UPSC NDA 2 2022 Mathematics Solved Paper

Concept:The adjugate of the given matrix is equal to its negative, which simplifies verifying the three statements.Explanation:Given $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & \sin\theta & -\cos\theta \end{pmatrix}$.The cofactor matrix is $\text{cof}(A) = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos\theta & -\sin\theta \\ 0 & -\sin\theta & \cos\theta \end{pmatrix}$.Taking the transpose gives the adjugate: $\text{adj }A = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos\theta & -\sin\theta \\ 0 & -\sin\theta & \cos\theta \end{pmatrix} = -A$.Hence $A + \text{adj }A = 0$, a null matrix. So statement 1 is correct.The determinant is $|A| = 1\cdot(-\cos^2\theta - \sin^2\theta) = -1$.Thus $A^{-1} = \frac{1}{|A|}\,\text{adj }A = -\text{adj }A$.Therefore $A^{-1} + \text{adj }A = 0$, making statement 2 correct.Substitute $\text{adj }A = -A$ into $A^{-1} = -\text{adj }A$: we get $A^{-1} = A$, so $A - A^{-1} = 0$. Statement 3 is correct.All three statements are true.Answer:All three statements are correct, corresponding to option D (1, 2 and 3).