Matrices and Determinants – System of Linear Equations – JEE Main 22 Jan 2026 Shift 2

Solution:

For a system of linear equations $AX = B$ to have a unique solution, the determinant of the coefficient matrix must be non-zero ($|A| \neq 0$).

$$ |A| = \left| \begin{matrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{matrix} \right| \neq 0 $$

Expanding the determinant along the first column:
$$ 1 \cdot [(n-2)(1) – (n+1)(n-1)] – 1 \cdot [(-n)(1) – (1)(n-1)] + 0 \neq 0 $$

$$ [ (n-2) – (n^2 – 1) ] – [ -n – n + 1 ] \neq 0 $$

$$ ( -n^2 + n – 1 ) – ( -2n + 1 ) \neq 0 $$

$$ -n^2 + n – 1 + 2n – 1 \neq 0 $$

$$ -n^2 + 3n – 2 \neq 0 $$

$$ n^2 – 3n + 2 \neq 0 $$

Factorizing the quadratic equation:
$$ (n-1)(n-2) \neq 0 $$

$$ \Rightarrow n \neq 1 \text{ and } n \neq 2 $$

Since $n$ is the number obtained on rolling a fair die, the sample space is $\{1, 2, 3, 4, 5, 6\}$.
The valid values of $n$ for which the system has a unique solution are $\{3, 4, 5, 6\}$.

The number of favorable outcomes is 4.
$$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{6} $$

Given that the probability is $\frac{k}{6}$, comparing terms gives $k = 4$.

The question asks for the sum of $k$ and all possible values of $n$ (for which the unique solution exists).

$$ \text{Sum} = k + (3 + 4 + 5 + 6) $$

$$ \text{Sum} = 4 + 18 = 22 $$

Ans. (4)