Algebra – Matrices and Determinants – JEE Main 02 April 2025 Shift 2

Solution:

For a system of linear equations to have infinitely many solutions, $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.

Evaluating $\Delta_z = 0$ to find $\lambda$:
$$ \Delta_z = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0 $$

$$ 2(18 – 35) – \lambda(27 – 28) + 5(15 – 8) = 0 $$
$$ 2(-17) – \lambda(-1) + 5(7) = 0 $$
$$ -34 + \lambda + 35 = 0 $$
$$ \lambda + 1 = 0 \Rightarrow \lambda = -1 $$

Evaluating $\Delta = 0$ to find $\mu$:
$$ \Delta = \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0 $$

Substitute $\lambda = -1$:
$$ \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0 $$

$$ 2(2\mu – (-5)) – (-1)(3\mu – (-4)) + 3(15 – 8) = 0 $$
$$ 2(2\mu + 5) + 1(3\mu + 4) + 3(7) = 0 $$
$$ 4\mu + 10 + 3\mu + 4 + 21 = 0 $$
$$ 7\mu + 35 = 0 $$
$$ 7\mu = -35 \Rightarrow \mu = -5 $$

Calculate the value of $(\lambda^{2}+\mu^{2})$:
$$ \lambda^{2}+\mu^{2} = (-1)^{2} + (-5)^{2} $$
$$ \lambda^{2}+\mu^{2} = 1 + 25 = 26 $$

Ans. (3)