Chapter 8, Problem 16PS

To determine

To find: the cost of a transformer, the cost per foot of wire, and the cost of a light

Answer to Problem 16PS

Cost of a transformer: $10

Cost per foot of wire: $0.2

Cost of a light: $1

Explanation of Solution

Let the cost of a transformer be $ x , the cost per foot of wire be $ y , and the cost of a light be $ z .

The given data can be represented as a system of linear equations as shown below:

  $\{\begin{array}{c}x+25y+5z=20\\ x+50y+15z=35\\ x+100y+20z=50\end{array}$

First find the coefficient matrix of this system,

  $A=\left[\begin{array}{ccc}1& 25& 5\\ 1& 50& 15\\ 1& 100& 20\end{array}\right]$

Determinant of the coefficient matrix A is:

  $\begin{array}{c}D=\left|\begin{array}{ccc}1& 25& 5\\ 1& 50& 15\\ 1& 100& 20\end{array}\right|\\ =1\left|\begin{array}{cc}50& 15\\ 100& 20\end{array}\right|-1\left|\begin{array}{cc}25& 5\\ 100& 20\end{array}\right|+1\left|\begin{array}{cc}25& 5\\ 50& 15\end{array}\right|\\ =1\left(1000-1500\right)-1\left(500-500\right)+1\left(375-250\right)\end{array}$

  $\begin{array}{c}=-500-0+125\\ =-375\end{array}$

Since $D\ne 0$ , Cramer’s rule can be applied.

Consider,

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}20& 25& 5\\ 35& 50& 15\\ 50& 100& 20\end{array}\right|\\ =20\left|\begin{array}{cc}50& 15\\ 100& 20\end{array}\right|-35\left|\begin{array}{cc}25& 5\\ 100& 20\end{array}\right|+50\left|\begin{array}{cc}25& 5\\ 50& 15\end{array}\right|\\ =20\left(1000-1500\right)-35\left(500-500\right)+50\left(375-250\right)\end{array}$

  $\begin{array}{c}=-10000-0+6250\\ =-3750\end{array}$

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}1& 20& 5\\ 1& 35& 15\\ 1& 50& 20\end{array}\right|\\ =1\left|\begin{array}{cc}35& 15\\ 50& 20\end{array}\right|-1\left|\begin{array}{cc}20& 5\\ 50& 20\end{array}\right|+1\left|\begin{array}{cc}20& 5\\ 35& 15\end{array}\right|\\ =1\left(700-750\right)-1\left(400-250\right)+1\left(300-175\right)\end{array}$

  $\begin{array}{c}=-50-150+125\\ =-75\end{array}$

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}1& 25& 20\\ 1& 50& 35\\ 1& 100& 50\end{array}\right|\\ =1\left|\begin{array}{cc}50& 35\\ 100& 50\end{array}\right|-1\left|\begin{array}{cc}25& 20\\ 100& 50\end{array}\right|+1\left|\begin{array}{cc}25& 20\\ 50& 35\end{array}\right|\\ =1\left(2500-3500\right)-1\left(1250-2000\right)+1\left(875-1000\right)\end{array}$

  $\begin{array}{c}=-1000+750-125\\ =-375\end{array}$

Now determine x ,

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{-3750}{-375}\\ =10\end{array}$

Now determine y ,

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{-75}{-375}\\ =0.2\end{array}$

Now determine z ,

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{-375}{-375}\\ =1\end{array}$

So, the cost of a transformer, the cost per foot of wire, and the cost of a light are:

Cost of a transformer: $10

Cost per foot of wire: $0.2

Cost of a light: $1