The given system of equations can be written in matrix form AX=B as follows

$ \left[ \begin{array}{ccc} 1& -1& 2\\ 3& 2& -3\\ 4& -4& 2 \end{array}\right] \left[ \begin{array}{ccc} x\\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{ccc} 2 \\ 2 \\ 2 \end{array}\right] \\ \; \\ \; \\ $

Let, A=LU such that

$ \left[ \begin{array}{ccc} 1& -1& 2\\ 3& 2& -3\\ 4& -4& 2 \end{array}\right] \;=\; \left[ \begin{array}{ccc} a& 0& 0\\ b& c& 0\\ d& e& f \end{array}\right] \left[ \begin{array}{ccc} 1& g& h\\ 0& 1& i\\ 0& 0& 1 \end{array}\right] \; \; \; \; \; \; \; i.e. \; LUX =B \\ \; \\ \; \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& -1& 2\\ 3& 2& -3\\ 4& -4& 2 \end{array}\right] \;=\; \left[ \begin{array}{ccc} a& ag& ah\\ b& bg+c& bh+ci\\ d& dg+e& dh+ei+f \end{array}\right] \\ \; \\ \; \\ \; \\ \; \\ \therefore a=1 , b=3 , d=4 \\ \; \\ \; \\ ag=-1 \; \therefore g\;=\;\dfrac{-1}{1} \;=\;-1 \\ \; \\ \; \\ bg+c=2 \; \; \therefore c=2-(-3)(-1) \;=\; 2+3 \;=\; 5 \\ \; \\ \; \\ dg+e\;=\; -4 \; \; \therefore e\;=\; -4-(-4)(-1) \;=\; -4+4 \;=\; 0 \\ \; \\ \; \\ ah=2 \; \; \therefore h=2 \\ \; \\ \; \\ bh+ci=-3 \; \; \; \therefore (3)(2)+(5)i=-3 \\ \; \\ \therefore 5i = -3-6 \;=\; -9 \; \; \; \therefore i=\dfrac{-9}{5} \\ \; \\ \; \\ dh+ei+f=2 \\ \; \\ \therefore 4(2)+0+f=2 \; \; \therefore f=2-8\;=\; -6 \\ \; \\ \; \\ \therefore L \;=\; \left[ \begin{array}{ccc} 1& 0& 0\\ 3& 5& 0\\ 4& 0& -6 \end{array}\right] \; \; and \; U \;=\; \left[ \begin{array}{ccc} 1& -1& 2\\ 0& 1& -9/5\\ 0& 0& 1 \end{array}\right] \\ \; \\ \; \\ \; \\ Now, \; \; LY=B \; \; Where Y\;=\;UX \\ \; \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& 0& 0\\ 3& 5& 0\\ 4& 0& -6 \end{array}\right] \left[ \begin{array}{ccc} y_1 \\ y_2 \\ y_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 2 \\ 2 \\ 2 \end{array}\right] \\ \; \\ \; \\ \therefore y_1=2 \\ \; \\ \; \\ 3y_1+5y_2=0 \; \therefore 5y_2\;=\;2-3(2) \;=\; 2-6 \;=\; -4 \; \; \therefore y_2= -4/5 \\ \; \\ \; \\ 4y_1-6y_3=2 \; \; \therefore 6y_3=4y_1-2=4(2)-2=6 \; \; y_3=1 \\ \; \\ \; \\ Now \; UX=Y \\ \; \\ \; \\ \; \\ \therefore \left[ \begin{array}{ccc} 1& -1& 2\\ 0& 1& -9/5\\ 0& 0& 1 \end{array}\right] \left[ \begin{array}{ccc} x \\ y \\ z \end{array}\right] \;=\; \left[ \begin{array}{ccc} 2 \\ -4/5 \\ 1 \end{array}\right] \\ \; \\ \; \\ \; \\ \therefore z=1 \\ \; \\ \; \\ y-\dfrac{9}{5}z=-\dfrac{4}{5} \; \; \; \therefore y\;=\;1 \\ \; \\ \; \\ x-y+2z=2 \; \; x=y-2z+2=1-2(1)+2=1 \\ \; \\ \; \\ \; \\ \therefore x=y=z=1 $