UPSC NDA II 2018 Mathematics Solved Paper

Concept:For an invertible square matrix $A$, the inverse satisfies $(A^{-1})^{-1} = A$ and $\det(A^{-1}) = \frac{1}{\det A}$. For scalar multiplication, $(\lambda A)^{-1} = \frac{1}{\lambda} A^{-1}$, not $\lambda A^{-1}$.Explanation:Property 1: Starting from $A A^{-1} = I$, take inverse on both sides: $(A A^{-1})^{-1} = I^{-1}$.Using $(AB)^{-1} = B^{-1} A^{-1}$, we get $(A^{-1})^{-1} A^{-1} = I$.Post‑multiply by $A$: $(A^{-1})^{-1} A^{-1} A = I A \Rightarrow (A^{-1})^{-1} I = A \Rightarrow (A^{-1})^{-1} = A$.Thus property 1 is correct.Property 2: Take determinant of $A A^{-1} = I$: $\det(A A^{-1}) = \det(I)$.Since $\det(AB) = \det A \cdot \det B$, we have $\det A \cdot \det(A^{-1}) = 1$.Hence $\det(A^{-1}) = \frac{1}{\det A}$. Property 2 is correct.Property 3: For any nonzero scalar $\lambda$, $(\lambda A)^{-1} = \frac{1}{\lambda} A^{-1}$.The given statement $(\lambda A)^{-1} = \lambda A^{-1}$ is false unless $\lambda = \pm 1$ (not generally true). Thus property 3 is incorrect.Answer:Only properties 1 and 2 hold. Therefore the correct option is A (1 and 2 only).