QuestionApply elimination and back substitution to solve: $$\begin{array}{r} 2 x-2 y-1 z=3 \\ -4 x+2 y+3 z=-5 \\ 2 x+2 y-4 z=2 \end{array}$$
The three pivots are: $\square$ $\square$ $\square$ . The three multipliers are: $\ell_{21}=$ $\square$ ,$\ell_{31}=$ $\square$ ,$\ell_{32}=$ $\square$ . The solution is $(x, y, z)=($ $\square$ ,$\square$, $\square$ ).