Proving the Efficiency of the Elimination Method in Solving Systems of Linear Equations

The elimination method is a powerful technique in linear algebra, used to solve systems of linear equations. This method is efficient, straightforward, and relies on the basic properties of equality and linear combinations. Here, we delve into a detailed exploration of this method, providing a step-by-step guide and examples to demonstrate its application.

Understanding the System of Equations

A system of linear equations can be represented in the general form:

[a_1x b_1y c_1qquad 1]

[a_2x b_2y c_2qquad 2]

Here, a_1, b_1, c_1, a_2, b_2,text{ and }c_2, are constants. The elimination method can be applied to solve such a system systematically.

The Elimination Method

Let's break down the process involved in the elimination method:

Step 1: Multiply the Equations

If necessary, multiply one or both equations by constants to make the coefficients of one of the variables, say y,, the same or opposites. For example, to eliminate y,, we might multiply equation 1 by b_2, and equation 2 by b_1, as follows:

[b_2a_1x b_1b_2y b_2c_1qquad 3]

[b_1a_2x b_1b_2y b_1c_2qquad 4]

Step 2: Subtract the Equations

Next, subtract one equation from the other to eliminate y,:

[b_2a_1x - b_1a_2x b_2c_1 - b_1c_2]

This simplifies to:

[(b_2a_1 - b_1a_2)x b_2c_1 - b_1c_2]

Step 3: Solve for the Remaining Variable

Solve for x, by dividing both sides by the coefficient of x, (assuming it's non-zero):

[x frac{b_2c_1 - b_1c_2}{b_2a_1 - b_1a_2}]

Step 4: Back Substitution

Once you have x,, substitute it back into either of the original equations to find y,.

Step 5: Conclusion

The elimination method is valid as long as the system is consistent and the equations are linear. It relies on the properties of equality and the linearity of equations, allowing us to systematically eliminate variables and ultimately find a solution for the system.

Example: Applying the Elimination Method

Consider the system:

[2x 3y 6quad 1]

[4x - 3y 12quad 2]

To eliminate y,, multiply equation 1 by 3:

[6x 9y 18quad 3]

Adding equation 2 to 3:

[6x - 3y 4x - 3y 18 12]

[1 30]

Solving for x, gives:

[x 3]

Substitute x 3, back into equation 1:

[2(3) 3y 6]

[6 3y 6]

[3y 0]

[y 0]

Thus, the solution to the system is:

[x 3 text{ and } y 0]

Conclusion

The elimination method is a reliable technique for solving systems of linear equations. By systematically eliminating variables, it guarantees that if a solution exists, it can be found. Understanding the steps involved and applying them correctly ensures the accuracy and efficiency of this method.