UPSC NDA 2 2016 Mathematics Solved Paper

Concept:For a system of linear equations $AX = B$, if $\det(A) \neq 0$, a unique solution exists. If $\det(A) = 0$, we check $( \operatorname{adj} A ) \cdot B$: zero gives infinitely many solutions; non-zero gives no solution.Explanation:Write the system in matrix form: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & -1 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 6 \\ k \end{bmatrix}$$Compute $\det(A) = 1(-5 + 2) - 1(5 - 6) + 1(-1 + 3) = -3 + 1 + 2 = 0$.Since $\det(A) = 0$, a unique solution is impossible. Statement 3 is false.Find $\operatorname{adj} A = \begin{bmatrix} -3 & -6 & 3 \\ 1 & 2 & -1 \\ 2 & 4 & -2 \end{bmatrix}$.For $k = 15$, $B = \begin{bmatrix} 8 \\ 6 \\ 15 \end{bmatrix}$. Compute $(\operatorname{adj} A) \cdot B$: $(-3)(8) + (-6)(6) + (3)(15) = -24 -36 + 45 = -15 \neq 0$ ⇒ no solution. Statement 1 is correct.For $k = 20$, $B = \begin{bmatrix} 8 \\ 6 \\ 20 \end{bmatrix}$. Compute $(\operatorname{adj} A) \cdot B$: $(-3)(8) + (-6)(6) + (3)(20) = -24 -36 + 60 = 0$ ⇒ infinitely many solutions. Statement 2 is correct.Answer:Only statements 1 and 2 are correct. Hence the correct code is A (1 and 2 only).