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.