UPSC NDA II 2023 Mathematics Solved Paper

Concept:A diagonal matrix raised to any positive integer power remains diagonal and symmetric, but its determinant is non-zero so it is non-singular.Explanation:Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.Statement 1: "$A^n$ will always be singular for any positive integer $n$".Singular means $\det = 0$. For $A$, $\det(A) = 2 \times 3 \times 4 = 24 \neq 0$. So $A$ is non-singular. Since $\det(A^n) = (\det A)^n \neq 0$, $A^n$ is also non-singular. Statement 1 is false.Statement 2: "$A^n$ will always be a diagonal matrix for any positive integer $n$".$A$ is a diagonal matrix. Multiplying diagonal matrices yields a diagonal matrix (each diagonal entry raised to power $n$). Thus $A^n$ is always diagonal. Statement 2 is true.Statement 3: "$A^n$ will always be a symmetric matrix for any positive integer $n$".A diagonal matrix is symmetric because $a_{ij} = 0 = a_{ji}$ for $i \neq j$. Since $A^n$ is diagonal, it is symmetric. Statement 3 is true.Therefore, only statements 2 and 3 are correct.Answer:Option B: 2 and 3 only.