Anonymous

(1) : x + y + z = - 5 (2) : x - y + 3z = - 1 (3) : 4x + y + z = - 2 You calculate (1) + (2) and you obtain the equation (4) (4) : (x + y + z) + (x - y + 3z) = (- 5) + (- 1) (4) : x + y + z + x - y + 3z = - 5 - 1 (4) : 2x + 4z = - 6 (4) : x + 2z = - 3 (4) : x = - 3 - 2z You calculate (3) + (2) (4x + y + z) + (x - y + 3z) = (- 2) + (- 1) 4x + y + z + x - y + 3z = - 2 - 1 5x + 4z = - 3 → recall (4): x = - 3 - 2z 5.(- 3 - 2z) + 4z = - 3 - 15 - 10z + 4z = - 3 - 6z = 12 → z = - 2 Recall (4): x = - 3 - 2z x = - 3 + 4 → x = 1 Recall (1): x + y + z = - 5 y = - 5 - x - z y = - 5 - 1 + 2 → y = - 4 With Matrix: (1) : x + y + z = - 5 (2) : x - y + 3z = - 1 (3) : 4x + y + z = - 2 m0 : matrix of coefficients: _1__1__1 _1__-1__3 _4__1__1 m0 = 1.[(- 1 * 1) - (3 * 1)] - 1.[(1 * 1) - (3 * 4)] + 1.[(1 * 1) - (- 1 * 4)] = 12 K : matrix of constants: _-5 _-1 _-2 To obtain m1, you get m0 and you substitute the first column with K _-5__1__1 _-1__-1__3 _-2__1__1 m1 = - 5.[(- 1 * 1) - (3 * 1)] - 1.[(- 1 * 1) - (3 * - 2)] + 1.[(- 1 * 1) - (- 1 * - 2)] = 12 To obtain m2, you get m0 and you substitute the second column with K _1__-5__1 _1__-1__3 _4__-2__1 m2 = 1.[(- 1 * 1) - (3 * - 2)] - (- 5).[(1 * 1) - (3 * 4)] + 1.[(1 * - 2) - (- 1 * 4)] = - 48 To obtain m3, you get m0 and you substitute the third column with K _1__1__-5 _1__-1__-1 _4__1__-2 m3 = 1.[(- 1 * - 2) - (- 1 * 1)] - 1.[(1 * - 2) - (- 1 * 4)] + (- 5).[(1 * 1) - (- 1 * 4)] = - 24 x = m1/m0 = 12/12 = 1 y = m2/m0 = - 48/12 = - 4 z = m3/m0 = - 24/12 = - 2

Anonymous

. x = 1, y = -4, z = -2 https://flic.kr/p/KwbAhR