91Ó°ÊÓ

Let \(A\) and \(B\) be two orthogonal matrices, thus \(A^T A = I\) and \(B^T B = I\). We want to prove that their product, \(AB\), is orthogonal. To prove this, we need to show that \((AB)^T (AB) = I\). Let's compute the transpose of the product, \((AB)^T\), and the product \((AB)^T (AB)\): \((AB)^T = B^T A^T\) (by the rule for the transpose of a product) Now let's calculate \((AB)^T (AB)\): \((AB)^T (AB) = (B^T A^T)(AB) = B^T (A^T A) B = B^T I B = B^T B\), because \(A^T A = I\) Since \(B^T B = I\), we have \((AB)^T (AB) = I\). Hence, the product of orthogonal matrices, \(AB\), is an orthogonal matrix.

Let's suppose that \(A\) is an orthogonal matrix, so \(A^T A = I\). We want to prove that its inverse, \(A^{-1}\), is also orthogonal. Recall that for any matrix and its inverse, we have the following identity: \(A A^{-1} = I\) Taking the transpose of both sides: \((A A^{-1})^T = I^T\) Using the transpose property, we can rewrite the left side: \((A^{-1})^T A^T = I\) Since \(A^T A = I\), we have \((A^{-1})^T A^T = A^T A\). Multiplying both sides by \(A^{-1}\) on the right, we get: \((A^{-1})^T = A^{-1}\) So, \((A^{-1})^T (A^{-1}) = I\), which shows that the inverse of an orthogonal matrix, \(A^{-1}\), is also an orthogonal matrix. In conclusion, we have proved that the products and inverses of orthogonal matrices are orthogonal, and hence, the orthogonal matrices form a group under multiplication, called the orthogonal group.