I'm struggling with this question and can't seem to come up with an example:

Give an example of a linear system (augmented matrix form) that has:

  • reduced row echelon form
  • consistent
  • 3 equations
  • 1 pivot variable
  • 1 free variable

The constraints that I'm struggling with is: If the system has 3 equations, that means the matrix must have at least 3 non-zero rows. And given everything else, how can I have only 1 pivot?


1 Answer 1


Hint: It's gotta have only three columns, one for each of the variables (1 pivot, 1 free) and one column for the constants in the equations.

  • $\begingroup$ if I only have 2 columns then I still don't see how I can have 3 pivots and satisfy the reduced row echelon form. The book that we're using states that a) if a row does not consist entirely of zeroes, then the first nonzero number in the row is a 1. If I have all 1s there it also can't be in the reduced echelon form because b) column with leading 1 must have 0 everywhere else .. $$ \left[\begin{array}{rrr|r} 1 & * \\ 1 & * \\ 1 & * \end{array}\right] $$ Also, wouldn't this be just 1 variable in total? $\endgroup$
    – user59547
    Jan 25, 2013 at 5:15
  • $\begingroup$ Notice that the condition is "if a row does not consist entirely of zeroes". In your case, some of your rows will have to consist of only zeroes. In particular, how many rows must be all zero? (Use the fact that there's only one pivot) $\endgroup$ Jan 25, 2013 at 5:36
  • $\begingroup$ Hmm.. looks like I missed this. I was under the assumption that a row that contains all zeroes does not count as 1 equation. $\endgroup$
    – user59547
    Jan 25, 2013 at 5:42
  • 1
    $\begingroup$ It counts as the equation $0x + 0y + ... = 0$. I know it sounds silly, but that's going to be what it means. On that note, I actually made a mistake (which I will now fix), as I didn't see that they were asking for an augmented matrix, in which case you need to throw in one more column. $\endgroup$ Jan 25, 2013 at 8:48

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