Chapter 2.5B, Problem 2GCE

To determine

To find: the augmented matrix and the reduced row echelon form for the given system of equations.

Answer to Problem 2GCE

  $\begin{array}{l}A=\left[\begin{array}{cccc}\begin{array}{l}1\\ 2\\ 4\end{array}& \begin{array}{l}1\\ -3\\ -8\end{array}& \begin{array}{l}1\\ 4\\ 4\end{array}& \begin{array}{l}6\\ 3\\ 12\end{array}\end{array}\right]\\ rref\left(A\right)=\left[\begin{array}{cccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}0\\ 0\\ 1\end{array}& \begin{array}{l}7\\ 1\\ -2\end{array}\end{array}\right]\end{array}$

Here, $A$ is augmented matrix and $reff\left(A\right)$ is reduced echelon form matrix.

Explanation of Solution

Given information:

  $\begin{array}{c}x+y+z-6=0\\ 2x-3y+4z-3=0\\ 4x-8y+4z-12=0\end{array}$

Calculation:

The given system of equations can be rewritten as follows:

  $\begin{array}{c}x+y+z=6\\ 2x-3y+4z=3\\ 4x-8y+4z=12\end{array}$

The augmented matrix for the given system of equations is shown below.

  $A=\left[\begin{array}{cccc}\begin{array}{l}1\\ 2\\ 4\end{array}& \begin{array}{l}1\\ -3\\ -8\end{array}& \begin{array}{l}1\\ 4\\ 4\end{array}& \begin{array}{l}6\\ 3\\ 12\end{array}\end{array}\right]$

Put $rref(A)$ command in the graphing calculator for matrix to get the reduced row echelon form for matrix $A$ . The reduced row echelon form for the given system of equations is as follows:

  $\left[\begin{array}{cccc}\begin{array}{l}1\\ 0\\ 0\end{array}& \begin{array}{l}0\\ 1\\ 0\end{array}& \begin{array}{l}0\\ 0\\ 1\end{array}& \begin{array}{l}7\\ 1\\ -2\end{array}\end{array}\right]$