PrecalculusHardSystems, Matrices, and Determinants ShareSystems, Matrices, and Determinants · Hard problemSolve by Cramer's Rule: {x+y+z=42x+y−z=3x−y+2z=1\begin{cases} x+y+z=4 \\ 2x+y-z=3 \\ x-y+2z=1 \end{cases}⎩⎨⎧x+y+z=42x+y−z=3x−y+2z=1A(−1,2,1)(-1, 2, 1)(−1,2,1)B(1,2,1)(1,2,1)(1,2,1)C(1,−2,1)(1, -2, 1)(1,−2,1)D(2,1,1)(2, 1, 1)(2,1,1)Check Show answer Walk through it step by step Practice more like this Review this concept