JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Matrices

For infinitely many solutions the third equation must be a linear combination of the first two. Adding the first two equations gives $2x+3y+6z=16$, which matches the coefficient pattern of the third equation exactly, forcing $\lambda =6$ and $\mu =16$. (Equivalently the coefficient determinant $\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 2& 3& \lambda \end{array}\right|=\lambda -6$ vanishes at $\lambda =6$, and consistency then gives $\mu =16$.) Therefore