Chapter EP, Problem 4.5.16EP

To determine

The solution of the system of equations by using the Cramer’s rule.

Answer to Problem 4.5.16EP

  $(-2.5,3.5)$ .

Explanation of Solution

Given information:

The system of equations:

  $\begin{array}{l}x+y=1\\ 2x-2y=-12\end{array}$

Calculation:

The given system of equations:

  $\begin{array}{l}x+y=1\\ 2x-2y=-12\end{array}$

By using the following steps to find the determinant for the given matrix, using the Cramer’s rule are as shown below:

  $\begin{array}{l}x=\frac{\left|\begin{array}{cc}m& b\\ n& g\end{array}\right|}{\left|C\right|}\text{ , }y=\frac{\left|\begin{array}{cc}a& m\\ f& n\end{array}\right|}{\left|C\right|}\text{ [Use the Cramer's rule]}\\ x=\frac{\left|\begin{array}{cc}1& 1\\ -12& -2\end{array}\right|}{\left|\begin{array}{cc}1& 1\\ 2& -2\end{array}\right|}\text{ , }y=\frac{\left|\begin{array}{cc}1& 1\\ 2& -12\end{array}\right|}{\left|\begin{array}{cc}1& 1\\ 2& -2\end{array}\right|}\text{ [Substitue the values from the given equations]}\\ x\text{=}\frac{(1)(-2)-(1)(-12)}{(1)(-2)-(1)(2)}\text{ , }y=\frac{(1)(-12)-(1)(2)}{(1)(-2)-(1)(2)}\text{ [Write the diagonal of each element]}\\ x=\frac{-2+12}{-2-2}\text{ , }y=\frac{-12-2}{-2-2}\text{ [Multiply]}\end{array}$

  $\begin{array}{l}x=-\frac{10}{4}\text{ , }y=\frac{14}{4}\text{ [Subtract]}\\ x=-2.5\text{ , }y=3.5\end{array}$

Therefore, the solution of the system of equation is $(-2.5,3.5)$ .