91Ó°ÊÓ

When working with an augmented matrix like the one given in the exercise, you aim to transform it into an echelon form. This transformation simplifies the process of solving that system of linear equations. Echelon form is a specific kind of matrix form that provides a clear, systematic approach to finding solutions for the variables involved.

In echelon form, the non-zero rows of the matrix have a ‘staircase’ appearance, where each leading entry (also called a pivot) in a row is located to the right of the leading entry in the row above it. Additionally, any rows consisting entirely of zeros are placed at the bottom of the matrix.

For example, with our matrix:
\[\begin{pmatrix} \frac{1}{3} & -\frac{1}{2} & | & 10 \ 0 & 2 & | & 10 \end{pmatrix}\]
- The first row can be seen as forming part of the "staircase," setting the stage for simplifying each corresponding linear equation in sequence.
- Each leading entry helps to easily target specific variables, which is crucial for simplifying complex systems of equations.

The goal when manipulating matrices into echelon form is to create a system that is easy to solve through processes like substitution.

A system of linear equations is essentially a collection of one or more linear equations involving the same set of variables. In these types of problems, we aim to find values for these variables that satisfy all the given equations simultaneously.

In the exercise, two equations were derived from the augmented matrix. Here is the breakdown:

• The first equation is \(\frac{1}{3}x - \frac{1}{2}y = 10\).
• The second equation is \(2y = 10\).

These equations correspond directly to the rows in the matrix:
- The first row gives us the equation concerning both \(x\) and \(y\).
- The second row conveniently isolates \(y\), making it straightforward to solve initially.

Systems of equations can either be consistent (a solution exists) or inconsistent (no solution exists). By representing the relations among variables, we can graph these equations to visually explore their intersections, representing potential solutions. Graphically, the intersection of the lines of such equations would point to the solutions.

Solving the system of linear equations involves finding values for each variable that satisfies all the given equations. Let's look at how we approached solving for \(x\) and \(y\) in this problem.

Initially, we solved the second equation \(2y = 10\) for \(y\):

• Divide both sides by 2, yielding \(y = 5\).

This simple maneuver gives us our first variable value. By knowing \(y = 5\), we can substitute this value back into the first equation to solve for \(x\):
Substitute \(y = 5\) into the first equation:

• \(\frac{1}{3}x - \frac{1}{2}(5) = 10\)
• Simplify the equation to \(\frac{1}{3}x - \frac{5}{2} = 10\)
• Add \(\frac{5}{2}\) to both sides getting, \(\frac{1}{3}x = \frac{25}{2}\)
• Multiply both sides by 3 leads to \(x = \frac{75}{2}\)

Through these steps, by either the substitution or the elimination method, each variable can be solved incrementally, starting with the easier equation and then tackling the more complex ones, until reaching a complete solution like \((x, y) = \left(\frac{75}{2}, 5\right)\).