91Ó°ÊÓ

An augmented matrix is a simple yet powerful tool in solving systems of linear equations. Think of it as a neat way to transform the equations into a matrix format that is easier to manipulate. In an augmented matrix, you write each equation's coefficients in a row. Then, you place a vertical line that acts as a visual divider before writing the constants from each equation on the right side. For instance, the equations \( x + 2y = 4 \) and \( 3x - y = 5 \) become:

• First row: coefficients of the first equation and constant: 1, 2 | 4
• Second row: coefficients of the second equation and constant: 3, -1 | 5

This forms the augmented matrix:\[\begin{bmatrix}1 & 2 & | & 4 \3 & -1 & | & 5\end{bmatrix}\] By using this format, we can easily keep track of all the information from the original system while working through subsequent steps of Gaussian elimination.

Row operations are the steps we use to simplify an augmented matrix. They help us move closer to a solution by making the equations easier to solve. There are three primary row operations:

• Swap two rows of the matrix.
• Multiply all elements of a row by a non-zero constant.
• Add or subtract a multiple of one row from another row.

These operations are crucial in the Gaussian elimination process. In our exercise, we used row operations to create zeros below the first pivot (which is 1 in the first row). Specifically, we subtracted three times the first row from the second row, turning the matrix into:\[\begin{bmatrix}1 & 2 & | & 4 \0 & -7 & | & -7\end{bmatrix}\] This simplifies the system, making it much easier to solve for the variables at the next stage.

The upper triangular form of a matrix is a key step in Gaussian elimination. At this stage, the matrix has been simplified so each equation below the first has zeros at the beginning, resembling steps going down - called triangular because of this pattern. After applying row operations, our matrix became: \[\begin{bmatrix}1 & 2 & | & 4 \0 & -7 & | & -7\end{bmatrix}\] In this form, it's easy to solve the equations by back-substitution, starting from the bottom equation and moving upwards. Solving \( -7y = -7 \) gives \( y = 1 \), which then allows us to substitute back into the top equation to find \( x = 2 \). Achieving upper triangular form simplifies finding solutions to the system.

A system of equations involves several equations that you need to solve at the same time. Each equation could represent a different constraint or condition in a problem you're trying to solve. Here, each equation is a line on a graph, and a solution represents the point where these lines intersect. In this exercise, the system consisted of:

• \( x + 2y = 4 \)
• \( 3x - y = 5 \)

To solve this system, we sought common values \( x \) and \( y \) that satisfy both equations. The method of Gaussian elimination leverages matrices to do this systematically, revealing the point of intersection \((x, y) = (2, 1)\), where both equations are true. This process is a systematic approach to tackle various systems, whether they involve two, three, or more equations.