Lecture Notes: Solving Systems of Equations Using Gaussian Elimination

Introduction

• Focus on using Gaussian elimination to solve a system of equations with three variables.
• Example problem setup:
  • Equations:
    1. ( x + y - z = -2 )
    2. ( 2x - y + z = 5 )
    3. ( -x + 2y + 2z = 1 )

Conversion to Augmented Matrix

• Convert system of equations into an augmented matrix:
  • [ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 2 & -1 & 1 & | & 5 \ -1 & 2 & 2 & | & 1 \end{bmatrix} ]

Row Echelon Form

• Goal: Convert matrix to row echelon form (diagonal of ones and zeros below it).
• Eliminating non-zero entries below the pivot:
  • Step 1: Make zeros below the first pivot (first column):
    • Add Row 1 to Row 3 and assign to Row 3.
  • Step 2: Make zeros below the second pivot (second column):
    • Multiply Row 1 by (-2) and add to Row 2.
  • Step 3: Make zeros below the third pivot (third column):
    • Add Row 2 to Row 3.
• Intermediate result:
  • [ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 0 & -3 & 3 & | & 9 \ 0 & 0 & 4 & | & 8 \end{bmatrix} ]

Final Steps

• Convert pivot elements to 1 by scaling:
  • Scale Row 2 by (-1/3) and Row 3 by (1/4).
• Final matrix:
  • [ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 0 & 1 & -1 & | & -3 \ 0 & 0 & 1 & | & 2 \end{bmatrix} ]

Back Substitution

• Convert back to system of linear equations:
  1. ( x + y - z = -2 )
  2. ( y - z = -3 )
  3. ( z = 2 )
• Solve using back substitution:
  • Solution: ( x = 1, y = -1, z = 2 )

Summary of Row Echelon Process

• The row echelon form has a diagonal of ones with zeros below.
• This form helps identify solutions directly.

Second Example

• System of equations given:
  • ( 2x + y - z = 1 )
  • ( 3x + 2y + z = 10 )
  • ( 2x - y + 2z = 6 )
• Use Gaussian elimination with back substitution without complete row echelon form.

Augmented Matrix

• Initial augmented matrix:
  • [ \begin{bmatrix} 2 & 1 & -1 & | & 1 \ 3 & 2 & 1 & | & 10 \ 2 & -1 & 2 & | & 6 \end{bmatrix} ]

Gaussian Elimination

• Eliminate entries to form zeros below pivots:
  • Use row operations to make zeros below each pivot.
  • Transform matrix to simpler form to solve directly.

Solve Using Back Substitution

• Equations after transformations:
  1. ( 2x + y - z = 1 )
  2. ( y + 5z = 17 )
  3. ( -13z = -39 )
• Solve for variables:
  • Solve for ( z ), then ( y ), and finally ( x ).
• Final solution: ( x = 1, y = 2, z = 3 )

Conclusion

• Gaussian elimination simplifies solving systems of equations by transforming matrices to row echelon form or a simpler form.
• Back substitution extracts solutions from the simplified matrix form.
• Ensures precision and careful attention to sign changes during operations.