Statement I Analysis: Symmetric Matrix and its Inverse

A matrix $A$ is symmetric if its transpose equals the matrix itself, i.e., $A^T = A$. The inverse $A^{-1}$ exists only if $A$ is non-singular ($det(A) \neq 0$).

For $A^{-1}$ to be symmetric, its transpose must equal itself: $(A^{-1})^T = A^{-1}$.

Using the property that the transpose of an inverse is the inverse of the transpose, we have $(A^{-1})^T = (A^T)^{-1}$.

Since $A$ is symmetric ($A^T = A$), we can substitute $A$ for $A^T$: $(A^{-1})^T = (A)^{-1} = A^{-1}$.

Thus, if a symmetric matrix $A$ is non-singular, its inverse $A^{-1}$ is also symmetric. Statement I is correct.

Statement II Analysis: Singular Matrix and its Adjoint

A matrix $A$ is singular if its determinant is zero, $det(A) = 0$. We are given $n > 1$.

The fundamental relationship between a matrix, its adjoint, and its determinant is:

$A \cdot adj(A) = det(A) \cdot I$

If $A$ is singular ($det(A) = 0$), this equation becomes:

$A \cdot adj(A) = 0 \cdot I = O$

(where $O$ is the zero matrix).

To determine if $adj(A)$ is singular, we examine its determinant. The property relating the determinant of the adjoint to the determinant of the original matrix is:

$det(adj(A)) = (det(A))^{n-1}$

Since $det(A) = 0$ and $n > 1$ (meaning $n-1 \ge 1$), we have:

$det(adj(A)) = (0)^{n-1} = 0$

Because the determinant of $adj(A)$ is 0, the adjoint matrix $adj(A)$ is singular. Statement II is correct.

Conclusion

Both Statement I and Statement II are correct.