Let us assume a 4-node network, described by $\dot x = A x + B u $, where
$$ A=\begin{pmatrix} 0 & 0 & 0 & 0 \\\ b & 0 & 0 & 0 \\\ c & 0 & 0 & e \\\ d & 0 & 0 & 0 \end{pmatrix}. $$
What is the intuitive difference from an engineering point of view between choosing the input to be a matrix
$$B=\begin{pmatrix} b_1 & 0 & 0 & 0 \\\ 0 & b_2 & 0 & 0 \\\ 0 & 0 & b_3 & 0 \\\ 0 & 0 & 0 & b_4 \end{pmatrix} $$ and the input to be a vector $$ B=\begin{pmatrix} b_1 \\\ b_2 \\\ b_3 \\\ b_4 \end{pmatrix}? $$
As @jnez71 proposed the first $\boldsymbol{B}$ does give you more control authority, because your input vector $\boldsymbol{u}=[u_1,u_2,u_3,u_4]^T$ has four components instead of a scalar component $u$ for the vector $\boldsymbol{b}$. Because of this, you have a higher degree of freedom for the control of your system.
In order to see this, we can look at a simplified system
$$\dot{x}_1=x_1 + u_1$$ $$\dot{x}_2=x_2 - u_2$$
we can quickly see that $u_1=-x_1-k_1x_1$ and $u_2=x_2+k_2x_2$ will drive the system to the origin if $k_1$ and $k_2$ are positive.
If we look at the same system but with only one input we will have
$$\dot{x}_1=x_1 + u$$ $$\dot{x}_2=x_2 - u.$$
If we apply $u=-x_1-k_1x_1$, we can stabilize the first equation for $k_1>0$. The problem is now that the second equation will now be directly affected by this choice of $u$.
$$\dot{x}_2=x_2+x_1+k_1x_1$$
Hence, we cannot freely choose $u$ without constraining the values of $k_1$ which will lead to an asymptotic behavior of the system. It turns out that for this example we will not be able to stabilize the system with such a control law.
I hope you can see how we lost control authority by having a vector instead of a matrix as the input matrix.