Chapter 7.3, Problem 40E

Solution

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

  $\begin{array}{c}x+2y-3z=1\\ -3x-5y+8z=-29\end{array}$

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

Given information:

The given equation is.

  $\begin{array}{c}x+2y-3z=1\\ -3x-5y+8z=-29\end{array}$

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.

  $\begin{array}{c}x+2y-3z=1\\ -3x-5y+8z=-29\end{array}$

So, its corresponding augmented matrix is:

  $\left[\begin{array}{cccc}1& 3& -3& 1\\ -3& -5& 8& -29\end{array}\right]$

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

  $\begin{array}{l}\left[\begin{array}{cccc}1& 3& -3& 1\\ -3& -5& 8& -29\end{array}\right]\left(3\right)R_1+R_2\to \left[\begin{array}{cccc}1& 3& -3& 1\\ 0& 1& -1& -26\end{array}\right]\\ \left[\begin{array}{cccc}1& 3& -3& 1\\ 0& 1& -1& -26\end{array}\right]\left(-2\right)R_2+R_1\to \left[\begin{array}{cccc}1& 0& -1& 53\\ 0& 1& -1& -26\end{array}\right]\end{array}$

This corresponds to the system of equation.

  $\begin{array}{c}x-z=53\\ y-z=-26\end{array}$

Having three variables but only two equation 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 $\left(z+53,z-26,z\right)$ .