Chapter 6.7, Problem 40PPS

To determine

To solve: The given system of equation

Answer to Problem 40PPS

Here, the solution to the system of equation is $\left(\frac{3}{4},\frac{-5}{2}\right)$ .

Explanation of Solution

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

Calculation:

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

Place the coefficient of the equation and the constant term into a matrix

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

  $\left[\begin{array}{l}2\text{ 1 | }-\text{1}\\ -\text{2 1 | }-\text{4}\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{l}2\text{ 1 | }-\text{1}\\ -\text{2 1 | }-\text{4}\end{array}\right]$

To make the first element in the first row a $1$ , multiply first row by $\frac{1}{2}$

  $\left[\begin{array}{l}2\text{ 1 | }-\text{1}\\ -\text{2 1 | }-\text{4}\end{array}\right]$

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ -2\text{ 1 | }-\text{4}\end{array}\right]$

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

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ -2\text{ 1 | }-\text{4}\end{array}\right]$

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ \text{0 2 | }-\text{5}\end{array}\right]$

To make the second element in the second row as $1$ , multiply second row by $\frac{1}{2}$

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ \text{0 2 | }-\text{5}\end{array}\right]$

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ \text{0 1 }\text{| }\frac{-5}{2}\end{array}\right]$

To make the second element in the first row as $0$ , multiply second row by $\frac{-1}{2}$ and add the result to the first row

  $\left[\begin{array}{l}1\frac{1}{2}\text{ | }\frac{-1}{2}\\ \text{0 1 }\text{| }\frac{-5}{2}\end{array}\right]$

  $\left[\begin{array}{l}1\text{ 0 | }\frac{3}{4}\\ \text{0 1 | }\frac{-5}{2}\end{array}\right]$

Hence the solution to the system of equation is $\left(\frac{3}{4},\frac{-5}{2}\right)$ .

Conclusion:

Therefore, the solution to the system of equation is $\left(\frac{3}{4},\frac{-5}{2}\right)$ .