Chapter 6.8, Problem 51SR

To determine

each system by using matrices.

Answer to Problem 51SR

The solution is $x=-2,y=8$

Explanation of Solution

Given:

  $\begin{array}{l}-3x-6y=-42\\ x+4y=30\end{array}$

Calculation:

Here the equations are:

  $\begin{array}{l}-3x-6y=-42\\ x+4y=30\end{array}$

The strategy is to solve the system by using matrices.

Here, the augmented matrices are:

  $\left[\begin{array}{cc}-3& -6\\ 1& 4\end{array}\right]\left[\begin{array}{c}-42\\ 30\end{array}\right]$

Now the matrix in row reduced form will be:

  $\begin{array}{l}\left[\begin{array}{cc}-3& -6\\ 1& 4\end{array}\right]\left[\begin{array}{c}-42\\ 30\end{array}\right]\\ \stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{cc}1& 4\\ -3& -6\end{array}\right]\left[\begin{array}{c}30\\ -42\end{array}\right]\\ \stackrel{R_2+3R_1}{\to }\left[\begin{array}{cc}1& 4\\ 0& 6\end{array}\right]\left[\begin{array}{c}30\\ 48\end{array}\right]\\ \stackrel{\frac{1}{2}{R}_2}{\to }\left[\begin{array}{cc}1& 4\text{ |}\\ 0& 1\text{ |}\end{array}\begin{array}{c}30\\ 8\end{array}\right]\\ \stackrel{R_1-4R_2}{\to }\left[\begin{array}{cc}1& 0\text{ |}\\ 0& 1\text{ |}\end{array}\begin{array}{c}-2\\ 8\end{array}\right]\end{array}$

Hence, the solution is $x=-2,y=8$

Conclusion:

Hence, the solution is $x=-2,y=8$