Chapter 8.1, Problem 78E

To determine

The Gauss-Jordan Elimination method.

Answer to Problem 78E

System of linear equations has No solution.

Explanation of Solution

Given information: use matrices to solve the system of equations (if possible). Use Gauss-Jordan elimination method.

  $\{\begin{array}{cccc}x& -3y& =& 5\\ -2x& +6y& =& -10\end{array}$

Calculation:

Now the system of equations is,

  $\begin{array}{cccc}x& -3y& =& 5\\ -2x& +6y& =& -10\end{array}$

The augmented matrix of above system of equations is,

  $\left[\begin{array}{cccc}1& -3& ⋮& 5\\ -2& 6& ⋮& -10\end{array}\right]$

Now use Gauss-Jordan elimination method.

  $1/2R_2\to R_2$

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

Add row one and row two.

  $R_1+R_2\to R_2$

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

The elements of $R_2$ row are zero; it means the system of linear equations is not consistent

Hence system of linear equations has No solution.