In solving systems of equations, one effective method involves transforming the system into a matrix and applying row operations to achieve a specific structure known as row echelon form. This form is characterized by having ones along the diagonal and zeros below it, which simplifies the process of solving for the variables.

To illustrate this, consider a system of equations such as:

-x + 2y = 4

x + 7y = 14

First, we convert this system into an augmented matrix, which captures the coefficients of the variables:

\[\begin{bmatrix}-1 & 2 & | & 4 \\1 & 7 & | & 14\end{bmatrix}\]

The goal is to manipulate this matrix using row operations—specifically, swapping rows, multiplying rows by non-zero constants, and adding rows—to achieve the desired row echelon form. The first step is to create a leading one in the top left position. In this case, we can swap the two rows to place the row with a leading one at the top:

\[\begin{bmatrix}1 & 7 & | & 14 \\-1 & 2 & | & 4\end{bmatrix}\]

Next, we focus on eliminating the entry below the leading one. To do this, we add the first row to the second row:

\[\begin{bmatrix}1 & 7 & | & 14 \\0 & 9 & | & 18\end{bmatrix}\]

Now, we need to convert the second row's leading coefficient into a one. This can be achieved by multiplying the entire second row by \(\frac{1}{9}\):

\[\begin{bmatrix}1 & 7 & | & 14 \\0 & 1 & | & 2\end{bmatrix}\]

At this point, we have achieved row echelon form, with ones along the diagonal and zeros below. The next step is to convert this matrix back into a system of equations:

x + 7y = 14

y = 2

From the second equation, we can directly solve for \(y\). Substituting \(y = 2\) back into the first equation allows us to solve for \(x\):

x + 7(2) = 14, which simplifies to x = 0.

This method of solving systems of equations is often referred to as Gaussian elimination. It emphasizes the importance of systematically applying row operations to simplify the matrix, ultimately leading to a straightforward solution through back substitution.