Gaussian Elimination is a systematic method for solving systems of linear equations. It works by transforming a given system into an equivalent one in row echelon form using a sequence of row operations. Once in this form, the system can be solved efficiently using back-substitution.
What is Gaussian Elimination?
Gaussian elimination consists of two main stages:
- Forward Elimination: Convert the system into an upper triangular form.
- Back-Substitution: Solve for unknowns starting from the last equation.
Definition of a Pivot
A pivot is the first nonzero entry in a row when moving from left to right. Pivots are used to eliminate the elements below them, transforming the system into an upper triangular form.
Step-by-Step Example
Consider the system of equations:
\[ \begin{aligned} 2x + 3y - z &= 5 \\ 4x + y + 2z &= 6 \\ -2x + 5y - 3z &= -4 \end{aligned} \]Step 1: Convert to Augmented Matrix
We represent this system as an augmented matrix:
\[ \begin{bmatrix} 2 & 3 & -1 & | 5 \\ 4 & 1 & 2 & | 6 \\ -2 & 5 & -3 & | -4 \end{bmatrix} \]Step 2: Normalize the First Pivot (Making it 1)
We normalize (make it 1) the first pivot by dividing Row 1 by 2:
\[ \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & | \frac{5}{2} \\ 4 & 1 & 2 & | 6 \\ -2 & 5 & -3 & | -4 \end{bmatrix} \]Step 3: Eliminate First Column Below Pivot
Subtract \(4\) times Row 1 from Row 2 and add \(2\) times Row 1 to Row 3:
\[ \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & | \frac{5}{2} \\ 0 & -5 & 4 & | -4 \\ 0 & 8 & -4 & | 1 \end{bmatrix} \]Step 4: Normalize the Second Pivot (Making it 1)
Divide Row 2 by \(-5\) to make the pivot 1:
\[ \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & | \frac{5}{2} \\ 0 & 1 & -\frac{4}{5} & | \frac{4}{5} \\ 0 & 8 & -4 & | 1 \end{bmatrix} \]Step 5: Eliminate Below the Second Pivot
Subtract \(8\) times Row 2 from Row 3:
\[ \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & | \frac{5}{2} \\ 0 & 1 & -\frac{4}{5} & | \frac{4}{5} \\ 0 & 0 & \frac{12}{5} & | \frac{-27}{5} \end{bmatrix} \]Step 6: Normalize the Third Pivot (Making it 1)
Divide Row 3 by \( \frac{4}{5} \):
\[ \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & | \frac{5}{2} \\ 0 & 1 & -\frac{4}{5} & | \frac{4}{5} \\ 0 & 0 & 1 & | \frac{-9}{4} \end{bmatrix} \]Step 7: Solve Using Back-Substitution
From the last row:
\[ z = \frac{-9}{4} \]Substituting into the second row:
\[ y - \frac{4}{5} \times \frac{-9}{4} = \frac{4}{5} \] \[ y + \frac{36}{20} = \frac{4}{5} \] \[ y = -1 \]Substituting into the first row:
\[ x + \frac{3}{2} \times (-1) - \frac{1}{2} \times \frac{-9}{4} = \frac{5}{2} \] \[ x - \frac{3}{2} + \frac{9}{8} = \frac{5}{2} \] \[ x = \frac{23}{8} \]Thus, the final solution is:
\[ x = \frac{23}{8}, \quad y = -1, \quad z = \frac{-9}{4} \]Conclusion
Gaussian Elimination is a systematic and efficient approach for solving linear systems. Understanding its step-by-step application provides a strong foundation for advanced linear algebra topics.