Chapter 7.3, Problem 38E

Solution

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

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

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

Given information:

The given equation is.

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

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.

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

So, its corresponding augmented matrix is:

  $\left[\begin{array}{cccc}1& 0& -1& 2\\ -2& 1& 3& -5\\ 2& 1& -1& 3\end{array}\right]$

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

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

This corresponds to the system of equation.

  $\begin{array}{c}x-z=2\\ y+z=-1\\ 0=0\end{array}$

The last equation is a true statement which tells us that there are infinitely which tells us that there are infinitely many solution since can use any real value for $z$ to obtain its corresponding $x\text{and}y$ values solving for $x\text{and}y$ in term of $z$ the solution of the system is $\left(z+2,-z-1,z\right)$ .