91Ó°ÊÓ

Since A is orthogonal, we know that \(A^T A = I\). Now, let's find \((A^T)^T\). We know that \((A^T)^T = A\). Then, let's multiply \(A^T\) by its transpose: \(A^T (A^T)^T = A^T A\) Since we already know that \(A^T A = I\), then: \(A^T (A^T)^T = I\) Hence, \(A^T\) is orthogonal.

Since A is orthogonal and invertible, we know that \(A^T = A^{-1}\) and \(A^T A = I\). To find the transpose of \(A^{-1}\), we first substitute \(A^T\) for \(A^{-1}\), which gives us \((A^T)^T\). We have already found \((A^T)^T = A\) in Step 1. Now, let's multiply \(A^{-1}\) by its transpose: \(A^{-1}(A^{-1})^T = A^{-1}A\) Since we already know that \(A^{-1}A = I\), then: \(A^{-1}(A^{-1})^T = I\) Therefore, \(A^{-1}\) is orthogonal.

We need to show that \((AB)^T = (AB)^{-1}\). Let's find the transpose of \(AB\): \((AB)^T = B^TA^T\) Since both A and B are orthogonal matrices, we know that \(A^TA = I\) and \(B^TB = I\). Now let's multiply \((AB)^{-1}\) by the transpose we found, \(B^TA^T\): \((AB)^{-1}(AB)^T = (AB)^{-1}(B^TA^T)\) We know that \((AB)^{-1} = B^{-1}A^{-1}\) for any invertible matrices A and B. Since both A and B are orthogonal, their inverses are their transposes, so we have \(B^{-1} = B^T\) and \(A^{-1} = A^T\). Substituting these into the expression above: \((B^TA^T)(B^TA^T) = B^T(A^TA)B^T\) Using the property we found earlier, \(A^TA = I\), we can simplify this expression: \(B^TIB^T = B^TB^T\) We also know that \(B^TB = I\), so \(B^TB^T = I\) Therefore, \((AB)^T = (AB)^{-1}\), which means that the product \(AB\) is orthogonal. In conclusion, we have shown that \(A^T, A^{-1}\), and \(AB\) are all orthogonal, proving the statement.