$R ={\begin{bmatrix}0.9697253054707993 & 0.04804422035332832 & -0.2394255308445735\\-0.01069682073773017 & 0.9878712527451343 & 0.1549063137056192\\ 0.2439639521643922 & -0.1476554803940568&0.9584776727467001\end{bmatrix}}$
$R.inv() = \underbrace {\begin{bmatrix}0.9697253054707994 & -0.01069682073773021 & 0.2439639521643922\\0.04804422035332829 & 0.9878712527451345 & -0.1476554803940568\\ -0.2394255308445735 & 0.1549063137056193 &0.9584776727467004\end{bmatrix}}_\text{inverse R}$
$R.t() = \underbrace {\begin{bmatrix}0.9697253054707994 & -0.01069682073773021 & 0.2439639521643922\\0.04804422035332829 & 0.9878712527451345 & -0.1476554803940568\\ -0.2394255308445735 & 0.1549063137056193 &0.9584776727467004\end{bmatrix}}_\text{transpose R}$
above inverse R = R transposed
consider Q:
$Q = {\begin{bmatrix}2 & 3 & 1\\5 & 4 & 3\\2 &3 &6\end{bmatrix}}$
$Q.inv() =\underbrace {\begin{bmatrix}-0.4285714285714285 & 0.4285714285714285 & -0.1428571428571428 \\0.6857142857142857 & -0.2857142857142857 & 0.02857142857142857\\-0.2 &-0 &0.2\end{bmatrix}}_\text{Q inverted}$
$Q.t() = \underbrace {\begin{bmatrix}2 & 5 & 2\\3 & 4 & 3\\1 &3 &6\end{bmatrix}}_\text{Q transposed}$
inverse Q $\neq$ Q transposed
Why isn't Q.inv() = Q.t() as with R?
C++ are same types well:
Mat Q is of type 64FC1 and should be accessed with Mat.at(y,x)
Mat R is of type 64FC1 and should be accessed with Mat.at(y,x)
The matrix $R$ is an orthogonal matrix, whereas the matrix $Q$ is not. Note that the matrices $R$ and $Q$ are not similar by the usual definition of matrix similarity.