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MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022)

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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 + ...

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  • July 11, 2022
  • 14
  • 2021/2022
  • Exam (elaborations)
  • Questions & answers

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  • mat1503
  • mat1503 assignment 03 solutions semester 01 2022
  • mat1503 assignment 03 solutions semester 01 2022 grade a
  • a i we shall do this by substitution thus

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MAT1503 - ASSIGNMENT 03
SOLUTIONS (SEMESTER 01 -
2022)

, QUESTION 1

(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




2

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