Gaussian Elimination to Solve Systems of Linear Equations

Introduction

• Focus on solving systems of equations using Gaussian elimination.
• Example system:
  • Equation 1: (x + y - z = -2)
  • Equation 2: (2x - y + z = 5)
  • Equation 3: (-x + 2y + 2z = 1)
• Goal: Convert the system into an augmented matrix and then to row echelon form.

Converting to Augmented Matrix

• Write coefficients in matrix form:

  | 1 | 1 | -1 | -2 | |---|----|-----|---| | 2 | -1 | 1 | 5 | | -1| 2 | 2 | 1 |

Steps to Row Echelon Form

1. Make bottom-left elements zero: Add Row 1 to Row 3

   • New Row 3: (r3 \leftarrow r1 + r3)

     | 1 | 1 | -1 | -2 | |---|---|----|---| | 2 | -1| 1 | 5 | | 0 | 3 | 1 | -1 |

2. Make the second column, second row zero: Modify Row 2

   • New Row 2: (r2 \leftarrow -2r1 + r2)

     | 1 | 1 | -1 | -2 | |---|---|----|----| | 0 | -3| 3 | 9 | | 0 | 3 | 1 | -1 |

3. Make third row, second column zero: Add Row 2 to Row 3

   • New Row 3: (r3 \leftarrow r2 + r3)

     | 1 | 1 | -1 | -2 | |---|---|----|---| | 0 | -3| 3 | 9 | | 0 | 0 | 4 | 8 |

4. Normalize the diagonal elements to 1

   • Divide Row 2 by -3 and Row 3 by 4

     | 1 | 1 | -1 | -2 | |---|----|----|----| | 0 | 1 | -1 | -3 | | 0 | 0 | 1 | 2 |

Converting Back to System of Equations

1. Equation from Row 3: (z = 2)
2. Equation from Row 2: (y - z = -3)
   • Substitute (z = 2 \rightarrow y - 2 = -3 \rightarrow y = -1)
3. Equation from Row 1: (x + y - z = -2)
   • Substitute (y = -1, z = 2 \rightarrow x - 1 - 2 = -2 \rightarrow x = 1)

Final Solution

• ((x, y, z) = (1, -1, 2))
• Achieved Row Echelon Form: Diagonal ones and zeros below

Another Example

• Equations:
  • (2x + y - z = 1)
  • (3x + 2y + z = 10)
  • (2x - y + 2z = 6)

Steps to Simplify

1. Perform row operations to make zeros below the pivot positions:

   • Modify Row 3: (r3 \leftarrow r3 - r1)

     | 2 | 1 | -1 | 1 | |---|---|----|----| | 3 | 2 | 1 | 10 | | 0 | 2 | -3 | -5 |

2. Adjust Row 2:

   • Modify Row 2: (r2 \leftarrow -3r1 + 2r2)

     | 2 | 1 | -1 | 1 | |---|---|----|----| | 0 | 1 | 5 | 17 | | 0 | 2 | -3 | -5 |

3. Modify Row 3 again:

   • (r3 \leftarrow -2r2 + r3)

     | 2 | 1 | -1 | 1 | |---|---|----|----| | 0 | 1 | 5 | 17 | | 0 | 0 | -13| -39|

Back Substitution

1. From Row 3: (-13z = -39) (\rightarrow z = 3)
2. Substitute in Row 2: (y + 5z = 17 \rightarrow y + 5(3) = 17 \rightarrow y = 2)
3. Substitute in Row 1: (2x + y - z = 1 \rightarrow 2x + 2 - 3 = 1\rightarrow x = 1)

Final Solution

• ((x, y, z) = (1, 2, 3))