JEE Main 2025 Apr 7 Shift 2, Mathematics Q19: Linear Equations And Cramer’s Rule

For infinitely many solutions, the coefficient determinant must vanish:

$\left|\begin{array}{c}15-1\\ 43-3\\ 241\lambda \end{array}\right|=0\Rightarrow -17\lambda -289=0\Rightarrow \lambda =-17$.

For consistency, the determinant formed by replacing the $x$-column with the constants must also vanish:

$\left|\begin{array}{c}15-1\\ 73-3\\ \mu 1-17\end{array}\right|=0\Rightarrow 540-12\mu =0\Rightarrow \mu =45$.

The system becomes $x+5y-z=1$, $4x+3y-3z=7$, $24x+y-17z=45$. Setting $z=t$ and solving the first two equations for $x,y$ in terms of $t$ gives $x=\frac{32+12t}{17}$, $y=\frac{t-3}{17}$.

Writing $t=3+17k$ for integer $k$ makes both $x=4+12k$ and $y=k$ integers automatically, with $z=3+17k$.

Then $x+y+z=7+30k$. The constraint $7\le x+y+z\le 77$ gives $0\le 30k\le 70\Rightarrow 0\le k\le \frac{7}{3}$, so $k=0,1,2$.

$\therefore$ there are 3 integer solutions satisfying the constraint.