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Is it possible to have RREF([A|b]) = $$ \begin{matrix} 1 & * & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{matrix} $$ Or $$ \begin{matrix} 0 & 1 & * & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{matrix} $$ Or $$ \begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{matrix} $$

I think the answer is no for all the matrices because we should have infinitely many solutions for Ax=b according to cramer's rule but actually there is no solution.

  • 0
    You can RREF anything, but it might not be nice, or full rank, or diagonal. – 2017-01-26
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    the matrices you've written are in RREF form, so they are the RREF form of themselves – 2017-01-26
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    @Exodd but is it possible to have matrices which have these matrices as RREF? – 2017-01-26
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    Not every system of equations has infinitely many solutions – 2017-01-26

1 Answers 1

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Yes, it is possible.

Check for the following conditions:

  • all non-zero rows are above zero rows.
  • the pivot of a nonzero row is always strictly to the right of the pivot of the row above it
  • every leading coefficient is $1$.

If we obtain those RREF, it means that we do not have any solution as we have $0=1$.

Let $$A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ and $$b=\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$$

and you obtain such RREF.

remark: apply Cramer's rule only when $A$ is nonsingular.