Let $V=M_2(\mathbb R )$ with $ \left\langle A,B \right\rangle =\operatorname{tr}(B^tA)$. Define $$ T\begin{pmatrix}a& b\\c& d\end{pmatrix}=\begin{pmatrix}3d& 2c\\-b& 4a\end{pmatrix}. $$ Find $T^\ast$.
My attempt: Using the usual isomorphism $[-]_\beta:M_2(\mathbb R)\cong \mathbb R^4$ defined by $\begin{pmatrix}a& b\\c& d\end{pmatrix}\mapsto (a,b,c,d)^t$, let $[T]_\beta$ be the representing matrix of $T$ w.r.t the standard basis $\beta$. Let $G$ be the Gramian. Then
$$ \left\langle TA,B \right\rangle = \left\langle G[TA]_\beta,G[B]_\beta \right\rangle_\mathbb{R^4}=\left\langle G[T]_\beta[A]_\beta,G[B]_\beta \right\rangle_\mathbb{R^4}=\left\langle G[A]_\beta,G[T]_\beta^t[B]_\beta \right\rangle_\mathbb{R^4}$$ Let $T^\ast$ be the operator with $[T^\ast ]_\beta=[T]_\beta ^t$. Then $$\left\langle G[A]_\beta,G[T]_\beta^t[B]_\beta \right\rangle_\mathbb{R^4}=\left\langle G[A]_\beta,G[T^\ast]_\beta[B]_\beta \right\rangle_\mathbb{R^4}=\left\langle G[A]_\beta,G[T^\ast B]_\beta \right\rangle_\mathbb{R^4}=\left\langle A,T^\ast B \right\rangle.$$ So if I understand correctly, all I need to do is find $[T]_\beta^t$ and I'm done. Is this correct?
Yes but note that since you were asked to find the operator $T^{*}$ and not its matrix representation, you need to go back and write the formula for $T^{*}$ similarly to how the formula for $T$ was given to you.