We get then a new matrix $$\mathbf{B} =\left( \begin{array}{ccc} b_{11} & b_{12}& b_{13} \\ b_{12}& b_{22}& 0 \\ b_{13}& 0& a_{33} \end{array} \right).$$ We repeat then assuming that $b_{12}$ is the largest non-diagonal matrix element and get a new matrix $$\mathbf{C} = \left( \begin{array}{ccc} c & -s & 0 \\ s & c & 0 \\ 0 & 0 & 1 \end{array} \right)\left( \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{12} & b_{22} & 0 \\ b_{13} & 0 & b_{33} \end{array} \right) \left( \begin{array}{ccc} c & s & 0 \\ -s & c & 0 \\ 0 & 0 & 1 \end{array} \right).$$ We continue this process till all non-diagonal matrix elements are zero (ideally). You will notice that performing the above operations that the matrix element $b_{23}$ which was previous zero becomes different from zero. This is one of the problems which slows down the jacobi procedure.