Proving the Relationship Between Augmented Matrices and Reduced Row Echelon Form

  • Thread starter Thread starter BK201
  • Start date Start date
  • Tags Tags
    Matrix Proof
BK201
Messages
4
Reaction score
0
Hi,

i have just started with learning linear algebra ,so please bear with me. It seems like a quite simple question:

Let [A b] an augmented matrix. Prove that if its reduced row echelon form (rref) is [R c] ,then R is the rref of A.
 
Last edited:
Physics news on Phys.org
What is the defintion of "reduce row echelon form"?
 
First ,thanks for replying.

rref is the simplest and most suitable form for Gaussian Elimination obtained by applying elementary row operations.(interchange ,scaling ,or row addition)

Following are requirements :
1.Every nonzero row lies above each zero row.
2.The leading entry of a nonzero row lies in a column to the right of the column containing the leading entry of any preceding row.
3.If a column contains the leading entry of some row ,then all the other entries in that column are zero.
4.The leading entry of each nonzero row is 1.

ex:

1 0 * * 0 0 *
0 1 * * 0 0 *
0 0 0 0 1 0 *
0 0 0 0 0 1 *
0 0 0 0 0 0 0 (* represents arbitrary real number)

A row with no other entry except zero is referred to zero row ,vice versa.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top