New to Linear Algebra: Confused on Reduced Row Echelon Form

In summary, the discussion is about reduced row echelon form in a math course and the confusion about the placement of zeroes in the matrix. The textbook mentions that the leading coefficients should have all zeroes except for the leading coefficient, but there is a matrix in the examples with a non-zero value. The expert explains that the only columns that matter are the ones with leading coefficients and any other values are allowed.
  • #1
sniper_spike
7
0
Hey guys, lurked here a bit, but now I'm in this new math course right. So anyway it seems like a completely new language to me. There's some discussions about reduced row echelon form in the textbook I'm using. I was taught (before going into the course) that the object was to get leading 1s in a diagonal, then 0's in every other spot. The textbook doesn't seem to corroborate this, instead it only puts zeros above the 1s. Then I check wiki and it states

'reduced row echelon form... satisfies the additional condition:

Every leading coefficient is 1 and is the only nonzero entry in its column,[2] as in this example'

but then at the same time it shows a matrix which has nonzeros other than 1 and says its also a reduced row echelon form.

https://en.wikipedia.org/wiki/Row_echelon_form
Could anyone help clear up some of these initial confusions I'm having?
 
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  • #2
Can you post the title and author of your textbook??

Anyway, about the wiki example you mention. You should only look at the leading coefficients. For example, the first row has a leading coefficient in the first column. Every other value in that row should be zero. The second row has a leading coefficient in the second column. Every other value in the second column should be zero. And finally, we have a leading coefficient n the third row at the fourth column. This implies that everything else in the fourth column should be zero. Other columns should necessarily have zeroes.

So in our example, the leading coefficients occur in column 1, 2 and 4. So these are the only columns which should have zeroes. The rest of the columns can be anything.
 
  • #3
What do you mean a leading coefficient in the third row at the fourth column? not third column?

Also its called Elementary Linear Algebra, Applications Version and its by Howard Anton and Chris Rorrers.

In fact it does state these properties of having all zeroes except for the leading coefficient, then in the list of matrices which are examples of reduced row echelon form there is a matrix with a nonzero.

the matrix is

0 1 -2 0 |1
0 0 0 1 |3
0 0 0 0 |0
0 0 0 0 |0

so what's the -2 doing there?

and also a generic form

1 0 * *
0 1 * *
0 0 0 0
0 0 0 0

once again a generic example is

0 1 * 0 0 0 * * 0 *
0 0 0 1 0 0 * * 0 *
0 0 0 0 1 0 * * 0 *
0 0 0 0 0 1 * * 0 *
0 0 0 0 0 0 0 0 1 *

with the *'s being replaced with real numbers

And somewhere I read that you can have a non zero if there is a zero below it. Doesn't make full sense to me.
 
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  • #4
sniper_spike said:
What do you mean a leading coefficient in the third row at the fourth column? not third column?

Also its called Elementary Linear Algebra, Applications Version and its by Howard Anton and Chris Rorrers.

In fact it does state these properties of having all zeroes except for the leading coefficient, then in the list of matrices which are examples of reduced row echelon form there is a matrix with a nonzero.

the matrix is

0 1 -2 0 |1
0 0 0 1 |3
0 0 0 0 |0
0 0 0 0 |0

The only columns we care about are the columns with a leading coefficient. The columns here with a leading coefficient are column 1 and column 4. So those columns should contain all zeroes except for the leading coefficient.
Column 3 contains a -2, but this is not a leading coefficient since there is a 1 in front of it. So column 3 contains no leading coefficients, so nonzero values are allowed.
 
  • #5
Makes sense. thanks a lot!
 

1. What is reduced row echelon form?

Reduced row echelon form (RREF) is a special type of matrix that has been manipulated using elementary row operations to have certain properties. In RREF, the leading coefficient (the first non-zero number) of each row is equal to 1, and the leading coefficient is the only non-zero entry in its column. This form is useful for solving systems of linear equations and finding the basis of a vector space.

2. How do I convert a matrix to reduced row echelon form?

To convert a matrix to RREF, you must use elementary row operations. These include multiplying a row by a non-zero constant, swapping two rows, and adding a multiple of one row to another. Apply these operations until the matrix satisfies the properties of RREF. It is important to note that the final result may not be unique, as there may be multiple ways to get the matrix in RREF.

3. What is the purpose of reduced row echelon form?

RREF is useful for solving systems of linear equations, as it makes it easier to see the solutions. It also helps to determine the rank of a matrix, which is the number of linearly independent rows or columns. Furthermore, RREF can be used to find the basis of a vector space, which is a set of linearly independent vectors that span the entire space.

4. Can any matrix be converted to reduced row echelon form?

Not all matrices can be converted to RREF. For a matrix to be converted, it must be a square matrix (same number of rows and columns) and have a non-zero determinant. If a matrix does not meet these criteria, it cannot be converted to RREF.

5. How can I use reduced row echelon form to solve a system of linear equations?

RREF can be used to solve a system of linear equations by using back substitution. First, convert the system of equations into an augmented matrix (a matrix with the coefficients and constants). Then, convert the augmented matrix to RREF. The solutions can be found by looking at the values of the variables corresponding to the pivot columns (columns with leading coefficients of 1) in the rightmost column of the RREF matrix.

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