Chapter 8.5, Problem 7E

To determine

To solve:

The determinants of the value by cramer’s rule

Answer to Problem 7E

The solution is $x=-3\text{and}-2$

Explanation of Solution

Given:

  $\begin{array}{l}-7x+11y=-1\\ 3x-9y=9\end{array}$

Concept used:

The system is $\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$

Has a solution $x=\frac{D_x}{D},y=\frac{D_y}{D}$

Where

  $D_x=\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|,D_y=\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|$

Calculation:

The given system is

  $\begin{array}{l}-7x+11y=-1\\ 3x-9y=9\end{array}$

The determinant of the coefficient matrix

  $\begin{array}{l}D=\left|\begin{array}{cc}-7& 11\\ 3& -9\end{array}\right|\\ D=63-33\\ D=30\end{array}$

This determinant is not zero

So apply the cramer’s rule

  $x=\frac{D_x}{D}=\frac{\left|\begin{array}{cc}-1& 11\\ 9& -9\end{array}\right|}{30}=\frac{9-99}{30}=\frac{-90}{30}=-3$

  $y=\frac{D_y}{D}=\frac{\left|\begin{array}{cc}-7& -1\\ 3& 9\end{array}\right|}{30}=\frac{-63+3}{-96}=\frac{-60}{30}=-2$

The solution is $x=-3\text{and}-2$