Chapter 8, Problem 28RE

To determine

Find the solution of the system of linear equation using Gaussian elimination with back-substitution method.

Answer to Problem 28RE

There is no solution.

Explanation of Solution

Given information:

The givensystem of linear equationsare

  $\begin{array}{l}x+2y+w=3\\ -3y+3z=0\\ 4x+4y+z+2w=0\\ 2x+z=3\end{array}$

Calculation:

A matrix derived from a system of linear equations is the augmented matrix of the system. Moreover, the matrix derived from the coefficients of the system is the coefficient matrix of the system.

Let the system of linear equations are

  $\begin{array}{l}ax+by+cz=p\\ dx+ey+fz=q\end{array}$

The augmented matrix is

  $\left[\begin{array}{ccccc}a& b& c& :& p\\ d& e& f& :& q\end{array}\right]$

Write an augmented matrix for the given system of linear equations.

  $\left[\begin{array}{cccccc}1& 2& 0& 1& :& 3\\ 0& -3& 3& 0& :& 0\\ 4& 4& 1& 2& :& 0\\ 2& 0& 1& 0& :& 3\end{array}\right]$

Use Gaussian elimination method.

  $\begin{array}{l}{R}_3\leftrightarrow R_3-4R_1\\ \left[\begin{array}{cccccc}1& 2& 0& 1& :& 3\\ 0& -3& 3& 0& :& 0\\ 0& -4& 1& -2& :& -12\\ 2& 0& 1& 0& :& 3\end{array}\right]\\ R_4\leftrightarrow R_4-2R_1\\ \left[\begin{array}{cccccc}1& 2& 0& 1& :& 3\\ 0& -3& 3& 0& :& 0\\ 0& -4& 1& -2& :& -12\\ 0& -4& 1& -2& :& -3\end{array}\right]\\ R_2\leftrightarrow R_2-R_3\\ \left[\begin{array}{cccccc}1& 2& 0& 1& :& 3\\ 0& 1& 2& 2& :& 12\\ 0& -4& 1& -2& :& -12\\ 0& -4& 1& -2& :& -3\end{array}\right]\end{array}$

Now the system of linear equations are

  $\begin{array}{l}x+2y+w=3---\left(i\right)\\ y+2z+2w=12---\left(ii\right)\\ -4y+z-2w=-12---\left(iii\right)\\ -4y+z-2w=-12=-3---\left(iv\right)\end{array}$

In the above result, the equations $\left(iii\right)$ and $\left(iv\right)$ contradict each other.

Hence there is no solution.