Atividade Álgebra Linear 3

Patr\u00edcia Martins Lisboa M: 20902660 1) [ 1 -1 0 ] [y] [0] [ 2 0 1 ] [y] [0] [ 1 -2 2 ] [z] [0] x - z = 0 2x + z = 0 x - 2y + 2z = 0 -3x = 0 x - z = 0 x - 2y + 2z = 0 x = 0 0 - z = 0 0 - z = 0 z = 0 y = 0 :: Lago e n\u00facleo \u00e9 dado por N(T) = [0 0] 2) T: R^2 -> R^3 T(x,y) = (x+y, x-y, 0) I) T(u+v) = (T(u)+T(v), T(u+v), 0) T(u+v) = (u2+v2+u1+v1, u2+v2-u1-y1, 0) T(u+v) = (v1+u1)+(v1+y2), (u2-u1)+(0)+(0) T(u+v) = T(u) + T(v) T(αu+βv) = (αu1+u2, (αu2-βv2, (αu2+β)0) 3) I) T(v) = [cos(45\u00b0) -sin(45\u00b0) 0] [sin(45\u00b0) cos(45\u00b0) 0] [0 0 1] v = [ 2 -√2 0 ] [√2/2 √2/2 0 ] T(v) = (√2/2) [ 2] [-√2/2 0 0] [0 0 1] 0 = [ -√2/2] v·T(v) = ||v|| ||T(v)|| cos(θ) 5(a)\nT(u) = u = (1 3)(x)\n (-1 0)(y)\n (-x)\n = (x + 3y)\n (-x)\n = -(x)\n\n { x + 3y = x\n -x = y \n 3y = 0 \n y = 0 }\n\n x = -y \n = 0\n\nu = (0)\n (0)\n\nb.\nT(v) = (0,0)\n (1 3)(x)\n (-1 0)(y)\n (0)\n (0)\n = (x + 3y)\n (-x)\n = (0)\n\n { x + 3y = 0 \n -x = 0 }\n\n { 0 + 3y = 0\n x = 0\n y = 0,\nv = (0)\n (0) }\n 5(c)\nT(4,0,-1) = (0,0)\n\nT = ( a b c )\n ( d e f )\n\n ( 0 - ( a b c ) ( 1 )\n ( 0 d e f ) ( 0 )\n 0 - 1)\n\n { a - c = 0\n T = d - f = 0 }\n\n { a = c\n d = f }\n\n T = ( a b a )\n ( d e a )\n\nConclusions\n a x + b y + a z = 0\n d x + e y + d z = 0\n\n: O sistema é indeterminado, logo não existe\n com os característicos indiciais.