This looks like a homework question. Because, there is not other way to represent the inverse of the transformation without using the provided rotation matrix and translation vector.
I guess the person who asked the question would like you to see that the form of the inverse looks "nice" because the last row of the transformation ins [0, 0, 0, 1]
You could derive this by hand for a generic 4x4 matrix. See here for a formula.
Another way to derive this is to go to first principles. The inverse of a matrix $A$ is a matrix $B$ such that $AB=I$.
Let us look at the rotation part. Rotations are members of the Special Orthogonal group $SO(3)$ and have the property that for $R\in SO(3)$, and $det(R)=+1$ $R^{-1} = R^T$.
Look at a rigid transformation with rotation only, i.e.
$\begin{pmatrix}R & 0 \\ 0^T & 1\end{pmatrix}$, its inverse is:
$\begin{pmatrix}R^T & 0\\ 0^T & 1\end{pmatrix}$ because:
$\begin{pmatrix}
R & 0 \\ 0^T & 1
\end{pmatrix}
\begin{pmatrix}R^T & 0\\ 0^T & 1\end{pmatrix} =
\begin{pmatrix}
RR^T & 0 \\ 0^T & 1
\end{pmatrix} = \begin{pmatrix}I & 0\\ 0^T & 1 \end{pmatrix} = I
$
Now, if we have a translation vector you should be able to see that the inverse is given by:
$\begin{pmatrix}
R^T & -R^Tt\\ 0^T & 1
\end{pmatrix}.$
Another way of deriving this is to forget about the matrix form and look at the effect of a rigid-body transformation on $3$D point. Let
$Y = RX + t$ be a transformed point with rotation matrix $R$ and translation vector $t$. The inverse transform is a rotation matrix and translation vector such that we get back the point $X$, i.e.:
$X = R^T(Y - t) = R^TY - R^T t$. Hence, the inverse rotation is simply $R^T$ and the inverse translation is $-R^T t$. Writing this in homogeneous coordinates, the inverse transform is:
$T^{-1} = \begin{pmatrix}R^T & -R^T t\\ 0^T & 1\end{pmatrix}$
