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.
Best Answer
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: $$ \left[ \begin{array}{cc|c} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array} \right] $$