedia.org/wiki/Parallelepiped#Volume

And apparently for higher dimensions, the volume is equal to the determinant of the Gram matrix (but I don't yet understand the Gram matrix very well)
• that was really neat and beautiful
• Hi, this might be kind of weird question out of the blue. But would taking the determinate of a vector be somewhat similar to taking the integral of a function (where both of them seeks for the area)?
• That is a good line of thought.
This is what i think....Having given it a thought.

A vector has magnitude and direction. It has a zero determinant.

Given a linear function above the x-axis(for simplity), the integral of the function is the area under the graph.

This linear function can also be thought of as
line kV1(scaler multiples of some vector v1) through the origin or
Vo + kV1 where the scaler multiples start off from a vector Vo.

This line can be seen as a matrix-vector product
T = AX taking an input Rn and mapping it to R1.
Meaning it takes a vector in Rn and squishes it to a line.
Now finding the determinant of A(the transformation matrix) is 0. det(A).

That is, the determinant of the transformation matrix is 0 and the determinant of the line (if viewed as a long vector) is also zero.

Nonetheless, the area below the line may not be zero but the determinant will always be zero.

The case gets 🤢 if the function is not linear. Probably some crazy curve.

The integral is still the area under the curve.

But what is the determinant? This is not a linear transformation mind you, and not a square matrix(n,n)... because it maps from some Rn to R1. And so no determinant. This transformation matrix A takes a vector in Rn and maps it onto a point. HA!
So
T = AX
where A is 1 by n matrix and X is n by 1 vector in Rn.
T = A(1,n) * X(n, 1)

Happy Learning_HL
• how do I use the parallelogram rule to construct the resultants for vectors
• Sketch the two vectors head to tail and draw a line from the tail of the first to the head of the second.
• At Sal says that V1.V1 is in the numerator, and also in the denominator, so they cancel out, but then V1.V1 is still in the final equation he uses. Shouldn`t V1.V1 have canceled out, and should we not just be left with v2.v2 - (v2.v1) ^2?
• solve quadrilateral abcd vertices a(4,4),b(2,0),c(-4,-2) and d(-2,2) prove that abcd is a parallelogram