Chapter 6.7, Problem 8CYU

To determine

To use: the augmented matrix to solve the given system.

Answer to Problem 8CYU

Hence the solution to the system of equation is $(2,4)$ .

Explanation of Solution

Given:

  $\begin{array}{l}x-y=-2\hfill \\ 2x+2y=12\hfill \end{array}$

Calculation:

Objective: To use the augmented matrix to solve the given system.

Place the coefficients of the equations and the constant term into a matrix

  $\begin{array}{l}x-y=-2\hfill \\ 2x+2y=12\hfill \end{array}\to \left[\begin{array}{ccc}1& -1& -2\\ 2& 2& 12\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{ccc}1& -1& -2\\ 2& 2& 12\end{array}\right]$

To make the first element in the second row a 0 , multiply first row by -2 and add the result to the second row

  $\left[\begin{array}{ccc}1& -1& -2\\ 2& 2& 12\end{array}\right]$

  $\left[\begin{array}{ccc}1& -1& -2\\ 0& 4& 16\end{array}\right]$

To make the second element in the second row a 1 , multiply second row by $\frac{1}{4}$

  $\left[\begin{array}{ccc}1& -1& -2\\ 0& 4& 16\end{array}\right]$

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

To make the second element in the first row a 0, multiply second row by 1 and add the result to the first row

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

  $\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 4\end{array}\right]$

Hence the solution to the system of equation is $(2,4)$ .

Conclusion:

Hence the solution to the system of equation is $(2,4)$ .