Compute the reduced row echelon form of the following matrix and circle the pivots.
$$ %\begin{bmatrix*}[r] \begin{bmatrix} 0&2&-1&0\\ -1&3&2&-1\\ 0&-4&2&0 \end{bmatrix} %\end{bmatrix*} $$Below is a geometrical representation of the following linear system.
$$ \left\{ \begin{array}{ccccc} -2 x_1 & +2 x_2 & +10 x_3 & = & 2\\ -3 x_1 & + x_2 & + 9 x_3 & = & 5\\ 4 x_1 & -2 x_2 & - 14 x_3 & = & -6 \end{array}\right. $$Using row reduction, we found a simpler equivalent system.
$$ \left\{ \begin{array}{ccccc} x_1 & & - 2x_3 & = & -2\\ & x_2 & + 3 x_3 & = & -1\\ \end{array} \right. $$In the plot below, we observe that this system has the same solution set as our original system, represented by the red line.