Consider the following statements : I. If n× n(n > 1) matrix is...

Let $A$ be an invertible symmetric matrix of order $n \times n$. Then $A^T = A$.

We know that $(A^{-1})^T = (A^T)^{-1}$.

Substituting $A^T = A$, we get $(A^{-1})^T = A^{-1}$.

Thus, the inverse of a symmetric matrix is also symmetric. The first statement is correct.

Let $A$ be a singular matrix of order $n \times n$. Then $|A| = 0$.

The determinant of the adjoint of $A$ is given by $|\text{adj}(A)| = |A|^{n-1}$.

Since $n > 1$ and $|A| = 0$, we have $|\text{adj}(A)| = 0^{n-1} = 0$.

Thus, the adjoint of a singular matrix is also a singular matrix. The second statement is correct.

Both statements are correct.

Answer: Both I and II