Title | Exercise 2-1 - Gauss-Jordan Elimination |
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Author | Tenzin Ngawang |
Course | Ordinary Differential Equations and Linear Systems for Mechanical Engineers |
Institution | British Columbia Institute of Technology |
Pages | 3 |
File Size | 106.7 KB |
File Type | |
Total Downloads | 18 |
Total Views | 134 |
Download Exercise 2-1 - Gauss-Jordan Elimination PDF
MATH 3499 – Differential Equations and Linear Systems for Mechanical Engineers Exercise 2-1 Topics: Gauss-Jordan elimination
1. Solve the following systems of equations using the method of Gauss-Jordan elimination. x + 3y = 4 a. 2x − y = 1 6x − 4 y = 2 b. 4x + 8y = 8 2 x − y = 17 c. 4x − 8 = 2 y
2 x + 4 y + 6z = 4 d. x + 5 y + 9z = 2 2 x + y + 3z = 7 x + y −z =2 e. x − 3 y + 2z = 1 3x − 5y + 3z = 4
f.
2x + y + z = 4 x − 2y − z = 3 3x + 3y − 2z = 1
4x − 16y + 2z = − 50 g. −3 x +12 y + 2 z = 48 −2 x + 8 y + 4 z = 40 2x + 4 y − z = 2 h. −6 x −12 y + 3 z = −6 8 x +16 y − 4 z = 8
i.
3x + 4y − z = − 9 4x + 3y + 2z = 0 − x − 6 y + 7 z = 19
MATH 3499 – Differential Equations and Linear Systems for Mechanical Engineers 2. Solve the following systems of equations using the method of Gauss-Jordan elimination.
−3 x + 4 y − 12 z = −66 a. −3 x + 6 y −18 z = −102 2 x − 2 y + 4 z = 22 2x + y + z = 4 b. x − 2 y − z = 3 3x + 3y − 2z = 1 i1 + i2 + i3 = 0 c. 6i1 − 10i3 = 8 6i − 2i = 5 2 1
3. Solve the following systems of equations using the method of Gauss-Jordan elimination.
−4 x − 9 y −10 z = −36 a. 3 x + 9 y +12 z = 45 −2 x − 2 y − z = −4 2x + y + 1= 0 b. 3x − 2y + 5 = 0 x −4 y +5 = 0 5 3 1 x− y = c. 2 + 1 = 7 x y
x = y − 2z d. 2 y = x + 3z + 1 z = 2 y − 2x − 3
MATH 3499 – Differential Equations and Linear Systems for Mechanical Engineers 1 1 1 x+ y+z =5 2 3 4 e. − − = − 11 x y z 3 2 1 x + y − z = −6
f.
2x + 4y + 6z = 4 x + 5 y + 9z = 2 2 x + y + 3z = 7
2 x − 3 y + 3z = 7 g. 3x + y − 2z = − 11 5x − 2y + 4z = 11 3a + 2 b − 4 c + 2 d = 3 5a − 3b − 5c + 6d = 8 h. 2a − b + 3c − 2d = 1 −2 a + 3b + 2 c − 3d = −2
CHALLENGE QUESTION...