Álgebra Linear- Matrizes Determinantes Matriz Inversa Matriz Adjunta

1ª avaliação de Álgebra Linear\nALUNA: Kariny Alanda Teixeira Costa\n23/06/21\n\n\\left\\{ \\begin{array}{l}\n2x_1 - 4x_2 - x_3 = 1 \\\\\n x_1 - 3x_2 + x_3 = 1 \\\\\n 3x_2 - 5x_3 = 1 \n\\end{array} \\right. \n\nL_2 = L_2 + (-2)L_1\nL_3 = L_3 + (-3)L_1\n\n\\left[ \\begin{array}{ccc|c}\n2 & -4 & 1 & 1 \\\\\n1 & -3 & 1 & 1 \\\\\n3 & -5 & 3 & 1\n\\end{array} \\right]\n\nd_{b}\n\\left[ \\begin{array}{ccc|c}\n1 & -3 & 1 & 1 \\\\\n0 & -4 & 2 & -1 \\\\\n0 & 4 & -6 & -2\n\\end{array} \\right]\nComo o sistema é indeterminado, admite várias soluções:\n\nA) (3, 1, 1)\nSubstituindo:\n\n1\\quad 2\\cdot3 - 4\\cdot1 - 1 = 6 - 4 - 1 = 1\n\nII\\quad 3\\cdot1 + 1 + 1 = 3 - 3 + 1 = 1\n\nIII\\quad 3\\cdot3 - 5\\cdot1 - 3\\cdot1 = 9 - 5 - 3 = 1\n\nB) \\left( \\frac{13}{2}, \\frac{5}{2}, 2 \\right)\nSubstituindo:\n\n1\\quad 2\\cdot \\frac{13}{2} - 4\\cdot \\frac{5}{2} - 2 = 13 - 10 - 2 = 1\n\n2\\quad \\frac{13}{2} - 3.5 + 2 = \\frac{13}{2} - 3.5 + 2 = -1\n\nIII\\quad 3\\cdot \\frac{13}{2} - 5\\cdot \\frac{5}{2} - 3\\cdot2 = 39 - 25 - 6 = -1 c) (17, 7, 5)\nSubstituindo:\n\nI\\quad 2\\cdot 17 - 4\\cdot 7 - 5 = 34 - 28 - 5 = 1\n\nII\\quad 17 + 3\\cdot 7 + 5 = 17 - 21 + 5 = -1\n\nIII\\quad 3\\cdot 17 - 5\\cdot 7 - 3\\cdot 5 = 51 - 35 - 15 = 1\n\n\\left\\{ \\begin{array}{l}\n4x + y + 2z + w = 6 \\\\\n3x + 7y - 2w = 1 \\\\\n7x + 3y - 52 + 8w = -3 \\\\\nx + y + 2z = 3 \n\\end{array} \\right. \n\n\\begin{vmatrix} 4 & 1 & 1 \\\\\n3 & 7 & 1 \\\\\n7 & 3 & -8 \\\\\n1 & 2 & 1 \\end{vmatrix}\n\n\\text{det } A = 4 \cdot \\begin{vmatrix}7 & 1 \\ 3 & 1 \\end{vmatrix} - 1 \\cdot \\begin{vmatrix}3 & 1 \\ 7 & -8 \\end{vmatrix} + 1 \\cdot \\begin{vmatrix}3 & 1 \\ 7 & 3 \\end{vmatrix}\n D_y = 4 \\begin{vmatrix} 3 & -1 & 1 \\ 7 & 5 & 8 \\ 1 & 2 & 1 \\end{vmatrix}\nD_y = 4(-24 + 10 + 36 - 15 - 8) - 6(-8 + 30 + 7 + 14 - 5 - 24)\n- (8 + 30 + 7 + 14 - 5 - 24)\n+ (1)(56 + 18 + 7 - 98 - 3 - 94) - 1(-35 - 9 - 7 - 49 + 3 + 15)\ndet A = 4(-1\ - 36) + 1(-44) - 1(-64)\nD_y = -184 - 84 + 72 + 80 = -260\n\ny = \\frac{-D_y}{424} = \\frac{260}{424} \\approx \\frac{65}{106}\n\n\\begin{vmatrix} 4 & c & a \\ 3 & -1 & 1 \\ 7 & -3 & 5 \\end{vmatrix}\n det A = a^4 - 3b^4 + 8.b^3a - 6a^2b^2 det A = a(b^3 + a^3 - 3b^2a - 2a^2b) - b(b^3 + a^2b^2 - b^2a - ab^2) + b(b^2a + b^3 - b^2a - a^2b) Para que A seja invertível, é necessário que exista uma matriz B tal que B.A = I\nA é invertível