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Why does the half of the absolute value of a matrix formed with its coordinates give

  1. Feb 25, 2012 #1
    Why does the half of the absolute value of a matrix formed with its coordinates give the area of a triangle?


    I don't see any similarity between that and the heron's formula.
     
  2. jcsd
  3. Feb 25, 2012 #2

    dx

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    Re: Why does the half of the absolute value of a matrix formed with its coordinates g

    You mean determinant.
     
  4. Feb 26, 2012 #3
    Re: Why does the half of the absolute value of a matrix formed with its coordinates g

    Yeah, why is that?
     
  5. Mar 1, 2012 #4
    Re: Why does the half of the absolute value of a matrix formed with its coordinates g

    The reason is because half the area of a parallelogram is the area of a triangle. Say you have k linearly independent vectors, the volume of the parallelepiped they span is given by the formula:[tex] V(x_1,...,x_k)= \sqrt{det(X^{tr}X)}[/tex]. This formula is easiest to understand in 3 dimensions. Say you have two vectors a and b which are elements of [tex]\mathbb{R}^3[/tex] then the volume formula says that [tex]V(a,b)^2=det\begin{bmatrix}
    \|a\|^2 & \langle a,b \rangle \\
    \langle b,a \rangle & \|b\|^2 \\
    \end{bmatrix} = \|a\|^2\|b\|^2-\langle a,b \rangle^2 = \|a\|^2\|b\|^2(1-\cos^2(\theta)) = \|a\|^2\|b\|^2\sin^2(\theta) [/tex] which is just the area of the parallelogram spanned by a and b. This is the well known formula that magnitude of the cross product of the two vectors is the area of the parallelogram spanned by the two vectors.
     
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