Question

3. Show that an orthogonal matrix is always invertible and find det (A).​

Answer 1

$\underline{\textsf{To prove:}}$

$\textsf{An orthogonal matrix is always invertible}$

$\underline{\textsf{Solution:}}$

$\underline{\textsf{Orthogonal matrix:}}$

$\textsf{A square matrix of order is said to be orthogonal if}$

$\mathsf{A\,A^T=A^T\,A=I}$

$\textsf{where I is the identity of order n}$

$\textsf{Let A be an orthogonal matrix}$

$\textsf{Then,}$

$\mathsf{AA^T=A^TA=I}$

$\textsf{Consider,}$

$\mathsf{AA^T=I}$

$\implies\mathsf{det(AA^T)=det(I)}$

$\text{Using,}$

$\boxed{\mathsf{det(AB)=det(A)\,det(B)}}$

$\implies\mathsf{det(A)\,det(A^T)=det(I)}$

$\text{Using,}$

$\boxed{\mathsf{det(A^T)=det(A)}}$

$\boxed{\textsf{Determinant of identity matrix is always 1}}$

$\implies\mathsf{det(A)\,det(A)=1}$

$\implies\mathsf{[det(A)]^2=1}$

$\textsf{Taking square root, we get}$

$\implies\mathsf{det(A)=\pm\,1}$

$\textsf{Since}\;\mathsf{det(A){\neq}0},\;A^{-1}\;\textsf{exists}$

$\textsf{Hence A is invertible}$

$\underline{\textsf{Answer:}}$

$\mathsf{det(A)=\pm\,1}$