Atividade 5

1- vp(x) = det A A = [1 -1 1] [3 0 -x] [0 2/3 0] det A = [1 -1 1] [3 0 -x] [0 2/3 0] det A = (1 -1) (0) + (1 - x) (0) - (1-0-0) + (0 -2) + (1 -x) (0) + (1 -3) (2/3) = 6 + 2x + 2 = 2x + 8 = P(x) A - P(x) = 2x + 8 P(7) = 2(7)+8 P(7) = 22 kg B - P(x) = 2x + 8 30 = 2x + 8 2x = 22 x = 11 anos 2- A = [1 2] [2 3] B = [0 3] [1 4] C = [-1 0] [2 -1] I - X = A + B - C [1 2]+[0 3]=[1+0 2+3]= [1 5] [2 3] [1 4] [2+ 1 3+4] [3 7] [1 5] | [-1 0]=[-(-)] 5-0] = [2 5] [3 7] | [2 -1] [3-2 7-(-)] [2 8] (verdadeiro) II - Y = B - A - C [0 3] | [1 2] = [0 -(1 3 - 2) = [-1 1] [1 4] | [2 3] [1-2] 4-3] [ -1 1] -1 1 | -1 0 = 0 1 -1 -1 |-3 -1 (verdadeiro) III - 2A - C = Z 1 2 2 2 = 2 4 2 2 3 2 4 6 2 4 | -1 0 = 3 4 4 6 |-1 -1 6 7 (verdadeiro) -> Alternativa correta: b) todas as afirmações são verdadeiras. 3- camisas meses botões 3 1 3 | 100 50 6 5 5 | 50 100 50 50 [3 3] [3 3] [ 50 ] 50 x 100 50 50 [100] [3 3] [50] [ 50] 100 [ 50] 50 300 + 50 + 150 150 + 100 + 150 = 600 + 250 + 250 300 + 500 + 250 [ 500 400 ] [1100 1050 ] 4A - A = [1 3] [2 2] 2 x 2 B = [1 3 1] [2 2 1] [1 1 3] 3 x 3 det A = [1 3] [2 2] det A = 1 . 2 - (2 . 3) = -4 det B = [1 3 1] [2 2 1] [1 3 1] det B = (1 . 2 . 1) + (3 . 1 . 1) + (1 . 2 . 1) - (1 . 2 . 1) + (1 . 1 . 1) + (3 . 2 . 3)) = 6 + 3 + 2 - (2 + 1 + 18) = -10 a^2 - 2b = (-4)^2 -2 (-10) = 16 + 20 = 36 4B - det [2 x x 1 2 x] [-1 -2 -1 -1 -2] [3 1 2 3 1] det = (2 - 2) (3 ) + ( x - 1) (3 ) + ( x - x ) (1 ) - ( x - 2) (3 ) + (2 - ) (1)) + ( x - ) (2 )) = -8 - 3x - x - (6 x - 2 - -2 x) -8 - 4x + 8x + 2 14 x - 6 1 |4x - 6| = 8 - leg_8 14x - 6| = 8 - 8 14x - 6| = 6 4x - 6 = 6 4x = 6 + 6 x = 12 4 = 3 4x - 6 = -6 4x = -6 + 6 x = 0 2) 5 = {3 ou 0} 5- \{ \begin{array}{rcl} x-2y+z & = & 3 \\ 2x+y+z & = & 1 \\ 3x-y+2z & = & 2 \end{array} \quad \begin{aligned} x-2y+z & =3 \\ 2x+y+z & =1 \\ x-2y+z & =3 \\ 3x-y+2z & =2 \end{aligned} \\ x-2y+z=3 \\ x=2y-2+3 \\ 2x+y+z=1 \\ 2(2y-2+3)+y+z=1 \\ 4y-2z+6+y+z=1 \\ 5y-z=-5 \\\\ 3x-y+2z=2 \\ 3(2y-2+3)-y+2z=2 \\ 6y-3z+9-y+2z=2 \\ 5y-z=-7 \\\\ \{ \begin{array}{rcl} 5y-z & = & -5 \\ 5y-z & = & -7 \end{array} \\ 5y-z=-7 \\\\ -z=7-5y \\\\ z=7+5y \\\\ 5y-(7+5y)=-5 \\ 5y-7-5y=-5 \\ 0y=2 \\ \emptyset \\\\\ 5=\{(x,y,z) \in \emptyset\} 6- \{ \begin{array}{rcl} x+2y+z & = & 8 \\ 2x-y+z & = & 3 \\ 3x+y-z & = & 2 \end{array} \quad \text{det} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & -1 \end{bmatrix} \\\\ \text{det}=1((-1)(-1))+(2)(-3)-((1)(-1)(3))+(1)(1)(-1))+(2)(2-(-1)))=1+6+2-(-3+1-4)=15 \\ \text{det} x = \begin{bmatrix} 8 & 2 & 1 \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix} \\ \text{det} x = (8(-1)(-1))+(2)(3)+((1)(3-1)((-1)(2))+((8-1)(1))+ (8.4+3)-6-8)=8+4+3-(-2+8-6)=15 \\\\ \text{det} y = \begin{bmatrix} 1 & 8 & 1 \\ 2 & 3 & 1 \\ 3 & 2 & -1 \end{bmatrix} \\ \text{det} y = ((1.3)((-1))+((8.2.1)(8.2))+((-1))((3.3.-9.2))-((1.3.3)(8.2.2))-((1.2.9))-2.3(1(-1)))=(8)(-1)(3))+ (((((1.2+ 8).3+4-9+2-16)= 30 \\\\ \text{det} z = \begin{bmatrix} 1 & 2 & 8 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{bmatrix} \\ \text{det} z = ((1(-1))((2))-((2.3.3))+(8.2.1)=((-8-1)(3))+\text{((1)((8.2-1/3.3-3))= 2) ==> + 18 + 6 - (-24 + 3 + 8) = 45 \\\\\ x=\frac{15}{15}=1 \quad y=\frac{30}{15}=2 \quad z=\frac{45}{15}=3 \\\\\ 5=\{1,2,3\}