University:
Eastern Kentucky University
Course:
CSC 310 | Data Structures
Academic year:

2020
Views:

27

Pages:

9
Author:

opisivanetmi

Chapter 3 31 Because x = 1/3 * y_{i} + (2 * lambda/3) * i_{2} me representation of x with ズ respect to 18, Indeed, we have q_{1} F * Because xi r =-26 perp + 1.5' f perp the representation of th respect to Eis, 2.3 is [1.5] Indeed g 1 = [[0, 2], [1, 2]] [ matrix -2\\ r,5 matrix ]=[ matrix 3\\ 1 matrix ] we have 3.2 ||x_{1}|| 1 =2r 3+1=6 118, I_{2} = sqrt(4 + 9 + 1) = sqrt(17) = 5.74 1 a = 3 I x 2 || 1 = 1 + 1 * 1 = 3 ||x 2 || 2 = sqrt(1 + 1 + 1) = sqrt(3) = 1 ,73z x + Pi infty =1 J.3 y_{i} = x_{1} / a * x_{i}*d = i/(j, 74) * [[2], [- 3], [1]] u_{2} = x_{2} - (y_{1}' * x_{2}) * f_{1} = [[1], [1], [1]] - 0/(x_{1} * 7varphi) * q_{1} = [[1], [1], [1]] f 2 =u 2 /0v 2 ||= i/(t, 7.7 * 0.2) * [[r], [i], [i]] hole that x, and x_{2} are alreasly orthi. gonal. Therefor F_{1} and F_{2} as these normalized vectors 3.4 of 12m, AA is sy mastue and tas rank . If man den A'A-In und A is non singular. Thus A ^ - 1 = A' and A * A' =AA^ - prime I n As an example, for the orthonormal vectors q, and q, i Problem 3.3 we have Q' * Q = [[l, theta], [0, i]] Q =[\ 1 &7 2 \ 2010.6190-00452 0.4762 1-00452 0.7762 0.1190 10.4702 21140 04048 (symmetric) 3.5 r(A_{1}) = 2 nullity (A,)=3-2=1 P( 4 2 )=3 nullity (A₂)=3-3=0 P( A angle )=3 nullily (A)=4-3-1 3.6 Range space yace [] []; null space[] Range space[][][][] Range space [ matrix 1\\ 0\\ 0 matrix ] - [[2], [- 1], [6]] * [[4], [3], [1]] will appre 3.7 x = [[1], [1]] is a solution Because the millity is 0. the solution is unique 2 the equation has no solution for 1.8% is a solution. The sull space spanned, as computed in Problems.6, by 2-11-05 Thus che fsmeral selution is x = [[- i], [theta], [0], [0], [i]] + alpha * [[- i], [2], [- i], [0]] for any & There is one fue parameter. 3.9 From (3.17) x = [[x_{1}], [- a_{1} + a_{2}], [- a_{1}], [- a_{2}], [- a_{2}]] ||x|| ^ 2 = alpha_{1} ^ 2 + (- 4 + x_{1} + 2x_{2}) ^ 2 + x_{1} ^ 2 + x_{2} ^ 2 =\$ a_{i} ^ 2 + 5a_{j} ^ 2 + 4a_{i}*x_{j} - partial a i -i6 a 2 +i xi=0 B x i +6x i + omega x j - overline x =0 Rightarrow j*x_{j} + alpha*alpha_{j} - omega*phi(ij) 20 2 x 2 =10 4 2 + omega infty 1 - i*\mathcal{x} = phi 5x(t) + (- (2)) / 11 * d_{1} - 4 = (2) a_{1} = 4/41 a 2 = 1/2 * (4 - 3a_{i}) * 1/2 (4- i2 i1 )= 18 i 1 Thus the solution x = [q/11 * (- p)/11 * (- 4/11) * (- 14/11)]' has che smallest 2-nosen. 310 x =[ matrix -i&-a\\ 2x\\ -s\\ .\\ . matrix ] exists= ||x|| ^ 2 = (- 1 - alpha) ^ 2 + 4alpha ^ 2 + alpha ^ 2 + 1 = 6x ^ 2 + 2x + 2 d/dalpha (B) = izz = 0 -116 Thus the solution [1] how has the smallout 2-почт 3.11 if and only if the ron matux [6 Ab A^ prime prime-1 bJ r ronsmgular or has full ron ranh. 5.12 Delta(i) pm ikf(5 overline L -A)= (J - z) ^ 3 * (J - i) = s ^ 4 - 7s ^ 3 + i * theta ^ 2 * s ^ 2 - 20s + delta