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x+2y+3z=4 19. 3x+6y+z=2 (no solution) 2x+4y+z=-2 [x-2y+3z=4 20. 3x-y-z=2 x+y-3z=-2

x+2y+3z=4 19. 3x+6y+z=2 (no solution) 2x+4y+z=-2 [x-2y+3z=4 20. 3x-y-z=2 x+y-3z=-2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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9, 10, 19, 20 using Gauss-Jordan Elimi

### Example 19: This system of linear equations has no solution. The equations are: 1. \(x + 2y + 3z = 4\) 2. \(3x + 6y + z = 2\) 3. \(2x + 4y + z = -2\) The relationships between the equations indicate that they are inconsistent, meaning they do not intersect at a common point in three-dimensional space. ### Example 20: This is a system with infinitely many solutions. The equations are: 1. \(x - 2y + 3z = 4\) 2. \(3x - y - z = 2\) 3. \(x + y - 3z = -2\) The solution is expressed in parametric form as \((z, 2z - 2, z)\), where \(z\) is a parameter. This means the three planes defined by these equations intersect along a line where \(z\) can be any real number, and \(x\) and \(y\) depend on \(z\).
Transcribed Image Text:### Example 19: This system of linear equations has no solution. The equations are: 1. \(x + 2y + 3z = 4\) 2. \(3x + 6y + z = 2\) 3. \(2x + 4y + z = -2\) The relationships between the equations indicate that they are inconsistent, meaning they do not intersect at a common point in three-dimensional space. ### Example 20: This is a system with infinitely many solutions. The equations are: 1. \(x - 2y + 3z = 4\) 2. \(3x - y - z = 2\) 3. \(x + y - 3z = -2\) The solution is expressed in parametric form as \((z, 2z - 2, z)\), where \(z\) is a parameter. This means the three planes defined by these equations intersect along a line where \(z\) can be any real number, and \(x\) and \(y\) depend on \(z\).
Here are transcriptions of the two systems of equations provided, along with their solutions: **Problem 9:** \[ \begin{cases} x + y - z = -2 \\ x - y + z = 12 \\ x - y - z = 0 \end{cases} \] **Solution for Problem 9:** \[ (5, -1, 6) \] **Problem 10:** \[ \begin{cases} x + 2y + z = 1 \\ 2x - y + z = 0 \\ -x + y - z = -1 \end{cases} \] **Solution for Problem 10:** \[ (-1, 0, 2) \]
Transcribed Image Text:Here are transcriptions of the two systems of equations provided, along with their solutions: **Problem 9:** \[ \begin{cases} x + y - z = -2 \\ x - y + z = 12 \\ x - y - z = 0 \end{cases} \] **Solution for Problem 9:** \[ (5, -1, 6) \] **Problem 10:** \[ \begin{cases} x + 2y + z = 1 \\ 2x - y + z = 0 \\ -x + y - z = -1 \end{cases} \] **Solution for Problem 10:** \[ (-1, 0, 2) \]
Expert Solution
Step 1
Total Equations are 3

x+2y+3z=4→(1)

3x+6y+z=2→(2)

2x+4y+z=-2→(3)

Converting given equations into matrix form
  1 2 3   4  
  3 6 1   2  
  2 4 1   -2  


R2←R2-3×R1
 = 
  1 2 3   4  
  0 0 -8   -10  
  2 4 1   -2  


R3←R3-2×R1

 
 = 
  1 2 3   4  
  0 0 -8   -10  
  0 0 -5   -10  



i.e.

x+2y+3z=4

-8z=-10

-5z=-10

z=2

and  z= 10/8 both inconsistent .

Thus system has no solution.
 
 
2).
 
Total Equations are 3

x-2y+3z=4→(1)

3x-y-z=2→(2)

x+y-3z=-2→(3)

Converting given equations into matrix form
  1 -2 3   4  
  3 -1 -1   2  
  1 1 -3   -2  


R2←R2-3×R1
 = 
  1 -2 3   4  
  0 5 -10   -10  
  1 1 -3   -2  


R3←R3-R1
 = 
  1 -2 3   4  
  0 5 -10   -10  
  0 3 -6   -6  


R2←R2÷5

 
 = 
  1 -2 3   4  
  0 1 -2   -2  
  0 3 -6   -6  


R1←R1+2×R2
 = 
  1 0 -1   0  
  0 1 -2   -2  
  0 3 -6   -6  


R3←R3-3×R2
 = 
  1 0 -1   0  
  0 1 -2   -2  
  0 0 0   0  


i.e.

x-z=0

y-2z=-2

i.e.

x=z

y=-2+2z

Solution By Gauss jordan elimination method
x=z,y=-2+2z and z=z
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