EGD125 Introductory Engineering Mathematics Worksheet PDF
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This document is a set of exercises for an introductory engineering mathematics course (EGD125). It covers topics like matrices, forces, moments, determinants, and linear systems. The problems involve calculations and derivations using the concepts.
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EGD125 – Introductory engineering mathematics Matrices/Forces and Moments Worksheet II (Week 7) Warm-up problem-solving exercise 1. (a) A bolt is located at point 𝐴 = (3, 3, 0). A spanner is used to tighten the bolt. A force F = (2 3 − 5) T is applied to the spanner at point 𝐵 =...
EGD125 – Introductory engineering mathematics Matrices/Forces and Moments Worksheet II (Week 7) Warm-up problem-solving exercise 1. (a) A bolt is located at point 𝐴 = (3, 3, 0). A spanner is used to tighten the bolt. A force F = (2 3 − 5) T is applied to the spanner at point 𝐵 = (1, 2, 2). Find the moment of −−→ force M = d × F applied to the spanner about the bolt, where d = 𝐴𝐵 is the vector from the point of rotation 𝐴 to the point 𝐵 where the force is applied. The direction of the moment vector M gives the axis of rotation. Mathematical exercises TEQSA PRV12079 | CRICOS No. 00213J EGD125 1 EGD125 – Introductory engineering mathematics 1. Find the indicated determinants. 3 1 1 0 1 3 (a) | 𝐴| if 𝐴 = 2 1 1 0 3 2® © ª (c) |𝐶 | if 𝐶 = −4 2 0 −1® ® 1 1 3 (b) det (𝐵) if 𝐵 = 0 3 2® © ª « 0 0 1 2¬ «−2 4 −1¬ 2. Use Cramer’s rule to solve the following linear system. 3𝑥 + 𝑦 = 2 , 2𝑥 + 𝑦 = 1 3. Write the following systems of equations in matrix form, then form the augmented matrix [ 𝐴|𝑏]. Where possible, use Gaussian elimination and backsubstitution to solve the following linear systems. (a) −3𝑥 − 𝑦 = 2 (b) 𝑝 + 𝑞 − 𝑟 = 1 3𝑥 + 2𝑦 = 1 2𝑝 − 𝑞 − 𝑟 = 0 3𝑝 − 2𝑟 = 2 4. Find the indicated inverse matrices. −1 1 3 1 1 1 1 (a) 𝐴 if 𝐴 = 3 2 0 2 1 0® © ª (b) 𝐵−1 if 𝐵 = 0 0 1 3® ® «1 0 1 2¬ 5. Use the inverse matrix of the coefficient matrix to solve the following linear systems. (a) 𝑥 + 3𝑦 = 0 (b) 𝑎 + 𝑓 + 𝑘 + 𝑝 = 5 3𝑥 + 2𝑦 = 4 2𝑓 + 𝑘 = 0 𝑘 + 3𝑝 = −2 𝑎 + 𝑘 + 2𝑝 = 1 TEQSA PRV12079 | CRICOS No. 00213J EGD125 2 EGD125 – Introductory engineering mathematics Problem-solving exercises 2. A beam that weighs 150 N is held up in a horizontal position by two forces F1 and F2 , as indicated in the diagram (not drawn to scale). 𝐹2 𝐹1 45◦ 60◦ 150 N (a) Write the two force vectors F1 and F2 in Cartesian form. (Denote the, as yet unknown, magnitudes 𝐹1 and 𝐹2 , and evaluate any trig functions to exact values). (b) Write the force balance between the weight of the beam 150 N, and the two forces F1 and F2 , in the form of two linear equations (one for horizontal force and one for vertical force) where the unknown force magnitudes are the variables in the equations. (c) Write the coefficient matrix, solution vector and right hand side vector if this system of linear equations is written in matrix form. (d) Solve the system of linear equations using Cramer’s rule, multiplication by the coefficient matrix inverse, or Gaussian elimination and back substitution. Additional Questions TEQSA PRV12079 | CRICOS No. 00213J EGD125 3 EGD125 – Introductory engineering mathematics 1. Find the indicated determinants. 1 3 1 0 1 3 0 (a) det (𝐷) if 𝐷 = 3 2 1 0 0 3 2® © ª (c) det (𝐹) if 𝐹 = −4 0 2 0 −1® ® 1 −2 0 0 1 0 2® ® (b) |𝐸 | if 𝐸 = 0 −1 « 0 0 0 1 0¬ 2. Use Cramer’s rule to solve the following linear systems. (a) 𝑥 + 3𝑦 = 0, 3𝑏 + 2𝑐 = 1, 3𝑥 + 2𝑦 = 1 −2𝑎 + 4𝑏 − 𝑐 = 0 (b) 𝑎 + 𝑏 + 3𝑐 = 0, 3. Write the following systems of equations in matrix form, then form the augmented matrix [ 𝐴|𝑏]. Where possible, use Gaussian elimination and backsubstitution to solve the following linear systems. (a) 𝑎 + 𝑐 + 3𝑑 = 1 (b) 𝑥 + 𝑦 + 3𝑧 = 1 (c) 𝑎 + 𝑏 + 𝑐 + 3𝑑 = 0 𝑎 + 3𝑐 + 2𝑑 = 0 3𝑦 + 2𝑧 = 2 𝑎 + 𝑏 + 3𝑑 + 2𝑒 = 0 −4𝑎 + 2𝑏 − 𝑑 = 1 −2𝑥 + 4𝑦 − 𝑧 = 2 −4𝑎 − 4𝑏 + 2𝑐 − 𝑒 = 0 𝑐 + 2𝑑 = 4 𝑐 + 2𝑒 = 0 𝑑=0 4. Find the indicated inverse matrices. 1 1 3 −2 −1 1 (a) 𝐶 −1 if 𝐶 = 0 3 2® (b) 𝐷 −1 if 𝐷 = −6 −1 2® © ª © ª «−2 4 −1¬ « 3 1 −1¬ 5. Use the inverse matrix of the coefficient matrix to solve the following linear systems. (a) 𝑥 + 𝑦 + 3𝑧 = 2 (b) −2𝑥 − 𝑦 + 𝑧 = 0 3𝑦 + 2𝑧 = 1 −6𝑥 − 𝑦 + 2𝑧 = 1 −2𝑥 + 4𝑦 − 𝑧 = 3 3𝑥 + 𝑦 − 𝑧 = 3 TEQSA PRV12079 | CRICOS No. 00213J EGD125 4