45_hwc1SolnsODDA

# 45_hwc1SolnsODDA - f A(0 = 0 f A e 1 = e 1 f A e 2 = 2 e 1 3 e 2 f A e 1 e 2 = 3 e 1 3 e 2 Thus the image is the quadrilateral with vertices(0 0(1

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which is the equation of a line in the u, v plane. (c) As in part (a) , we use ( * ) to express the equation for S , here x 2 + y 2 = 1, in terms of u and v . The result is: ((3 u - 2 v ) / 3) 2 + ( v/ 3) 2 = 1 , which simpliﬁes to 9 u 2 + 5 v 2 - 6 uv = 9 , which is the equation of an ellipse in the u, v plane. 6.3 Let A be the matrix A = h 1 2 0 3 i , and let f A be the corresponding linear transformation. (a) Graph the image of the unit square under f A , and compute the area of this image. (b) Let S be the triangle with vertices (1 , 1), (1 , 4) and (3 , - 5). What is the area of S ? Graph the image of S under f a , and compute the area of this image. SOLUTION We ﬁrst work out the magniﬁcation factor of this linear transformation. By Theorem 16, this is | 1 × 3 - 2 × 0 | = 3 . (a) The area of the unit square is, of course, 1, and sine the magniﬁcation factor is 3, the area of its image is 3. To graph the image, we work out the vertices of the image: There are:
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Unformatted text preview: f A (0) = 0 f A ( e 1 ) = e 1 f A ( e 2 ) = 2 e 1 + 3 e 2 f A ( e 1 + e 2 ) = 3 e 1 + 3 e 2 . Thus the image is the quadrilateral with vertices (0 , 0) (1 , 0) (2 , 3) (3 , 3) , and you can now easily graph it. (b) The area of the triangle S is 3, as you can work out using the fact that area = 1 2 base × height . Since the magniﬁcation factor is 3, the area of the image is 9. To graph the image, we work out the vertices of the image: There are: f A ( e 1 + e 2 ) = 3 e 1 + 3 e 2 f A ( e 1 + 4 e 2 ) = 9 e 1 + 12 e 2 f A (3 e 1-5 e 2 ) =-7 e 1-15 e 2 . Thus the image is the triangle with vertices (3 , 3) (9 , 12) (-7 ,-15) , 21 /september/ 2005; 22:09 46...
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## This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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