hwsolB08

# hwsolB08 - A = [-1 3 4-3 2 1-2 4 1 ] (ii) The shortcut here...

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Sheet1 Page 1 HW 8 Solution ____________ (i) The columns of A are obtained by right-multiplying A by the three unit vectors. We have [1 -1 0] + [2 1 0] = [3 0 0] therefore A*[3 0 0] = A*[1 -1 0] + A*[2 1 0] = [-4 -5 -6] + [ 1 -4 0] = [-3 -9 -6] Thus the first column of A is A*[1 0 0] = [-1 -3 -2] Also, (-2)*[1 -1 0] + [2 1 0] = [0 3 0] Therefore T A*[0 3 0] = -2*A*[1 -1 0] + A*[2 1 0] = (-2)*[-4 -5 -6] + [1 -4 0] = [9 6 12] Thus the second column of A is A*[0 1 0] = [3 2 4] Also, A*[0 0 2] = A*[2 1 0] + A*[-2 -1 2] = [1 -4 0] + [ 7 6 2] = [8 2 2] Thus the third column of A is [4 1 1]

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Unformatted text preview: A = [-1 3 4-3 2 1-2 4 1 ] (ii) The shortcut here is to note that each matrix in the product is a rotation matrix, i.e., is of the form [ cos(t) -sin(t) sin(t) cos(t) ] for some t. From left to right, we have the angles t = pi/6 , pi/2 , theta , pi/3 and pi/4 The product of these matrices is a rotation by a total angle Sheet1 Page 2 pi/6 + pi/2 + theta + pi/3 + pi/4 = 5*pi/4 + theta The product matrix is an identity if and only if the total angle of rotation is 0 (or multiples of 2*pi). Thus we simply have theta = -5*pi/4 (same as 3*pi/4)...
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## This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB08 - A = [-1 3 4-3 2 1-2 4 1 ] (ii) The shortcut here...

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