UPSC NDA I 2019 Mathematics Solved Paper

Concept:For two square matrices of order $n$, the determinant of a scalar multiple follows: $\det(m \cdot X) = m^n \det(X)$ for any scalar $m$.Also, for an invertible matrix $X$, $\det(X^{-1}) = \frac{1}{\det(X)}$.Explanation:We are given $A$ and $B$ as $3 \times 3$ matrices with $\det A = 4$ and $\det B = 3$.We need $\det(3AB^{-1})$.First, apply the scalar multiplication property to the product $AB^{-1}$:$\det(3AB^{-1}) = 3^3 \cdot \det(AB^{-1})$ because the order is 3. That is $27 \cdot \det(AB^{-1})$.Next, use the product rule for determinants: $\det(AB^{-1}) = \det A \cdot \det(B^{-1})$.Since $\det(B^{-1}) = \frac{1}{\det B} = \frac{1}{3}$, we get $\det(AB^{-1}) = 4 \times \frac{1}{3}$.Now substitute: $\det(3AB^{-1}) = 27 \times \left(4 \times \frac{1}{3}\right) = 27 \times \frac{4}{3} = 9 \times 4 = 36$.Answer:$36$ (Option C).