The matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 2&1&2\\ 1&3&2 \end{array}} \right]\) can be decomposed into the product of a lower triangular matrix \(L = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ {{l_{21}}}&1&0\\ {{l_{31}}}&{{l_{32}}}&1 \end{array}} \right]\) and an upper triangular matrix \(U = \left[ {\begin{array}{*{20}{c}} {{u_{11}}}&{{u_{12}}}&{{u_{13}}}\\ 0&{{u_{22}}}&{{u_{23}}}\\ 0&0&{{u_{33}}} \end{array}} \right]\) as A = LU.

If \(b = {\left[ {\begin{array}{*{20}{c}} 1&1&1 \end{array}} \right]^T}\), then the solution \(z = {\left[ {\begin{array}{*{20}{c}} {{z_1}}&{{z_2}}&{{z_3}} \end{array}} \right]^T}\) of the system Lz = b is