The values of λ and μ such that the system of equations x + y + z = 6 ,   3 x + 5 y + 5 z =…

The values of λ and μ such that the system of equations x+y+z=6, 3x+5y+5z=26 and x+2y+λz=μ has no solution, are:
  1. λ=3,μ=5
  2. λ=3,μ≠10
  3. λ≠2,μ=10
  4. λ=2,μ≠10

Solution

For the system of equations $a_1x + b_1y + c_1z = d_1$, $a_2x + b_2y + c_2z = d_2$ and $a_3x + b_3y + c_3z = d_3$, We have $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$, $D_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$, $D_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}$ and $D_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$. And, the system has no solution, if $D = 0$ and at least one of $D_x$, $D_y$ & $D_z$ is non-zero. Thus, for the given system of equations, $x + y + z = 6$, $3x + 5y + 5z = 26$ and $x + 2y + \lambda z = \mu$, We have $D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix} = 0$, $\Rightarrow 1(5\lambda - 10) - 1(3\lambda - 5) + 1(6 - 5) = 0$, $\Rightarrow 5\lambda - 10 - 3\lambda + 5 + 1 = 0$, $\Rightarrow 2\lambda = 4$, $\Rightarrow \lambda = 2$. And, $D_z = \begin{vmatrix} 1 & 1 & 6 \\ 3 & 5 & 26 \\ 1 & 2 & \mu \end{vmatrix} \neq 0$, $\Rightarrow 1(5\mu - 52) - 1(3\mu - 26) + 6(6 - 5) \neq 0$, $\Rightarrow 5\mu - 52 - 3\mu + 26 + 6 \neq 0$, $\Rightarrow 2\mu \neq 20$, $\Rightarrow \mu \neq 10$. $\therefore$ For no solution $\lambda = 2$ and $\mu \neq 10$.

Asked in: JEE Main 2021 (22 Jul Shift 1)

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