UPSC NDA 2 2016 Mathematics Solved Paper

Concept:For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero and the product of the adjoint of that matrix and the constant vector must be zero.Explanation:Write the given equations in matrix form:$2x + 3y + 5z = 9$$7x + 3y - 2z = 8$$2x + 3y + \lambda z = \mu$$\Rightarrow A = \begin{bmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{bmatrix},\; B = \begin{bmatrix} 9 \\ 8 \\ \mu \end{bmatrix}$First, find $\det(A)$:$\det(A) = 2(3\lambda + 6) - 3(7\lambda + 4) + 5(15)$$= 6\lambda + 12 - 21\lambda - 12 + 75$$= -15\lambda + 75$Set $\det(A)=0$ for infinitely many solutions:$-15\lambda + 75 = 0 \Rightarrow \lambda = 5$Now compute $(\operatorname{adj}A)B$ for $\lambda=5$.The adjoint matrix is:$\operatorname{adj}A = \begin{bmatrix} 3\lambda+6 & -3\lambda+15 & -21 \\ -7\lambda-4 & 2\lambda-10 & 39 \\ 15 & 0 & -15 \end{bmatrix}$Substitute $\lambda=5$:$\operatorname{adj}A = \begin{bmatrix} 21 & 0 & -21 \\ -39 & 0 & 39 \\ 15 & 0 & -15 \end{bmatrix}$Multiply by $B$:$(\operatorname{adj}A)B = \begin{bmatrix} 21\cdot9 + 0\cdot8 + (-21)\mu \\ -39\cdot9 + 0\cdot8 + 39\mu \\ 15\cdot9 + 0\cdot8 + (-15)\mu \end{bmatrix} = \begin{bmatrix} 189 - 21\mu \\ -351 + 39\mu \\ 135 - 15\mu \end{bmatrix}$Set each entry to zero for consistency:From the third entry: $135 - 15\mu = 0 \Rightarrow \mu = 9$Check first entry: $189 - 21(9) = 189 - 189 = 0$ (consistent)Second entry: $-351 + 39(9) = -351 + 351 = 0$ (consistent)Thus, $\lambda = 5$ and $\mu = 9$ yield infinitely many solutions.Answer:Option B: $\lambda = 5$ and $\mu = 9$