Is matrix multiplication cheaper than inverse?
Matrix multiplication and matrix inverse have the same asymptotic running time. If we denote the running time of multiplying two $n \times n$ matrices by $T_1(n)$, and that of inverting an $n \times n$ matrix by $T_2(n)$, then this means that there exist constants $A,B$ such that
$$ T_1(n) \leq AT_2(n) \\ T_2(n) \leq BT_1(n). $$
However, the constants $A,B$ can be significant. (Also, numerical stability might be an issue.)

Checking your claim is, however, very easy in practice. Choose your favorite matrix library (e.g. LAPACK), and you can determine empirically $T_1(n),T_2(n)$ for the values of $n$ that you are interested in, on the hardware platform that you are interested in.