Left-looking Gaussian elimination

I read about left-looking Gaussian elimination which is a variant of Gaussian elimination we use (called right-looking Gaussian elimination) but I could not understand it. Could any one give me an example about it. This is an example of $3\times 3$ Gaussian elimination we use (right-looking because we are always modifying the bottom lower portion which is to the right of the pivot). $$\label{guass1} \left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{array} \right] \left[\begin{array}{ccc}x_{1}\\x_{2} \\x_{3} \\\end{array}\right] = \left[\begin{array}{ccc}b_{1}\\b_{2} \\b_{3} \\\end{array}\right]$$ $$\label{guass2} \left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ 0 & a_{22}^{(2)} & a_{23}^{(2)} \\ 0 & a_{32}^{(2)} & a_{33}^{(2)} \\ \end{array} \right] \left[\begin{array}{ccc}x_{1}\\x_{2} \\x_{3} \\\end{array}\right] = \left[\begin{array}{ccc}b_{1}\\b_{2}^{(2)} \\b_{3}^{(2)} \\\end{array}\right]$$ $$\label{gauss3} \left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ 0 & a_{22}^{(2)} & a_{23}^{(2)} \\ 0 & 0 & a_{33}^{(3)} \\ \end{array} \right] \left[\begin{array}{ccc}x_{1}\\x_{2} \\x_{3} \\\end{array}\right] = \left[\begin{array}{ccc}b_{1}\\b_{2}^{(2)} \\b_{3}^{(3)} \\\end{array}\right]$$ For the left-looking variant, I am quoting this from a reference,

delay the updates for $a_{ij}$ until column j is about to be pivotal, that is, $$\label{gauss4} a_{ij}^{(j)} = a_{ij} - \sum_{k=1}^{j-1} l_{ik}u_{kj}$$ The entries $l_{ik}$ are to the left of $a_{ij}$ in the matrix pattern, so this form is called left-looking.

I figured out how from the formula $$\left[ \begin{array}{ccc} u_{11} & a_{12}^{(2)} & a_{13} \\ l_{21} & a_{22}^{(2)} & a_{23} \\ l_{31} & a_{32}^{(2)} & a_{33} \\ \end{array} \right] \left[\begin{array}{ccc}x_{1}\\x_{2} \\x_{3} \\\end{array}\right] = \left[\begin{array}{ccc}b_{1}\\b_{2} \\b_{3}\\\end{array}\right]$$ where \begin{eqnarray*} u_{12} = a_{12}^{(2)} = a_{12} \quad,\quad a_{22}^{(2)} = a_{22} - l_{21} u_{12} & , & a_{32}^{(2)} = a_{32} - l_{31}u_{12}\\ \end{eqnarray*} In the next step, we modify the third column $$\left[ \begin{array}{ccc} u_{11} & u_{12} & a_{13}^{(3)} \\ l_{21} & u_{22} & a_{23}^{(3)} \\ l_{31} & l_{32} & a_{33}^{(3)} \\ \end{array} \right] \left[\begin{array}{ccc}x_{1}\\x_{2} \\x_{3} \\\end{array}\right] = \left[\begin{array}{ccc}b_{1}\\b_{2} \\b_{3}\\\end{array}\right]$$ where \begin{eqnarray*} a_{13}^{(3)} = u_{13} = a_{13} & , & a_{23}^{(3)} = a_{23} - l_{21}u_{13} \quad,\quad a_{33}^{(3)} = a_{33} - l_{31}u_{13} - l_{32}u_{23}\\ \end{eqnarray*} which yields the same results as the previous example.