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An augmented matrix is a helpful tool for solving systems of linear equations. It combines the coefficients of each variable in the equations and the constants from the right-hand side into a single rectangular array. For example, consider the system of equations:

• \( x + y + z = 4 \)
• \( x + 3y + 3z = 10 \)
• \( 2x + y - z = 3 \)

This system can be transformed into the following augmented matrix:\[\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 1 & 3 & 3 & | & 10 \ 2 & 1 & -1 & | & 3 \end{bmatrix}\]The vertical bar separates the coefficients of the variables from the constant terms. Essentially, this matrix represents the entire system's information in a compact form. With the augmented matrix, different methods like Gauss-Jordan elimination can be efficiently applied to solve the system.

Back substitution is a method used to solve linear equations after transforming the system into an upper triangular matrix. In simpler terms, it's a way to find the unknowns in a step-by-step manner, starting from the last equation and moving upwards.Once the system is in the form:

• \( x + y + z = 4 \)
• \( 0 + y + z = 3 \)
• \( 0 + 0 + z = 1 \)

You can directly solve for the third variable first, which is \( z = 1 \). Moving upwards, substitute \( z = 1 \) into the second equation to find \( y = 2 \). Finally, use \( y = 2 \) and \( z = 1 \) in the first equation to determine \( x = 1 \).This sequential solving process ensures that all solutions are accurate and satisfy the original system of equations.

Gauss-Jordan elimination is an efficient method used to solve systems of linear equations by transforming the augmented matrix into a reduced row echelon form. The ultimate goal is to simplify the matrix so each row leads to a straightforward solution of the variables.This is accomplished through a series of operations:

• Swapping rows if needed to get a non-zero leading coefficient.
• Making the leading coefficient in each row 1, often by dividing the entire row by the value of the leading coefficient.
• Eliminating all other numbers in the column of the leading 1 by adding or subtracting multiples of other rows.

In the original problem, the steps showed successive elimination of coefficients below leading 1s, leading to a matrix form where each variable neatly lines up with only one non-zero coefficient:\[\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 1 & 1 & | & 3 \ 0 & 0 & 1 & | & 1 \end{bmatrix}\]With this form, back substitution can be easily applied to find the solution for \( x, y, \) and \( z \), confirming the system of equations as consistent and solvable.