MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022)

MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022) (a) (i) We shall do this by substitution. Thus 2 (t + 2) + 4 (−2t − 1) + 6t = 2t + 4 − 8t − 4 + 6t = 0 Also 4 (t + 2) + 5 (−2t − 1) + 6t = 4t + 8 − 10t − 5 + 6t = 3 and finally 7 (t + 2) + 8 (−2t − 1) + 9t = 7t + 14 − 16t − 8 + 9t = 6 Thus, x = t+2, y = −2−1, z = t satisfy all three equation. This solution is general since it has the parameter t, enabling the generation of any other solution from this solution. NB. Only questions 1 and 3 will be marked. Questions 2 and 4 are for self assessment. (ii) x1 − x 2+ 4x5 = 2 x3 − x 5 = 2 x4 − x 5 = 3 Let µ = x5 so that x3 = µ + 2and x4 = µ + 3.Then x1 = x2 − 4µ + 2 Let v = x2 so that x1 = ν − 4µ + 2 Thus (x1, x2, x3, x4, x5) = (ν − 4µ + 2,ν, µ + 2,µ + 3,µ) = (2, 0, 2, 3, 0) + (ν − 4µ,ν, µ, µ, µ) = (2, 0, 2, 3, 0) + (ν,ν, 0, 0, 0) + (−4µ,0, µ, µ, µ) = (2, 0, 2, 3, 0) + ν (1,1, 0, 0, 0) + µ (−4,0, 1, 1, 1) (b) 2x + 8y + 4z = 2 2x + 5y + z = 5 4x + 10y − z = 1 The augmented matrix is   2 2 4   8 4 2 2 8 4 5 1 5  →  0 −3 −3 10 −1 1 0 −6 −9  2 2  −3 R2 − R 1 → R R3 − 2R1 → R   2  8 4 2 − 13 · R2 → R2  0 1 1 −11 R3 − −→ R3 0 2 3   2 0 −4 10 R  0 1 1 −1  R13 −− 82RR22 → → RR13 0 0 1 3 So z = 3(From row 3) From row 2: y + z = −1 ⇒ y + 3 = −1 ∴ y = −4 From row 1 : 2x − 4z = 10 ⇒ 2x − 4 · 3 = 10 ⇒ 2x = 10 + 12= 22 ∴ x = 11 solution (x, y, z) = (11, −4, 3)