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Understanding linear equations is crucial in solving systems using methods like Gaussian elimination. A linear equation is an equation of the form \( ax + by + cz = d \), where \( a \), \( b \), \( c \), and \( d \) are constants. The variables \( x \), \( y \), and \( z \) can change and are what you're solving for. Linear equations represent straight lines in a coordinate system. When you have multiple linear equations together, it forms a system of linear equations. The solution to these systems is the set of variable values that make all the equations true simultaneously. In Gaussian elimination, our goal is to solve these equations step-by-step by transforming them into a simpler form, making it easier to find the solutions.

In order to solve a system of linear equations efficiently, it's very useful to convert them into an augmented matrix. An augmented matrix contains all the coefficients of the variables, as well as the constants from the equations, in a compact form. For example, from the exercise, the system of equations:\[ \begin{align*} x+y+4z &= -1 \ 2x+y+2z &= 3 \ 3x-6z &= 12 \end{align*} \]is rewritten in matrix form as:\[ \begin{bmatrix} 1 & 1 & 4 & -1 \ 2 & 1 & 2 & 3 \ 3 & 0 & -6 & 12 \end{bmatrix} \].
An augmented matrix simplifies the process of performing operations needed for Gaussian elimination, as it allows us to focus on numbers and computations rather than maintaining the equation format. The vertical line separates the coefficients from the constants on the right side of the equations, helping us keep track of the results from row operations.

Row operations are mathematical actions applied to the rows of an augmented matrix to transform it into a form that makes solving easier. There are three basic types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row from another row.

These operations are crucial in converting an augmented matrix into row-echelon form, where the matrix resembles a staircase. In the exercise solution, we used row operations to eliminate variables below the pivot positions in each column. For instance, subtracting 2 times the first row from the second row helps introduce a zero beneath the first pivot, simplifying further calculations. These operations are vital as they ensure we systematically simplify the matrix while maintaining its equivalence to the original system of equations.

Once the augmented matrix is in row-echelon form, the next step is back substitution. This is when you start solving for the variables beginning from the last row and working upwards. The last row in our solved example showed \(0 = 0\), indicating no new information, and any value for \(z\) will satisfy it. The second row equation, \(y + 6z = -5\), was then solved for \(y\) in terms of \(z\), giving \(y = -5 - 6z\).
Finally, substitute this expression for \(y\) into the first row equation, \(x + y + 4z = -1\), and solve for \(x\) in terms of \(z\), resulting in \(x = 4 - 2z\).
With back substitution, each solution depends on previously found solutions, hence why we start from the bottom. It allows us to express the variables typically in terms of the leading variable in the free column, which in this case was \(z\). This provides the solution for the entire system, explaining it in terms of one variable, \(z\), which can vary freely.