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An augmented matrix is a compact way to represent a system of linear equations, which combines the coefficients of the variables and the constants from the equations into one matrix. In the context of solving linear systems, the left side of the matrix contains the coefficients that accompany each variable. The right side, separated by a vertical bar, represents the constants from the right-hand side of the equations.

To visualize this, consider the linear system that corresponds to the given augmented matrix from the exercise:

• The first row represents the equation for variable x, leading to the equation \(x - 2z = 7\).
• The second row corresponds to the variable y which translates to \(y + 4z = 3\).
• The third row, having all zeros before the vertical bar and also after the bar, is the identity \(0 = 0\), which is always true and has no variables.

This arrangement is efficient for both manually solving the system and programming computational methods, such as Gaussian elimination or matrix operations.

A consistent system of equations has at least one set of values for the variables that satisfies all equations in the system. In contrast, an inconsistent system has no solution; there are no values that simultaneously satisfy all equations.

To determine consistency, we look at the augmented matrix or the equivalent equations. If the matrix form does not create a row where there are non-zero coefficients leading to an impossible statement (like \(0=1\)), it indicates that the system may have either one unique solution or infinitely many solutions. For this exercise, as the last row indicates an identity \(0 = 0\), we conclude that the system is consistent. Furthermore, the presence of this row also suggests that not all variables are bound by unique constraints, hinting at the possibility of infinite solutions.

When a system of equations has more variables than independent equations, it often results in infinite solutions. This means there are countless combinations of variable values that satisfy the system. In the given matrix, we are left with two independent equations for three variables (x, y, and z), leading to one variable that can be freely chosen.

To express infinite solutions, one usually selects a variable (often the one with the least constraints) as a 'free variable', which can take any value. The other variables are then expressed in terms of this free variable. In our exercise, z was chosen as the free variable. This choice results in the solution set being written in terms of z:

• \(x = 7 + 2z\)
• \(y = 3 - 4z\)

So for any value of z, there are corresponding values of x and y that solve the system, hence the term 'infinite solutions'.

Parametric equations express the solutions of a system in terms of one or more parameters, which in the case of linear systems, are usually the free variables. Using parametric equations allows us to describe an infinite set of solutions in a concise way.

In the system given, z is the free variable and constitutes the parameter for the parametric equations. The solutions for x and y are written as equations involving z:

• \(x = 7 + 2z\)
• \(y = 3 - 4z\)

These parametric equations provide a clear description of the relationship between the variables and demonstrate how each variable depends on the parameter, z, for our exercise. It is a particularly effective approach when dealing with systems that have infinite solutions.