Question

If $$ A = \frac{1}{2} \begin{pmatrix} -1 & 1 & 1 & 1 \\ 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \end{pmatrix}, $$ then the inverse of $ AA^T $ is:

• A. \( I \)
• B. \( 2I \)
• C. \( 3I \)
• D. \( 4I \)

Hint:

Orthogonal matrices satisfy \( A^T = A^{-1} \) and \( A^2 = I \) when symmetric.

Answer 1

The Correct Option is A

Calculate \( AA^T \): Since \( A \) is symmetric, \( A^T = A \). \[ AA^T = A^2 \] Calculate \( A^2 \) for given \( A \) or use properties of orthogonal matrices. Matrix \( A \) here is orthogonal scaled by \( \frac{1}{2} \), satisfying \[ A^2 = I \] Hence, \[ (AA^T)^{-1} = (A^2)^{-1} = I \]