Each augmented matrix is in row-echelon form. Assume that...

Each augmented matrix is in row-echelon form. Assume that the variables are $x, y,$ and $z$ and use back-substitution to obtain the solution of the associated system of linear equations. $$ \left[\begin{array}{rrr|r} 1 & -\frac{1}{3} & -\frac{1}{3} & \frac{8}{3} \\ 0 & 1 & -\frac{5}{4} & \frac{7}{4} \\ 0 & 0 & 1 & -3 \end{array}\right] $$

Each augmented matrix is in row-echelon form. Assume that the variables are $x, y,$ and $z$ and use back-substitution to obtain the solution of the associated system of linear equations.
$$
\left[\begin{array}{rrr|r}
1 & -\frac{1}{3} & -\frac{1}{3} & \frac{8}{3} \\
0 & 1 & -\frac{5}{4} & \frac{7}{4} \\
0 & 0 & 1 & -3
\end{array}\right]
$$

Key Concepts

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations where each row corresponds to an equation and the columns represent the coefficients of the variables along with the constant terms. This format is particularly useful for applying algorithmic methods such as row-reduction to solve the system.

Row-Echelon Form

Row-echelon form is a stage in the process of solving a system of linear equations where the matrix has been transformed so that all nonzero rows are above rows of zeros, and each leading coefficient (pivot) in a nonzero row is to the right of the leading coefficient of the row above it. This structured form greatly facilitates methods like back-substitution for finding solutions.

System of Linear Equations

A system of linear equations is a collection of equations that share common variables and seeks values for these variables that satisfy all the equations simultaneously. The use of matrices in such systems allows for efficient computational techniques to determine whether unique, infinite, or no solutions exist.

Back-Substitution

Back-substitution is a method used to solve a system of linear equations once the augmented matrix has been transformed into an upper triangular (row-echelon) form. Starting from the bottom row, which contains only one variable, each variable is solved in turn and then substituted back into the equations above to find the remaining variables.

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