Show that if a matrix is orthogonal and its determinant is $\mathbf{+}\mathbf{1}\mathbf{,}$then each element of the matrix is equal to its own cofactor. Hint: Use (6.13) and the definition of an orthogonal matrix.

Matrix is orthogonal and its determinant is$+1$

An orthogonal matrix, also known as an orthonormal matrix, is a real square matrix whose columns and rows are orthonormal vectors in linear algebra. To determine if a given matrix is orthogonal, first determine its transpose. Then, use the transpose and multiply the provided matrix. The given matrix is orthogonal if the product is an identity matrix; otherwise, it is not.

Let A is an orthogonal matrix with determinant equal to$+1$

Orthogonal matrix:

$\begin{array}{r}{\mathrm{AA}}^{\mathrm{\top }}=1\\ \therefore {\mathrm{A}}^{\mathrm{\top }}={\mathrm{A}}^{-1}\\ \because {\mathrm{A}}^{-1}=\frac{1}{\det A}{C}^{\mathrm{\top }}.\end{array}$

Here C are the cofactors of the corresponding elements of matrix A

$C_{ij}=\text{cofactor of}{R}_{ij}$

Since we have assumed determinant of matrix is $\mathrm{A}\text{is}+1$. So, from equation (1):

$\begin{array}{r}{A}^{\mathrm{\top }}=C^{\mathrm{\top }}\left\{\because A^{-1}=A^{\mathrm{\top }}\right\}\\ A^{-1}=C^{\mathrm{\top }}\end{array}$

The transpose of a matrix simply interchanges the rows and columns of the original matrix${\left(A^{\mathrm{\top }}\right)}^{\mathrm{\top }}=A$

Thus, take transpose of equation (2):

$\begin{array}{r}{\left(A^{\mathrm{\top }}\right)}^{\mathrm{\top }}={\left(C^{\mathrm{\top }}\right)}^{\mathrm{\top }}\\ A=C\end{array}$

So, the elements of matrix A is equal to their cofactors:

$R_{ij}=C_{ij}$

Hence proved. That if a matrix is orthogonal and the determinant of matrix is $+1$, then each matrix component is equal to its cofactor.