Q28 / 75 · JEE Main 2026 (08 April Shift 2) · Mathematics
If the system of linear equations:
$x+y+z=6$,
$x+2y+5z=10$,
$2x+3y+\lambda z=\mu$
has infinitely many solutions, then the value of $\lambda+\mu$ equals:
$x+y+z=6$,
$x+2y+5z=10$,
$2x+3y+\lambda z=\mu$
has infinitely many solutions, then the value of $\lambda+\mu$ equals:
- A$12$
- B$16$
- Correct answerC$22$
- D$28$
Solution
For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix $\Delta$ must be zero, and $\Delta_x = \Delta_y = \Delta_z = 0$.
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = 0$
Expanding the determinant:
$1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = 0$
$\lambda - 6 = 0 \Rightarrow \lambda = 6$
Now, setting $\Delta_z = 0$:
$\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{vmatrix} = 0$
Expanding the determinant:
$1(2\mu - 30) - 1(\mu - 20) + 6(3 - 4) = 0$
$\mu - 16 = 0 \Rightarrow \mu = 16$
Therefore, $\lambda + \mu = 6 + 16 = 22$.
Answer: $22$
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = 0$
Expanding the determinant:
$1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) = 0$
$\lambda - 6 = 0 \Rightarrow \lambda = 6$
Now, setting $\Delta_z = 0$:
$\Delta_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{vmatrix} = 0$
Expanding the determinant:
$1(2\mu - 30) - 1(\mu - 20) + 6(3 - 4) = 0$
$\mu - 16 = 0 \Rightarrow \mu = 16$
Therefore, $\lambda + \mu = 6 + 16 = 22$.
Answer: $22$
Hence, the correct answer is option: 'C'.