Unformatted text preview: CS3251, Homework 1 Solution 1. Since v = (3, 1)T we have v = 10 and the unit vector is u1 = v/ v = ( 3 , 1 )T . 10 10 2 1 2 Similarly, w = (2, 1, 2)T so w = 3 and the unit vector is u2 = w/ w = ( 3 , 3 , 3 )T . We want to find U1 u1 and U2 u2 . This implies 0 = U1 u1 = (3x + y)/ 10, where U1 = (x, y)T , and 0 = U2 u2 = (2s + t + 2r)/3, where U2 = (s, t, r)T . We also want U1 and U2 to be unit vectors, which implies x2 + y 2 = 1 and s2 + t2 + r2 = 1. 1 1 Using the equations above we obtain U1 = ( 1 ,  3 )T and U2 = ( 2 , 0,  2 )T . 10 10 2. 2 3 5 1 1 2 4 2 0 1 3 6 6 12 4 2 = 24+32 54+12 = 14 22 9 7 0 0 3 1 = 1 2 1 = 13+21+41 23+0+1 3 2 + 6 (1) 6 2 + 12 (1) = = 8 7 6 6 5 [0,6] 4 [4,4] y 4 y x=4, y=2 [4,2]
3 2 2 [2,1] 1 [1,1] 2 0 2 4 6 8 4 2 0 2 4 x x 1 3. 2x + y x + 2y + z y + 2z + t z + 2t = = = = 0 2x + y 0 + 3y/2 + z 0 y + 2z + t 5 z + 2t = = = = 0 0 0 5 2x + y + 3y/2 + z + 4z/3 + t z + 2t = = = = 2x + y 0 0 + 3y/2 + z 0 + 4z/3 + t 5 5t/4 = = = = 0 0 0 5 Thus, x = 1, y = 2, z = 3 and t = 4. 4. Ax = 2 3 4 1 1 0 2 1 x1 x2 = 1 17 = A = 2 3 1 4 1 17 . E21 A = 2 3 1 4 1 17 2 3 1 0 5 15 Thus we obtain the system U x = c or, equivalently, 2 3 0 5 from which we get x1 = 5 and x2 = 3. 5. We have 0 0 A= 0 0 1 0 0 0 0 1 0 0 0 0 , 1 0 0 0 A2 = 0 0 0 0 0 0 1 0 0 0 0 1 , 0 0 0 0 A3 = 0 0 0 0 0 0 0 0 0 0 1 0 . 0 0 x1 x2 = 1 15 Thus A4 = 0 and k = 4. 2 ...
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This note was uploaded on 01/31/2012 for the course COMS 3251 taught by Professor Anargyrospapageorgiou during the Fall '11 term at Columbia.
 Fall '11
 AnargyrosPapageorgiou

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