UPSC NDA I and II 2020 Math Solved Paper

Concept:For a non-singular square matrix of order $n$, the product $A(\mathrm{adj}\,A)$ equals $|A|I$, and the determinant of the adjoint is $|\mathrm{adj}\,A| = |A|^{n-1}$.Explanation:Given $A$ is a non-singular matrix of order 3.Since $A$ is non-singular, its inverse exists and is given by $A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A$.Pre-multiply both sides by $A$: $A A^{-1} = \frac{A}{|A|}\,\mathrm{adj}\,A$ → $I|A| = A(\mathrm{adj}\,A)$ → $A(\mathrm{adj}\,A) = |A|I$.Post-multiply both sides by $A$: $A^{-1}A = \frac{1}{|A|}(\mathrm{adj}\,A)A$ → $I|A| = (\mathrm{adj}\,A)A$ → $(\mathrm{adj}\,A)A = |A|I$.Thus $A(\mathrm{adj}\,A) = |A|I = (\mathrm{adj}\,A)A$, so statement 1 is correct.Now, take determinant on both sides of $A(\mathrm{adj}\,A) = |A|I$.We have $|A(\mathrm{adj}\,A)| = \big||A|I\big|$.Using $|kA| = k^n|A|$ for a scalar $k$ and $n\times n$ matrix $A$, the left side becomes $|A|\cdot|\mathrm{adj}\,A|$, and the right side is $|A|^3 \cdot |I| = |A|^3$.So $|A|\cdot|\mathrm{adj}\,A| = |A|^3$.Since $|A|\neq 0$, divide both sides by $|A|$: $|\mathrm{adj}\,A| = |A|^2$.Hence $|\mathrm{adj}\,A| \neq |A|$ (unless $|A|=0$ or $|A|=1$, but the statement claims equality in general). Therefore statement 2 is incorrect.Answer:Only statement 1 is correct.So the correct option is A (1 only).