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In linear algebra, an augmented matrix is a powerful tool for solving systems of linear equations. It combines the coefficient matrix with the constant terms from the equations into a single matrix representation, making it easier to perform row operations. The augmented matrix for a system of two variables usually looks like this:

• The left side includes the coefficients of the variables.
• The right side lies after the divider and contains the constant terms.

For example, if you have a system of equations like: \[\begin{align*}1x + 4y & = -2 \0x + 0y & = 0 \\end{align*}\]it's represented in augmented matrix form as: \[\begin{bmatrix} 1 & 4 & | & -2 \ 0 & 0 & | & 0 \\end{bmatrix}\] Understanding this matrix form helps you visualize how row operations mimic elementary equation operations, which is crucial for advancing to the next steps in solutions.

The Row Echelon Form (REF) of a matrix simplifies the process of solving linear equations. In REF, a matrix is systematically organized with crucial properties:

• The first non-zero number in each row, called a pivot, is always to the right of the pivot in the row above.
• All rows of zeroes (if any) are at the bottom of the matrix.
• Below each pivot, elements are zeros.

For the given exercise, the augmented matrix is already in row echelon form:\[\begin{bmatrix}1 & 4 & | & -2 \0 & 0 & | & 0\end{bmatrix}\]Here, the first row has a pivot at the first position (the number 1), and all elements below this are zero, fulfilling the row echelon form criteria. Recognizing this form is crucial as it sets the stage for back-substitution, allowing easier solving of the system of equations.

Back-substitution is a method used to find the solutions to a system of equations once it’s in row echelon form or reduced row echelon form. After a matrix has been manipulated into row echelon form, back-substitution helps solve for variables starting from the last non-zero row upward. Let’s dive into how it works:

• You look at the equation corresponding to the last non-zero row in the matrix.
• Solve this equation for one of the variables; typically, the last variable in the equation.
• Substitute this solution back into the previous equations to find the remaining variables.

In our case, the equation from the first row is \(x + 4y = -2\). Since the second row is all zeroes, it doesn’t affect back-substitution. Simply solve for \(x\) in terms of \(y\):\[x = -2 - 4y\]This means that once you pick a value for \(y\), which is a free variable, you can find \(x\). Back-substitution systematically unravels the variables, leading you to the solution of the problem.

Free variables arise when the system of equations is underspecified, meaning there are infinitely many solutions. These occur in a matrix form in specific situations, such as:

• When a column lacks a pivot position (leading 1), making the variable not bound to a specific equation.
• When rows consist entirely of zeroes, indicating dependency among equations.

In the exercise provided, the variable \(y\) is considered a free variable. The second column doesn't have a pivot, which leaves \(y\) without constraints except through another variable:\[x + 4y = -2\]The lack of explicit information for \(y\) means it can take any value, typically denoted by a parameter, like \(t\), when expressing general solutions:

• \(x = -2 - 4t\)
• \(y = t\)

Here, \(t\) can be any real number \(\mathbb{R}\), describing an infinite solution set. Recognizing free variables helps you understand the dimensions and scope of possible solutions.