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Unformatted text preview: AME 301 Homework #4 Solutions | Fall 2010 1. The rotation matrix is Q = cos sin sin cos : (a) Using the ‘standard basis’ unit vectors x 1 = e 1 = 1 and x 2 = e 2 = 1 as input vectors, the corresponding output vectors are y 1 = Qx 1 = cos sin and x 2 = Qx 2 = sin cos : (b) INSERT SKETCH OF THE 4 VECTORS FOR THETA = 30 DEGREES. (c) By inspection (or by row reduction of A or A T ), the row vectors (and column vectors) are linearly independent. Hence, dim(column space) = 2. A basis for the column space is any two linearly independent vectors in R 2 , say the two columns of A ( a 1 , a 2 ) or the standard basis ( e 1 , e 2 ). (d) We know that dim(null space) + dim(row space) = 2. Since dim(row space = 2), we must have dim(null space) = 0. Hence, the null space contains only the zero vector . (e) Calculate Q 1 using Gauss-Jordan Q I = cos sin 1 0 sin cos 0 1 Multiply row 1 by sin = cos and add to row 2. Noting that cos +sin 2 = cos = 1 = cos , we obtain Q I = cos sin 1 1 = cos sin = cos 1 Next, eliminate q 12 by multiplying row 2 by cos sin and adding to row 1 . Simplifying the result using trigonometric relations, we obtain Q 00 I 00 = cos cos 2 sin cos 1 = cos sin = cos 1 The nal step is to multiply row 1 00 by 1 = cos and row 2 00 by cos . We obtain Q 000 I 000 = 1 0 cos sin 0 1 sin cos 1 Since Q 000 = I , we have I 000 = Q 1 where Q...
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This note was uploaded on 08/25/2011 for the course AME 301 taught by Professor Wu during the Fall '08 term at Arizona.
- Fall '08