Matrix word aplication problem

[wat2000]

a small school has 100 students who occupy three classrooms. A,B, and C. After the first period of the school day, half the students in room A move to room B, one-fifth of the students in room B move to room C, and one-third of the students in room C move to room A. Nevertheless, the total number of students in each room is the same for both periods. How many students occupy each room. I know the answer is 50 20 and 30 but I can't figure out how to put in a 4x4 matrix to solve it.

 

[HallsofIvy]

Let A, B, and C also represent the number of students in those class rooms respectively. We are told that 1/2 of the students in room A move to room B which, since there is no mention of students moving from room A to room B, means that 1/2 remain in room A. We are told that 1/5 of the students in room B move to room C which, since there is no mention of student moving from room B to room A, means that 4/5 of the student remain in room B. Finally, we are told that 1/3 of the students in room C move to room A which, since nothing is said of students moving from room C to room B, means 2/3 of them remain in room C.

Now, after the move, the number of students in room A is the (1/2)A remaining plus the number who moved from room C, (1/3)C. That is the total number of students in room A is now (1/2)A+ (1/3)C and we are told that this is the same as in period 1: (1/2)A+ (1/3)C= A.

The number of students in room B is the (1/2)A coming from room A and the (4/5)B who remained. The total number of students is (1/2)A+ (4/5)B and we are told that is the same number as in period 1: (1/2)A+ (4/5)B= B.

The number of students in room C is the (1/5)B coming from room B and the (2/3)C who remained. The total number of students is (1/5)B+ (2/3)C and were told that this is the same number as in period 1: (1/5)B+ (2/3)C= C

So you need to solve the three equations (1/2)A+ (1/3)C= A which is the same as (-1/2)A+ (1/3)C= 0 (or (1/2)A= (1/3)C); (1/2)A+ (4/5)B= B which is the same as (1/2)A- (1/5)B= 0 (or (1/2)A= (1/5)B); and (1/5)B+ (2/3)C= C which is the same as (1/5)B- 1/3)C= 0 (or (1/5)B= (1/3)C).

The fourth equation you have is [itex]A+ B+ C+ D= 50[/tex]. You would put that into a 4 by 4 matrix by writing the "augmented coefficient matrix" with each equation giving a row, the coefficients of A, B, and C being the first 3 columns and the "right hand side" of the equations being the fourth column.

Personally I think it would be simplest to solve the three equations directly without changing to matrix form.