PrecalculusChallengeSystems, Matrices, and Determinants ShareSystems, Matrices, and Determinants · Challenge problemSolve {3x+2y−z=12x−2y+4z=−2−x+12y−z=0\begin{cases} 3x+2y-z=1 \\ 2x-2y+4z=-2 \\ -x+\tfrac12 y-z=0 \end{cases}⎩⎨⎧3x+2y−z=12x−2y+4z=−2−x+21y−z=0A(1,+2,−2)(1,+2,-2)(1,+2,−2)B(2,−2,−2)(2,-2,-2)(2,−2,−2)C(1,−2,−2)(1,-2,-2)(1,−2,−2)D(1,−2,−3)(1,-2,-3)(1,−2,−3)Check Show answer Walk through it step by step Practice more like this Review this concept