Q6rs - (2,6). Computation. You need to find two vectors...

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Quiz VI 1. Compute the determinant of A = 0 0 0 2 1 0 0 3 0 1 0 4 0 0 1 5 . Computation. There is only one product in this determinant which is not zero: it is a 21 a 32 a 43 a 14 = (1)(1)(1)(2). The number of swaps to get [2341] back to [1234] is odd – you need at least three swaps. So the determinant is -2. Cofactor expansion is easier: just use the top row and you get 0 - 0 + 0 - 2det(I) =-2. 2. Use determinants to find the area of the triangle with vertices (1,4), (2,3) and
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Unformatted text preview: (2,6). Computation. You need to find two vectors that form sides of the triangle. So subtract the first vector from the last two. That gives you ⃗v 1 = [ 1-1 ] and ⃗v 2 = [ 1 2 ] . The absolute value of the determinant of the matrix with columns ⃗v 1 and ⃗v 2 is 3 and that’s the area of the parallelogram defined by ⃗v 1 and ⃗v 2 . So the area of the triangle is 3/2....
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This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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