JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Matrices

For the system to have no solution, the coefficient matrix must be singular (determinant = 0) but the augmented matrix must have a different rank.

The coefficient matrix is:
$\left|\begin{array}{ccc}1& 1& 1\\ 3& 5& 5\\ 1& 2& \lambda \end{array}\right|=1(5\lambda -10)-1(3\lambda -5)+1(6-5)=5\lambda -10-3\lambda +5+1=2\lambda -4$

For no solution, we need $2\lambda -4=0$, so $\lambda =2$.

With $\lambda =2$, the third equation becomes $x+2y+2z=\mu$.

From the first two equations: $R_2-3R_1$ gives $2y+2z=8$, so $y+z=4$.
From $x+y+z=6$ and $y+z=4$, we get $x=2$.

Substituting into the third equation: $2+2(4)=10$.

For no solution, we need $\mu \ne 10$.

Therefore, $\lambda =2,\mu \ne 10$.