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Gaussian Elimination: A Step-by-Step Guide

Gaussian Elimination: A Step-by-Step Guide

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.

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