Q46 / 160 · AP EAMCET 2016 · Mathematics

The equations $x-y+2 z=4$ $3 x+y+4 z=6$ $x+y+z=1$ have

  • A
    unique solution
  • Correct answer
    B
    infinitely many solutions
  • C
    no solution
  • D
    two solutions

Solution

Given equations, $x-y+2 z=4$ $\begin{aligned} & 3 x+y+4 z=6 \\ & x+y+z=1 \end{aligned}$ $\begin{aligned} & \text { Let } \Delta=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 1 & 4 \\ 1 & 1 & 1 \end{array}\right| \\ & =1(1-4)+1(3-4)+2(3-1) \\ & =-3-1+4=0 \\ & \text { and } \Delta_1=\left|\begin{array}{ccc} 4 & -1 & 2 \\ 6 & 1 & 4 \\ 1 & 1 & 1 \end{array}\right| \\ & =4(1-4)+1(6-4)+2(6-1) \\ & =-12+2+10=0 \end{aligned}$ Now, $\Delta=0$ and $\Delta_1=0$ $\therefore \quad$ These equations have infinitely many solutions.

Hence, the correct answer is option: 'B'.

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