# Using Gauss Jordan Elimination for a problem

• Sep 30th 2011, 05:24 PM
Using Gauss Jordan Elimination for a problem

Three people play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an amoutn equal to what the winner already has. After threegames, each has lost just once and each has \$24. With how much money did each begin?

I am supposed to find the answer using Gauss Jordan Elimination and an augmented matrix, but I'm having difficulty as to how to do it. At the moment I think the matrix will be a 4x4 matrix with each entry in the last column being 24, each column representing a game and each row representing each person. But other than that I'm nto sure how to do the problem.
• Sep 30th 2011, 10:35 PM
CaptainBlack
Re: Using Gauss Jordan Elimination for a problem
Quote:

Three people play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an amoutn equal to what the winner already has. After threegames, each has lost just once and each has \$24. With how much money did each begin?

I am supposed to find the answer using Gauss Jordan Elimination and an augmented matrix, but I'm having difficulty as to how to do it. At the moment I think the matrix will be a 4x4 matrix with each entry in the last column being 24, each column representing a game and each row representing each person. But other than that I'm nto sure how to do the problem.

Please post what you have done (particularly the set of linear equations you will have found to represent the problem)

CB
• Oct 1st 2011, 09:17 AM
Re: Using Gauss Jordan Elimination for a problem
At the time of that post that was all I had figured out. But now here's what I've got:

Each person's money will be either x, y, or z.

In the first round:
Player One: x-y-z
Player Two: 2y
Player Three: 2z

In the second round:
Player One: 2(x-y-z)
Player Two: 2y - (x-y-z) - 2z
Player Three: 4z

In the third round:
Player One: 4(x-y-z)
Player Two: 2(2y - (x-y-z) - 2z)
Player Three: 4z - (2y - (x-y-z) - 2z) - (2(x-y-z))

So my equations are:
4x-4y-4z = 24
-2x + 6y +2z = 24
-x-y+7z = 24

So my matrix is:

4 -4 -4 24
-2 6 2 24
-1 -1 7 24

I then used the Gauss Jordan elimination, but my answers ended up beign negative and not making sense. I don't have much time right now, but I'll try using different row operations to see if it works, but maybe my equation system is wrong.
• Oct 1st 2011, 10:03 AM
CaptainBlack
Re: Using Gauss Jordan Elimination for a problem
Quote:

At the time of that post that was all I had figured out. But now here's what I've got:

Each person's money will be either x, y, or z.

In the first round:
Player One: x-y-z
Player Two: 2y
Player Three: 2z

In the second round:
Player One: 2(x-y-z)
Player Two: 2y - (x-y-z) - 2z
Player Three: 4z

In the third round:
Player One: 4(x-y-z)
Player Two: 2(2y - (x-y-z) - 2z)
Player Three: 4z - (2y - (x-y-z) - 2z) - (2(x-y-z))

So my equations are:
4x-4y-4z = 24
-2x + 6y +2z = 24
-x-y+7z = 24

So my matrix is:

4 -4 -4 24
-2 6 2 24
-1 -1 7 24

I then used the Gauss Jordan elimination, but my answers ended up beign negative and not making sense. I don't have much time right now, but I'll try using different row operations to see if it works, but maybe my equation system is wrong.

CB
• Oct 1st 2011, 07:19 PM