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Pivot Variables in Reduced Row Echelon Form 
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#1
Jul212, 01:36 AM

P: 101

Why do we solve for only the pivot variables when we are trying to solve a system of equations using reduced row echelon form?
Thank you 


#2
Jul212, 01:54 AM

P: 4,579

The simple explanation is that the pivot variables end up telling us what is independent and what is not. When we get a pivot variable, we are essentially finding a column that has independence from the other columns in the matrix. If this happens in a square matrix and there is a pivot in every column, then we know that the matrix is invertible and for a system Ax = b, this means that x has a unique solution. If this is not the case (pivot in every column) it means that there are dependent variables somewhere. When this is the case for a square matrix, it means that there are infinitely many solutions for a particular Ax=b, or possibly even no solutions at all for an inconsistent system depending on what the b vector is. If we however take a general matrix and reduce it and find that it has a pivot count less than or equal to the number of rows, then it means that the vectors are linearly dependent and the basis for the space can be written in terms of a lesser number of vectors. This is important when we want to figure out the dimension of a system, and it's used in many ways including in RankNullity, and describing how to actually construct a proper basis from a set of vectors (and find the real dimension of the space). 


#3
Jul212, 10:15 AM

P: 101




#4
Jul312, 12:06 AM

P: 4,579

Pivot Variables in Reduced Row Echelon Form
Now in terms of the pivot variables, when we have a situation where there are less pivots than columns, we get the situation where at least for that system of equations, we have a situation where the other variables are linearly dependent on the other variables. As an example, lets look at a few simple systems in REF: [1 2 3 4] [0] [2 4 6 8] [0] which reduces to [1 2 3 4] [0] [0 0 0 0] [0] This translates into x + 2y + 3z + 4w = 0. This means that we pick one independent scalar variable and three dependent scalar variables. For this system, everything is linearly dependent on the vector (1,2,3,4) or any multiple thereof. Now lets look at the next example: [1 2 3 4] [0] [2 4 6 7] [0] which reduces to [1 2 3 0] [0] [0 0 0 1] [0] This means we have two linearly independent vectors (1,2,3,0) and (0,0,0,1) with two pivots, but in terms of the linearly dependent variables inside the system we have for our first vector one independent scalar variable and two dependent variables (since x + 2y + 3z = 0, pick one independent, the others are dependent). So in the first one we had one pivot with only one independent vector, the second had two pivots with two independent vectors, and the pattern does continue from there. If we have situations where we get pivots in every column (even if there are more rows than columns) we should have a full set of independent vectors equal in number to the number of columns and in a square matrix, this is a basis. The thing is that in the rows > columns scenario, you need to take into account inconsistent systems because in general, they will occur if you are not careful. 


#5
Jul312, 02:47 AM

P: 101




#6
Jul312, 02:53 AM

P: 4,579

The main point is to understand how a system with a certain number of pivot points with respect to the number of columns in the matrix affects the number of independent and dependent variables as well as subsequently the number of independent vectors, which is probably the more important aspect. Knowing these gives you the reduction of the entire system, and it tells you indirectly (if you wish to know) how you can take a vector if it's completely zeroed out and represent it in terms of a linear combination of the other vectors that aren't zeroed out. The ones that get zeroed out (all zeroes) are vectors that are a linear combination of the ones that haven't been zeroed out. If you only get one nonzero vector it means that everything is just a multiple of that vector. So for solving systems, if you have linearly dependent vectors in your system, you can get the independent ones and then if your system is still consistent, figure out how to represent by a parameter the infinite number of solutions because geometrically what is happening is that instead of everything meeting at a point (when you get a unique solution), something is parallel somewhere meaning that you get the infinite number of solutions. 


#7
Jul312, 03:43 AM

P: 101




#8
Jul312, 04:35 AM

P: 4,579

I didn't mean about the vectors themselves in the RREF which are either independent or dependent. Again sorry for the confusion. 


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