If A is an invertible symmetric matrix, then A^-1 is

The correct answer is Option (1) → a symmetric matrix

Since A is symmetric, \(A^T = A\). The transpose of an inverse is the inverse of the transpose:

    \((A^{-1})^T = (A^T)^{-1}\)

    \((A^{-1})^T = A^{-1}\)

So \(A^{-1}\) equals its own transpose, that is, \(A^{-1}\) is symmetric. It need not be the identity: for example, \(A = 2I\) is symmetric and \(A^{-1} = \frac{1}{2}I\).