# Help putting system of equations into Reduced Row Echelon Form?

• August 31st 2011, 05:38 AM
csgirl504
Help putting system of equations into Reduced Row Echelon Form?
I am having a lot of trouble solving systems of equations by putting them in RREF.

I don't understand how you know what to do after each step. I keep working the same problem and I'm not getting anywhere.

The system is :

2x + y + 3z = 0
y + 3z = 5
x + 3/2y + 9/2z = 5

which would make my augmented matrix:

2 1 3 :0
0 1 3 :5
1 3/2 9/2 :5

I've started it in so many ways, but then I go through steps and can't get anywhere.
Can someone please demonstrate a step-by-step way of getting this in RREF and if possible give some tips that explain the best way to do this? Thanks
• August 31st 2011, 07:20 AM
bondesan
Re: Help putting system of equations into Reduced Row Echelon Form?
Simply RREF means that you must find a way to transform the coefficients matrix into the identity matrix of the same order (if your system is of order 3x3, then you must end up with an 3x3 identity). There are cases in which you cannot fully transform the coefficients matrix into the identity matrix, but you can get close. Try reading something about elementary operations.

First of all, you must start with 1 in the first entry ( $a_{11}$). But why? Because you want to cancel the numbers below the first entry, that is, turn all the other numbers of the same column into zeros (and having 1 in the first entry makes it more simple to do).

In your case, you can either divide the first line by 2 or you can exchange the third line with the first.

Suppose that you exchange lines for preventing the division. Now you have

1 3/2 9/2 :5
0 1 3 :5
2 1 3 :0

As you want all zeros, you must eliminate the first entry in the third line. You multiply the first line by 2 and subtract it with the third line (but you mantain the first line). So:

1 3/2 9/2 :5
0 1 3 :5
0 -2 -6 :-10

Now you have to make zeros appear in the second column, but leaving 1 in the second entry of the second row $(a_{22})$, and then the same thing with the third column (leaving $(a_{33}=1)$. Can you do this by yourself now?
• August 31st 2011, 07:30 AM
csgirl504
Re: Help putting system of equations into Reduced Row Echelon Form?
Thank you for the explanation! I divided the first row by 2 and then subtracted the first row from the third, and then second from third. So I now have:

1 1/2 3/2 0
0 1 3 5
0 0 0 0

I don't think I can reduce it any further, so how will I know what the solutions are? Am I supposed to stick a letter in there as a parameter or something?
• August 31st 2011, 09:51 AM
bondesan
Re: Help putting system of equations into Reduced Row Echelon Form?
You still can reduce the first row, dividing the second row by 2 and subtracting the first row by it. You will end up with two equations and three variables, which means you're dealing with a system with infinite solutions. In this case, you will have a family of solutions that have the same structure (like something as (constant,constant,variable)). Can you understand this?