Q46 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics

For the system $S$ of linear equations $x+y+z=3,2 x+2 y-z=3, x+y+\lambda z=1$ the incorrect option among the following statements is

  • A
    $S$ has infinitely many solutions, if $\lambda=-1$
  • B
    $S$ has no solution, if $\lambda=-2$
  • C
    $S$ has no unique solution for any real $\lambda$
  • Correct answer
    D
    $S$ is consistent for all $\lambda \in R$

Solution

We have, $x+y+z=3$ $\begin{aligned} & 2 x+2 y-z=3 \\ & x+y+\lambda z=1 \\ \therefore \quad A & =\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 2 & -1 \\ 1 & 1 & \lambda\end{array}\right] \\ \therefore \quad|A| & =\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 2 & -1 \\ 1 & 1 & \lambda\end{array}\right| \\ & =\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & 2 & -1 \\ 0 & 1 & \lambda\end{array}\right|\left[\text { on applying } C_1 \rightarrow C_1-C_2\right] \\ & =0\end{aligned}$ $\therefore$ Wrong statement is $S$ is consistent for all $\lambda \in \mathbf{R}$.

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

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