Chapter 7.3, Problem 35E

Solution

To identify: The system of equations by finding the reduced row echelon form for the augmented matrix.

  $\begin{array}{c}x+2y-z=3\\ 3x+7y-3z=12\\ -2x-4y+3z=-5\end{array}$

The value of $x,y,z$ is $\left(-2,3,1\right)$ respectively.

Given information:

The given equation is.

  $\begin{array}{c}x+2y-z=3\\ 3x+7y-3z=12\\ -2x-4y+3z=-5\end{array}$

Explanation:

Consider the given information.

To obtain the augmented matrix of a system, write out the coefficients as entries, aligning those with the same variables, and placing 0 for variables that are not present. The last column corresponds to the constants.

  $\begin{array}{c}x+2y-z=3\\ 3x+7y-3z=12\\ -2x-4y+3z=-5\end{array}$

So, its corresponding augmented matrix is:

  $\left[\begin{array}{cccc}1& 2& -1& 3\\ 3& 7& -3& 12\\ -2& -4& 3& -5\end{array}\right]$

Apply elementary row operation to obtain a row echelon from of the matrix.

  $\begin{array}{l}\left[\begin{array}{cccc}1& 2& -1& 3\\ 3& 7& -3& 12\\ -2& -4& 3& -5\end{array}\right]\left(-3\right)R_1+R_2\to \left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 1& 0& 3\\ -2& -4& 3& -5\end{array}\right]\left(2\right)R_1+R_3\to \left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\\ \left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\left(1\right)R_3+R_1\to \left[\begin{array}{cccc}1& 2& 0& 4\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\left(-2\right)R_2+R_1\to \left[\begin{array}{cccc}1& 2& 0& 4\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\\ \left[\begin{array}{cccc}1& 2& 0& 4\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\left(-2\right)R_2+R_1\to \left[\begin{array}{cccc}1& 0& 0& -2\\ 0& 1& 0& 3\\ 0& 0& 1& 1\end{array}\right]\end{array}$

This corresponds to the system of equation.

  $\begin{array}{c}x=-2\\ y=3\\ z=1\end{array}$

So, the value of $x,y,z$ is $\left(-2,3,1\right)$ respectively.