Gaussian Elimination for Linear Systems

Gaussian Elimination: Linear Systems

Gaussian elimination is a classic technique used in the study of linear systems. It provides a systematic approach to solving a wide range of problems involving linear equations. In this article, we discuss how Gaussian elimination can be applied to solve linear systems.

Writing the Augmented Matrix

To start, we need to represent the linear system of equations in matrix form. This is done by writing the coefficients of the variables in a rectangular array known as the matrix. For example, consider the system of equations:

[ 3x + y = 1 ] [ 2x - y = 0 ] [ x - z = 2 ]

We can represent this as the augmented matrix:

[ \begin{bmatrix} 3 & 1 & | & 1 \ 2 & -1 & | & 0 \ 1 & 0 & | & 2 \end{bmatrix} ]

Performing Elementary Row Operations

The goal in Gaussian elimination is to transform the original matrix into upper triangular form. This involves performing a series of elementary row operations, which are defined as follows:

1. Interchange any two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of one row to another row.

By applying these operations, we aim to eliminate all but the diagonal entries below each element above it. For instance, we would first apply operation 3 to make the off-diagonal elements of the first column zero. Then, we would use operation 1 if necessary to move up a row with a non-zero entry. Next, we'd multiply any existing non-zero row by a suitable factor and then add or subtract multiples of other rows to reduce higher order terms. Finally, we'd perform back substitution to solve for the variables.

Solving Linear Systems

Let's illustrate this process using an example. Consider the following linear system of equations:

[ \begin{align*} 2x + y &= 6 \ 4x - y &= 8 \ x - y &= 7 \end{align*} ]

To solve this system using Gaussian elimination, we start by writing the equations as the augmented matrix:

[ \begin{bmatrix} 2 & 1 & | & 6 \ 4 & -1 & | & 8 \ 1 & -1 & | & 7 \end{bmatrix} ]

First, we use operation 1 and interchange rows 2 and 3 to get:

[ \begin{bmatrix} 2 & 1 & | & 6 \ 1 & -1 & | & 7 \ 4 & -1 & | & 8 \end{bmatrix} ]

Next, we multiply row 1 by 2 and subtract it from row 2:

[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & -3 & | & -14 \ 4 & -1 & | & 8 \end{bmatrix} ]

Now, divide row 2 by 3:

[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & 1 & | & \frac{-14}{3} \ 4 & -1 & | & 8 \end{bmatrix} ]

Finally, add the second row to the third row:

[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & 1 & | & \frac{-14}{3} \ 0 & 0 & | & \frac{-8}{3} \end{bmatrix} ]

At this point, our matrix is now upper triangular, so we can perform back substitution. From the bottom row, we have (\frac{-8}{3}). Substituting this into the second row, we get:

[ \frac{-14}{3} - \left(\frac{-8}{3}\right)\cdot\frac{1}{4} = \frac{-20}{12} = -\frac{5}{3} ]

Substituting (-\frac{5}{3}) into the top row, we find:

[ \frac{11}{4} - \left(-\frac{5}{3}\right)\cdot\frac{1}{2} = \frac{13}{6} ]

Thus, the solution to the system is (\left(\frac{13}{6}, -\frac{5}{3}, \frac{-8}{3}\right)).

In conclusion, Gaussian elimination provides a powerful tool for solving linear systems. By carefully applying elementary row operations, we can convert our matrix into upper triangular form, allowing us to easily solve for the unknown variables through back substitution.