REF

  • A matrix is in 2.1 REF if

    • rows with all zero entries (zero rows) if any are all below all non-zero rows
    • for any two non-zero rows, the first non-zero entry (leading entry) in the upper row is to the left of the leading entry in the lower rows
  • Examples

    • solve the following system by converting to REF$$ \begin{align} x+y=5 \

x+z=7 \ y+z=5 \end{align} \begin{bmatrix} 1 & 1 & 0 & |5 \ 1 & 0 & 1 & |7 \ 0 & 1 & 1 & |5 \end{bmatrix}R_{2}-R_{1},R_{3}+R_{2}\implies\begin{bmatrix} 1 & 1 & 0 & |5 \ 0 & -1 & 1 & |2 \ 0 & 0 & 2 & |7 \end{bmatrix}(REF)

\begin{align} x+y=5\implies x=\frac{7}{2} \ y-z=-2\implies y=\frac{3}{2} \ z=\frac{7}{2} \end{align}

- (z was strictly defined as $2z=7$ and using back sub we can solve the rest of the variables) - ## REF matrices are not unique - ###### there can be other solutions -