Answer to Question #298031 in Economics of Enterprise for Shahidul

Question

Four families decided to go watch a movie on a Friday evening.The Rahmans (R) have 7 family members, the Khans (K) have 4, Sarkars (S) 5,and the Hasans (H) 2.

They buy a movie-ticket for each member.During the intermission, R goes to the concession-stand and buys 0 large popcorns.K buys 1. S and the H buy 4 and 0 respectively. Once the movie ends, the families go to an ice-cream parlour nearby. R orders 4 scoops of ice-cream and 1 coffees. K orders 3 scoops of ice-cream and 3 coffees. S orders 1 scoops of ice-cream and 1 coffees. H orders 0 scoops of ice-creams and 2 coffees.At the end of the night, the total expenditure of the families are as follows: The Rahmans have spent Tk. 6491; the Khans have spent Tk. 4565; the Sarkars have spent Tk. 4807; and the Hasans have spent Tk. 1706.Using the information above, convert the Friday night expenditures of the four families into a system of linear equations. Convert this system of equations into the matrix form.

What is the determinant of the coefficient matrix?

Answer 1

let the ticket price be donated by T, the price of popcorn denoted by P, the price of ice cream denoted by I and the price of coffee denoted by C. The general system will be as follows;

7T+0P+4I+C=6491

4T+P+3I+3C=4565

5T+4P+I+C=4807

2T+0C+0I+2C=1706

Its matrix system is given by;

$\begin{bmatrix} 7 & 0&4&1\\ 4 & 1&3&3\\ 5&4&1&1\\ 2&0&0&2 \end{bmatrix}$ $\begin{bmatrix} T \\ P\\ I\\ C\\ \end{bmatrix}$ =$\begin{bmatrix} 6491 \\ 4565\\ 4807\\ 1706\\ \end{bmatrix}$

We shall solve this system by diagonalizing the extend matrix, by doing this we shall calculate the determinant of the matrix

calculate the determinant of the coefficient matrix, first note that when we change the position of the first and the last rows of the last matrix obtained and we also change the position of the second and third rows of the same matrix, we will obtain a traditional diagonal matrix, whose determinant is the multiplication of the elements that occupy the main diagonal, that is

Starting from the original coefficient matrix for the last matrix obtained, we multiply the rows by ;

7$\begin{bmatrix} 1&3&3 \\ 4&1&1\\ 0&0&2\\ \end{bmatrix}$ +4 $\begin{bmatrix} 4&1&3 \\ 5&4&1\\ 2&0&2\\ \end{bmatrix}$ -1$\begin{bmatrix} 4&1&3 \\ 5&4&1\\ 2&0&0\\ \end{bmatrix}$ =7$\begin{bmatrix} 1 \\ \end{bmatrix}$ $\begin{bmatrix} 1&1 \\ 0&2\\ \end{bmatrix}$-3$\begin{bmatrix} 4&1 \\ 0&2\\ \end{bmatrix}$ +4$\begin{vmatrix} 4\\ \end{vmatrix}$ $\begin{bmatrix} 4&1 \\ 0&2\\ \end{bmatrix}$ -1$\begin{bmatrix} 5&1 \\ 2&2\\ \end{bmatrix}$ +3$\begin{bmatrix} 5&4\\ 2&0\\ \end{bmatrix}$ -1$\begin{bmatrix} 4 \\ \end{bmatrix}$ $\begin{bmatrix} 4&1 \\ 0&0 \end{bmatrix}$ -1$\begin{bmatrix} 5&1 \\ 2&0\\ \end{bmatrix}$+3$\begin{bmatrix} 5&4 \\ 2&0\\ \end{bmatrix}$ =114

=$\frac{1}{114}$

starting from the original coefficient matrix for the last matrix obtained, we multiply the rows by $\frac{1}{2}$, $\frac{1}{2}$ ,$-\frac{2}{7}$ $-\frac{7}{5}$  and therefore, to find the determinant of the coefficient matrix from the determinant of the last matrix, we must divide the latter by these same fractions:

$\begin{vmatrix} 7&0&4&1\\ 4&1&3&3\\ 5&4&1&1\\ 2&0&0&2\\ \end{vmatrix}$ = 1/$\frac{1}{2}$ .$\frac{1}{2}$ .$-\frac{2}{7}$ .$-\frac{7}{5}$ =25