Orthogonal basis but not unitary matrix inverse?

Question

For an orthgonal matrix, which has orthogonal+normalized sets of basis, it is said that $A^T$=$A^-1$. But what about matrix with orthogonal but not normalized set of basis? Does it have some kind of characteristics? Thinking of constants that will pop out after normalization, would it be right to think that inverse of the matrix will be same as transpose for matrix with orthogonal but not normalized basis too? Since $B=cA=>cA^T=cA^-1?$ like we could choose any column or row to multiply c back into matrix that will eventually make $B^T=B^-1$

Answer 1

If the columns of $\ A\ $ are orthogonal, then $$ A^TA=\text{diag}\big(a_1^Ta_ 1,a_2^Ta_2,\dots,a_n^Ta_n\big) $$ where $\ a_i\ $ is its $\ i^\text{th}\ $ column. If these are all non-zero, then \begin{align} \text{diag}\left(\frac{1}{a_1^Ta_ 1},\frac{1}{a_2^Ta_ 2},\right.&\left.\dots,\frac{1}{a_1^Ta_ 1}\right)A^TA=I_{n\times n}\\ &=A^T\text{diag}\left(\frac{1}{a_1^Ta_ 1},\frac{1}{a_2^Ta_ 2},\dots,\frac{1}{a_1^Ta_ 1}\right) \end{align} What, therefore, are the inverses of $\ A\ $ and $\ A^T\ $? Do either of them necessarily have orthogonal columns or orthogonal rows?