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    Consider a shear linear transformation T in R^2 given by the matrix \begin{bmatrix} 1/2 & 1 \\0 & 1 \end{bmatrix}. Let "^" be a triangle with vertices at (-1,0), (1,0), and (0,1). What is the area of the triangle obtained from "^" if we apply the shear transformation six times in a row?

    Can anybody help me with this?

    Thank you in advance...
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    There are two things you need to know here:

    1. Areas are scaled by a factor of the determinant. So if S is some shape with area A and you apply your matrix M, then the area of M(S) is (\det M)A.

    2. The determinant of a power of a matrix is the power of the determinant of the matrix. That is, \det (M^n) = (\det M)^n

    You can take it from here
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    (Original post by nuodai)
    There are two things you need to know here:

    1. Areas are scaled by a factor of the determinant. So if S is some shape with area A and you apply your matrix M, then the area of M(S) is (\det M)A.

    2. The determinant of a power of a matrix is the power of the determinant of the matrix. That is, \det (M^n) = (\det M)^n

    You can take it from here

    So is this how you solve it...

    det(T) = \begin{vmatrix} 1/2 & 1 \\0 & 1 \end{vmatrix} = 1/2

    (1/2)^6 = 1/64

    The area of the triangle is...

    (base)(height)(1/2) = (2)(1)(1/2) = 1

    det(t) x area of the triangle = (1/64)(1) = 1/64...is this correct?
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    (Original post by Artus)
    is this correct?
    Yup that's fine
 
 
 
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