# Pivot positions and reduced row echelon form

I have the definition of reduced row echelon form (the relevant part) as The leading entry in each non-zero row is 1and each leading $1$ is the only non $0$ entry in its column.

I then have the definition of a pivot position as a location in a matrix that corresponds to a leading $1$ in the RRE form.

So, Suppose matrix A is $11\times 9$. Is it possible that matrix $A$ has a pivot position in every row?

This is no right? Because by definition you can't have more than one pivot position per column.

So this brings me to the point:

My practice test says:

Is $Ax = b$ consistent for a particular given $b$?

The way I would do this is simply solve the augmented matrix and make sure it doesn't have any impossible equations, such as $0=4$.

However, the way the practice says to do it is to:

To answer this question, we look at if the rightmost column of the augmented matrix $[A\quad b]$ is a pivot column.

But this doesn't make any sense to me, because a pivot column is only going to have 1 non 0 variable in it, and the point of an augmented matrix is to solve the system of equations for a column (vector) of multiple variables.

I hope this makes sense, let me know if I need to clarify anything, thanks.

The pivot column in the hint can refer to a column that has a leading entry. You don't need to transform a matrix $A$ to its reduced row echelon form to see whether it has solutions. A row echelon form is enough.

Even if you transform it to its reduced row echelon form, if the last column is a pivot column, the system has no solution. You can still get all zeros for the other entries except the last one. In this case, the last equation is just $0=1$.

Here is an example for you:

$$\left[ \begin{array}{cc|c} 1&2&3\\ 1&1&0\\ 1&3&1 \end{array} \right]$$

The row echelon form of the above augmented matrix is: $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&-1&-3\\ 0&0&-5 \end{array} \right]$$ You can already see that the last column is a pivot column because it has a leading entry.

If you prefer, you can find its reduced row echelon form: $$\left[ \begin{array}{cc|c} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array} \right]$$

• Hmm, ok. Lets say the question was How to do you find the pivot columns of augmented matrix [A b] Would you reduce this to RRE? In your example, it seems like there is a pivot in every column, but you can't tell that until you get it into RRE, if its just in RE then it looks like there is a pivot in the first 2, but not the last. You can only find pivot positions in RRE form right? – Jane Doe Dec 10 '15 at 16:18
• @BNSlug: You can find the pivot positions in RE form. In my example, the second matrix is in RE form. You can see that $1, -1, -5$ are all leading entries in their own rows. – KittyL Dec 10 '15 at 17:58