JEE Main 202330 Jan, Morning shiftMathematicsEasyMatrices and Determinants245× since 2002Let the system of linear equations x+y+kz=2x+y+kz=2x+y+kz=2 2x+3y−z=12x+3y-z=12x+3y−z=1 3x+4y+2z=k3x+4y+2z=k3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(k+1)x+(2k-1)y=7(k+1)x+(2k−1)y=7 (2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10 has :AOption A. unique solution satisfying x−y=1x-y=1x−y=1BOption B. infinitely many solutionsCOption C. no solutionDOption D. unique solution satisfying x+y=1x+y=1x+y=1CheckSkip · show solution
