The names of complexities normally refer to $b$, not the bitlength $n$.

$\mathcal{O}(log\ n) = \mathcal{O}(b)$ is polynomial time, $\mathcal{O}(n) = \mathcal{O}(2^b)$ is exponential time.

Matrix multiplication of a matrix of length $n$ would involve more operations than multiplying all elements of your group with all others. It has complexity of $\mathcal{O}(n^3) = \mathcal{O}(2^{3b})$ which is exponential time - it is not considered effective. In $\mathcal{O}(n^2)$, you could create the multiplication table of your group, which is basically a lookup table. If you have that, you're basically done with the DLP, so it should be obvious, that solving the DLP should be less complex than matrix multiplication.

The names of complexities normally refer to $b$, not the bitlength $n$.

$\mathcal{O}(log\ n) = \mathcal{O}(b)$ is polynomial time, $\mathcal{O}(n) = \mathcal{O}(2^b)$ is exponential time.

Matrix multiplication of a matrix of length $n$ would involve more operations than multiplying all elements of your group with all others. It has complexity of $\mathcal{O}(n^3) = \mathcal{O}(2^{3b})$ which is exponential time - it is not considered effective. In $\mathcal{O}(n^2)$, you could create the multiplication table of your group, which is basically a lookup table. If you have that, you're basically done with the DLP, so it should be obvious, that solving the DLP should be less complex than matrix multiplication.

Also, this does not seem to be consistent. In some cases (like crypto) it makes sense to say that $\mathcal{O}(n)$ is exponential time since the thing that grows is the bitsize, and thus $n$ grows by $2^b$. In other cases, when we're talking outside of crypto $\mathcal{O}(n)$ might be considered polynomial, since n is the thing that grows (like sorting algorithms).