Chapter 6.7, Problem 9CYU

To determine

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

Answer to Problem 9CYU

Hence the solution to the system of equation is $(-1,6)$ .

Explanation of Solution

Given:

  $\begin{array}{l}3x-4y=-27\\ x+2y=11\end{array}$

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

  $\begin{array}{l}3x-4y=-27\\ x+2y=11\end{array}$

  $\to \left[\begin{array}{ccc}3& -4& -27\\ 1& 2& 11\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{ccc}3& -4& -27\\ 1& 2& 11\end{array}\right]$

To make the first element in the first row a 1, multiply first row by $\frac{1}{3}$

  $\left[\begin{array}{ccc}3& -4& -27\\ 1& 2& 11\end{array}\right]$

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 1& 2& 11\end{array}\right]$

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

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 1& 2& 11\end{array}\right]$

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 0& \frac{10}{3}& 20\end{array}\right]$

To make the second element in the second row a 1 , multiply second row by $\frac{3}{10}$

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 0& \frac{10}{3}& 20\end{array}\right]$

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 0& 1& 6\end{array}\right]$

To make the second element in the first row a 0 , multiply second row by $\frac{4}{3}$ and add the result to the first row.

  $\left[\begin{array}{ccc}1& \frac{-4}{3}& -9\\ 0& 1& 6\end{array}\right]$

  $\left[\begin{array}{lll}1\hfill & 0\hfill & -1\hfill \\ 0\hfill & 1\hfill & 6\hfill \end{array}\right]$

Hence the solution to the system of equation is $(-1,6)$

Conclusion:

Hence the solution to the system of equation is $(-1,6)$ .