Determinants – System of Equations – JEE Main 29 Jan 2025 Shift 2

Solution:

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) must be zero, and the determinants $\Delta_x, \Delta_y, \Delta_z$ must also be zero.

Calculating $\Delta$:
$$ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix} $$
$$ \Delta = 1(20 – 16) – 1(10 – 4) + 1(4 – 2) $$
$$ \Delta = 4 – 6 + 2 = 0 $$

Since $\Delta = 0$ is satisfied, we now set $\Delta_x = 0$ to find the values of $m$:
$$ \Delta_x = \begin{vmatrix} 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 \end{vmatrix} = 0 $$

Expanding along the first row:
$$ 1(20 – 16) – 1(10m – 4m^2) + 1(4m – 2m^2) = 0 $$
$$ 4 – 10m + 4m^2 + 4m – 2m^2 = 0 $$
$$ 2m^2 – 6m + 4 = 0 $$
$$ m^2 – 3m + 2 = 0 $$
$$ (m-1)(m-2) = 0 $$

The values of $m$ are $1$ and $2$.
$$ \alpha = 1, \quad \beta = 2 $$

We need to calculate the sum:
$$ S = \sum_{n=1}^{10}(n^{\alpha}+n^{\beta}) = \sum_{n=1}^{10}(n^1 + n^2) $$
$$ S = \sum_{n=1}^{10} n + \sum_{n=1}^{10} n^2 $$

Using the summation formulas $\sum n = \frac{n(n+1)}{2}$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:
$$ \sum_{n=1}^{10} n = \frac{10(11)}{2} = 55 $$
$$ \sum_{n=1}^{10} n^2 = \frac{10(11)(21)}{6} = 385 $$

$$ S = 55 + 385 = 440 $$

Ans. (1)