$\large{ (i) \;\;\; \text{Let us write the system as } \\ AX=B \;\;\;\;\; \begin{bmatrix} 1 & -1 & \;\;\;2 \\ 3 & \;\;\;2 & -3 \\ 4 & -4 & \;\;\;2 \end{bmatrix} \;\; \begin{bmatrix} x \\ y \\z \end{bmatrix} = \;\; \begin{bmatrix} 2 \\2\\2 \end{bmatrix} \\ (ii) \;\;\; \text{Now, we write LU=A and find L and U.} \\ \begin{bmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix} \;\;\; \begin{bmatrix} 1 & u_{12} & u_{13} \\ 0 & 1 & u_{23} \\ 0 & 0& 1 \end{bmatrix} \;\;\; = \;\; \begin{bmatrix} 1 & -1 & \;\;\;2 \\ 3 & \;\;\;2 & -3 \\ 4 & -4 & \;\;\;2 \end{bmatrix}\\ \therefore \begin{bmatrix} l_{11} & l_{11} u_{12} & l_{11}u_{13} \\ l_{21} & l_{21} u_{12}+l_{22} & l_{21}u_{13}+ l_{22}u_{23} \\ l_{31} & l_{31}u_{12}+l_{32} & l_{31}u_{13}+l_{32}u_{23}+l_{33} \end{bmatrix} = \begin{bmatrix} 1 & -1 & \;\;\;2 \\ 3 & \;\;\;2 & -3 \\ 4 & -4 & \;\;\;2 \end{bmatrix} \\ \text{Equating the corresponding elements in the two matrices, we get,} \\ l_{11} = 1 , \;\;\;\; l_{11}u_{12}= -1 \\ \therefore u_{12}= \; -1 \\ l_{11} = 1 , \;\;\;\; l_{11}u_{13}= 2 \\ \therefore u_{13}= \; 2 \\ l_{21} = 3 , \;\;\;\; l_{21}u_{12}+l_{22 } = 2 \\ \therefore u_{22}= \; 5 \\ l_{21} = 3 , \;\;\;\; l_{21}u_{13}+l_{22 } u_{23}= -3 \\ \therefore u_{23}= \; - \frac 95 \\ l_{31} = 4 , \;\;\;\; l_{31}u_{12}+l_{32 } = -4 \\ \therefore l_{32}= \; 0 \\ l_{31} = 3 , \;\;\;\; l_{31}u_{13}+l_{32 }u_{23}+l_{33 } = 2 \\ \therefore l_{33}= \; -6 \\ \therefore L= \begin{bmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix} \;\;\; = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 5 & 0 \\ 4 & 0 & -6 \end{bmatrix} \\ and \; \; \; U = \begin{bmatrix} 1 & u_{12} & u_{13} \\ 0 & 1 & u_{23} \\ 0 & 0& 1 \end{bmatrix} \;\;\; = \;\; \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & - \frac 95 \\ 0 & 0& 1 \end{bmatrix} \\ (iii) \;\;\;\; Now, \; LV=B, \;\; gives \;\;\; \begin{bmatrix} 1 & 0 & 0 \\ 3 & 5 & 0 \\ 4 & 0 & -6 \end{bmatrix} \;\;\; \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \;\;\; = \begin{bmatrix} 2 \\2\\2 \end{bmatrix} \\ \therefore v_1= 2 \\ 3 v_1 + 5 v_2 =2 \\ \therefore v_2= -4 \\ 4 v_1 - 6 v_3 =2 \\ \therefore v_3 = 1 \\ (iv) \;\;\;\; UX=V , \;\;\; gives \;\;\;\; \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & - \frac 95 \\ 0 & 0& 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \;\;\;=\;\;\; \begin{bmatrix} 2 \\ - \frac 45 \\ 1 \end{bmatrix} \\ \therefore x-y+2z =2 \;\;\;\;\;\; y- \frac 95 z = - \frac 45, \;\;\;\;\;\; z=1 \\ \therefore y = 1 \;\; \\ and \;\;\; x=1 \\ \therefore x=1 \;\;\;\; y=1 \;\;\;\; z=1 } $