UPSC NDA I 2016 Mathematics Solved Paper

Concept:The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$.Explanation:Find the intersection points of the lines $y = 3x$, $y = 6x$, and $y = 9$.Point A: Solve $y = 3x$ and $y = 9$.$3x = 9$ gives $x = 3$; so $A = (3, 9)$.Point B: Solve $y = 3x$ and $y = 6x$.$3x = 6x$ gives $x = 0$; so $B = (0, 0)$.Point C: Solve $y = 6x$ and $y = 9$.$6x = 9$ gives $x = \frac{3}{2}$; so $C = \left(\frac{3}{2}, 9\right)$.Now apply the area formula with $A(3,9)$, $B(0,0)$, $C\left(\frac{3}{2},9\right)$:Area $= \frac{1}{2} \begin{vmatrix} 3 & 9 & 1 \\ 0 & 0 & 1 \\ \frac{3}{2} & 9 & 1 \end{vmatrix}$.Expand along row 2 (row of $B$): $= \frac{1}{2} \left| 0+0-1\left(3\cdot1 - 1\cdot\frac{3}{2}\right) \right|$ simplifies to $\frac{1}{2} \left| -\left(3 - \frac{3}{2}\right) \right| = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$? Let's recalculate carefully.Expanding along $R_2$: The minor is $\begin{vmatrix} 3 & 1 \\ \frac{3}{2} & 1 \end{vmatrix} = 3\cdot1 - 1\cdot\frac{3}{2} = 3 - \frac{3}{2} = \frac{3}{2}$. The cofactor for element $a_{22}=0$ is $(-1)^{2+2}=1$, so term is $0\cdot...=0$. The cofactor for $a_{21}=0$ is $(-1)^{2+1} = -1$, term is $0\cdot...=0$. The cofactor for $a_{23}=1$ is $(-1)^{2+3} = -1$, term is $1 \cdot (-1) \cdot \frac{3}{2} = -\frac{3}{2}$. So determinant = $0+0+\left(-\frac{3}{2}\right) = -\frac{3}{2}$. Then area = $\frac{1}{2} \left| -\frac{3}{2} \right| = \frac{3}{4}$. That's not matching $\frac{27}{4}$. Wait, the existing solution gave $\frac{27}{4}$. Let me check the determinant correctly: $\begin{vmatrix} 3 & 9 & 1 \\ 0 & 0 & 1 \\ \frac{3}{2} & 9 & 1 \end{vmatrix}$.Expanding along row 2: elements are 0,0,1. The minors:• For (0, first column): minor = $\begin{vmatrix} 9 & 1 \\ 9 & 1 \end{vmatrix} = 9*1 - 1*9 = 0$, cofactor = $(-1)^{2+1}= -1$, product = 0.• For (0, second col): minor = $\begin{vmatrix} 3 & 1 \\ \frac{3}{2} & 1 \end{vmatrix} = 3*1 - 1*\frac{3}{2} = 3 - 1.5 = 1.5$, cofactor $(-1)^{2+2}=1$, product = 0 because element 0.• For (1, third col): minor = $\begin{vmatrix} 3 & 9 \\ \frac{3}{2} & 9 \end{vmatrix} = 3*9 - 9*\frac{3}{2} = 27 - 13.5 = 13.5 = \frac{27}{2}$, cofactor $(-1)^{2+3}= -1$, product = 1 * (-1) * (27/2) = -27/2.So determinant = $0+0+\left(-\frac{27}{2}\right) = -\frac{27}{2}$. Area = $\frac{1}{2} * \left| -\frac{27}{2} \right| = \frac{27}{4}$. Yes, the earlier expansion I did missed the 9s. So the correct minor for element $a_{23}$ is the 2x2 formed by rows 1 and 3, columns 1 and 2: $\begin{vmatrix} 3 & 9 \\ \frac{3}{2} & 9 \end{vmatrix} = (3)(9) - (9)\left(\frac{3}{2}\right) = 27 - \frac{27}{2} = \frac{27}{2}$. So area = $\frac{27}{4}$ square units.Answer:$\frac{27}{4}$ square units.