Chapter 8, Problem 20RE

To determine

To solve: The system of the linear equations using the matrices if possible and use Gaussian elimination with the back-substitution.

Explanation of Solution

Given:

The linear equations are given as:

  $\{\begin{array}{l}-x+2y=3\\ 2x-4y=-6\end{array}$

The augmented matrix for the given linear equation:

  $\left[\begin{array}{l}-12:3\\ 2-4:-6\end{array}\right]$

Now for using the Gaussian elimination with the back substitution, the above matrix needs to be converted into the row echelon form by the elementary row transformation.

  $\begin{array}{l}\left[\begin{array}{l}-12:3\\ 2-4:-6\end{array}\right]\\ R_1\to -R_1\\ \left[\begin{array}{l}1-2:-3\\ 2-4:-6\end{array}\right]\\ R_2\to R_2-2R_1\\ \left[\begin{array}{l}-1-2:-3\\ 00:0\end{array}\right]\end{array}$

Hence, the system of the equation will be:

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

The second equation will not give any value.

Now, solving the first equation:

  $\begin{array}{l}x-2y=-3\\ x=2y-3\end{array}$

Now, assuming y = k, for some constant k .

Therefore, the solution of system of equation:

  $\begin{array}{l}x=2k-3\\ y=k\end{array}$