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The augmented matrix is a key concept in linear algebra when dealing with systems of linear equations. It is a compact way of writing down a system of equations as a matrix. To form an augmented matrix, align the coefficients of each variable into a matrix and add an extra column for the constants on the right side of the equations. For example, for the system:
\[ \begin{cases} x + y - z = 6 \ 3x - 2y + z = -5 \ x + 3y - 2z = 14 \end{cases} \]
The augmented matrix looks like this:
\[ \begin{bmatrix} 1 & 1 & -1 & \vert & 6 \ 3 & -2 & 1 & \vert & -5 \ 1 & 3 & -2 & \vert & 14 \end{bmatrix} \]
This matrix offers a convenient format for performing row operations, which helps in solving the system.

Row operations are a set of techniques used on matrices to simplify and solve systems of linear equations. They consist of three main types: swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting rows. These operations are aimed at transforming the matrix into row-echelon form or reduced row-echelon form. Here’s how you can use row operations:

• **Swapping Rows:** Change the positions of two rows to position a pivot element.
• **Multiplying Rows:** Multiply all elements of a row by a non-zero scalar to adjust coefficients.
• **Adding/Subtracting Rows:** Add or subtract multiples of one row to another to eliminate variables.

Let's look at an example from our system of equations: To eliminate the first element of the second row, we can use the operation:
Step: \( R2 \to R2 - 3R1 \)
On applying the operation:
\[ \begin{bmatrix} 1 & 1 & -1 & \vert & 6 \ 0 & -5 & 4 & \vert & -23 \ 1 & 3 & -2 & \vert & 14 \end{bmatrix} \]
Notice how the first column in the second row becomes zero. These operations simplify solving the system.

An inconsistent system is a system of linear equations that has no solution. This occurs when the equations contradict each other. In matrix terms, an inconsistent system often leads to a row in row-echelon form that looks like \( [0 \, 0 \, 0 \, | \, c] \) where \( c eq 0 \). This contradicts and implies no solution. For example, in our given problem, after using row operations, we reached:
\[ \begin{bmatrix} 1 & 1 & -1 & \vert & 6 \ 0 & -5 & 4 & \vert & -23 \ 0 & 0 & 0 & \vert & \frac{6}{5} \end{bmatrix} \]
The last row \( [0 \, 0 \, 0 \, | \, \frac{6}{5}] \) provides a contradiction:\( 0 = \frac{6}{5} \). Since this is impossible, the system is inconsistent and has no solution.