JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Matrices

For infinitely many solutions the coefficient determinant and an auxiliary determinant must both vanish. Replacing the third column by the constants, ${\Delta }_z=\left|\begin{array}{ccc}2& \lambda & 5\\ 3& 2& 7\\ 4& 5& 9\end{array}\right|=2(18-35)-\lambda (27-28)+5(15-8)=-34+\lambda +35=\lambda +1$, so $\lambda =-1$. The coefficient determinant $\Delta =\left|\begin{array}{ccc}2& \lambda & 3\\ 3& 2& -1\\ 4& 5& \mu \end{array}\right|=4\mu -3\lambda \mu -4\lambda +31$ must also vanish; with $\lambda =-1$ this is $7\mu +35=0$, so $\mu =-5$. The third equation then equals $(-1)$ times the first plus $2$ times the second, confirming consistency. Therefore ${\lambda }^2+{\mu }^2=1+25=26$.