UPSC NDA II 2018 Mathematics Solved Paper

Concept:Elementary row and column operations do not change the determinant value.We can simplify the determinant by adding rows and then using column operations.Explanation:Let the given determinant be $\Delta = \begin{vmatrix} x & y & y+z \\ z & x & z+x \\ y & z & x+y \end{vmatrix}$.Apply $R_1 \to R_1 + R_2 + R_3$:$\Delta = \begin{vmatrix} x+y+z & x+y+z & 2(x+y+z) \\ z & x & z+x \\ y & z & x+y \end{vmatrix}$.Factor $(x+y+z)$ from the first row:$\Delta = (x+y+z) \begin{vmatrix} 1 & 1 & 2 \\ z & x & z+x \\ y & z & x+y \end{vmatrix}$.Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - 2C_1$:$\Delta = (x+y+z) \begin{vmatrix} 1 & 0 & 0 \\ z & x-z & (z+x)-2z \\ y & z-y & (x+y)-2y \end{vmatrix}$.Simplify the elements:$\Delta = (x+y+z) \begin{vmatrix} 1 & 0 & 0 \\ z & x-z & x-z \\ y & z-y & x-y \end{vmatrix}$.Now expand along the first row (since first row has 1, 0, 0):$\Delta = (x+y+z) \times [1 \cdot ((x-z)(x-y) - (x-z)(z-y)) - 0 + 0]$.Factor $(x-z)$ inside the bracket:$\Delta = (x+y+z)(x-z)[(x-y) - (z-y)]$.Simplify the bracket: $(x-y) - (z-y) = x-y-z+y = x-z$.Thus $\Delta = (x+y+z)(x-z)(x-z) = (x+y+z)(x-z)^2$.Since $(x-z)^2 = (z-x)^2$, we get $\Delta = (z-x)^2 (x+y+z)$.Answer:Option D: $(z-x)^2 (x+y+z)$.