Chapter 10.3, Problem 29E

To determine

To find: the system of linear equations is inconsistent or dependent.

Answer to Problem 29E

The system has no solution to the above system.

Explanation of Solution

Given:

  $\{\begin{array}{l}x+y+z=2\\ y-3z=1\\ 2x+y+5z=0\end{array}$

Calculation:

Consider the following system of equations:

  $\begin{array}{l}x+y+z=2\\ y-3z=1\\ 2x+y+5z=0\end{array}$

First translate the system of equations above into an augmented matrix as shown below:

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

Next use elementary row operations to convert the matrix into row echelon form.

To do this, start with the following row operation:

  $\begin{array}{l}\text{Row 3}-2\times \text{Row }1=\text{New Row 3}\\ \left[\begin{array}{cccc}2& 2& 2& 4\\ 0& 1& -3& 1\\ 2& 1& 5& 0\end{array}\right]2\times \text{Row 1}\\ \left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& -3& 1\\ 0& -1& 3& -4\end{array}\right]\text{Row 3}-\text{2}\times \text{Row 1=New Row 3}\end{array}$

The next required row operation is shown below:

  $\begin{array}{l}\text{Row 3+Row 2}=\text{New Row 3}\\ \left[\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& -3& 1\\ 0& 0& 0& -3\end{array}\right]\end{array}$

This last matrix is in row-echelon form, so stop the Gaussian elimination process.

Now, translate the last row back into equation form,

  $0x+0y+0z=-3$ or $0=-3$ , which is false.

No matter what values pick for x, y, and z, the last equation will never be a true statement.

This means that the system has no solution to the above system.

Conclusion:

Therefore, the system has no solution to the above system.