To solve a system of equations involving three variables (x, y, z), we can utilize matrices and convert them into row echelon form. This process involves organizing the coefficients of the equations into a matrix and performing row operations to achieve a specific format where the leading coefficients (or pivots) are 1, and all entries below these pivots are 0.

Start by constructing a matrix from the coefficients of the equations. For example, if we have the equations:

1. \(2x + 4y + 6z = 24\)

2. \(x + 5y + 12z = 60\)

3. \(3x + 6y + 15z = 20\)

We can represent this as:

\[\begin{bmatrix}2 & 4 & 6 & | & 24 \\1 & 5 & 12 & | & 60 \\3 & 6 & 15 & | & 20\end{bmatrix}\]

The first step is to swap rows if necessary to position a leading 1 in the top left corner. After swapping rows, we can perform row operations to eliminate the entries below the leading 1. For instance, to eliminate the 2 below the leading 1, we can replace row 2 with the result of adding a multiple of row 1 to it. This process continues until we achieve a matrix in row echelon form.

Once in row echelon form, we can back substitute to find the values of x, y, and z. For example, if we reach a form like:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\0 & 1 & 3 & | & 16 \\0 & 0 & 1 & | & -\frac{16}{3}\end{bmatrix}\]

We can interpret this as:

1. \(x + 5y + 12z = 60\)

2. \(y + 3z = 16\)

3. \(z = -\frac{16}{3}\)

Substituting the value of z into the second equation allows us to solve for y:

y + 3(-\frac{16}{3}) = 16

y - 16 = 16

y = 32

Next, substitute the values of y and z back into the first equation to solve for x:

x + 5(32) + 12(-\frac{16}{3}) = 60

x + 160 - 64 = 60

x + 96 = 60

x = -36

Thus, the solution to the system of equations is:

x = -36, y = 32, z = -\frac{16}{3}.

This methodical approach of using matrices and row operations not only simplifies the process of solving systems of equations but also provides a clear pathway to finding the values of multiple variables efficiently.