PrecalculusMediumSystems, Matrices, and Determinants ShareSystems, Matrices, and Determinants · Medium problemUsing that inverse, solve {x+2y+3z=6y+4z=4z=1\begin{cases} x+2y+3z=6 \\ y+4z=4 \\ z=1 \end{cases}⎩⎨⎧x+2y+3z=6y+4z=4z=1A(0,3,1)(0, 3, 1)(0,3,1)B(4,0,1)(4, 0, 1)(4,0,1)C(−3,0,1)(-3, 0, 1)(−3,0,1)D(3,0,1)(3,0,1)(3,0,1)Check Show answer Walk through it step by step Practice more like this Review this concept