Chapter 6.7, Problem 10CYU

To determine

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

Answer to Problem 10CYU

Hence the solution to the system of equation is $(-2,-1)$ .

Explanation of Solution

Given:

  $\begin{array}{l}x+4y=-6\\ 2x-5y=1\end{array}$

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

  $\begin{array}{l}x+4y=-6\\ 2x-5y=1\end{array}$

  $\to \left[\begin{array}{ccc}1& 4& -6\\ 2& -5& 1\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{ccc}1& 4& -6\\ 2& -5& 1\end{array}\right]$

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

  $\left[\begin{array}{ccc}1& 4& -6\\ 2& -5& 1\end{array}\right]$

  $\left[\begin{array}{ccc}1& 4& -6\\ 0& -13& 13\end{array}\right]$

To make the second element in the second row a 1 , multiply second row by $\frac{-1}{13}$

  $\left[\begin{array}{ccc}1& 4& -6\\ 0& -13& 13\end{array}\right]$

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

To make the second element in the first row a 0 , multiply second row by

  $-4$ and add the result to the first row.

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

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

Hence the solution to the system of equation is $(-2,-1)$

Conclusion:

Hence the solution to the system of equation is $(-2,-1)$ .