Consider the $4\times 4$ matrix $M$ whose general element $\mu\nu$ is given by $$ M_{\mu\nu} = -2 b_1\delta_{\mu 1} \delta_{\nu 1} + \delta_{\nu\{\lambda}b_\lambda \delta_{\mu\}1} $$
$\mu$ and $\nu$ take the values $1,2,3,4$, so for example $b$ is a 4-component vector. Curly braces denote a sum over even permutations of the indices, e.g. $A_{\{\mu \nu} b_{\rho\}}=A_{\mu\nu}b_\rho + A_{\rho\mu}b_\nu +A_{\nu\rho}b_\mu$. The Einstein summation convention is employed on repeated indices, e.g. $b_\lambda b_\lambda = \sum_{\lambda=1}^4 b_\lambda^2$.
In matrix form, $$ M=\left(\begin{array}{cccc} b_1 & b_2 & b_3 & b_4 \\ b_2 & b_1 & 0 & 0\\ b_3 & 0 & b_1 & 0\\ b_4 & 0 & 0 & b_1 \end{array}\right) $$
The problem is to find an expression for the element $\mu\nu$ of the inverse matrix $M^{-1}$, similar to the expression for $M_{\mu\nu}$: $$ (M^{-1})_{\mu\nu}=\;? $$
Here is the answer:
$$M^{-1}=C\left(\begin{matrix} -b_1^2 & b_1b_2 & b_1b_3 & b_1b_4 \\ b_1b_2 & -b_1^2+b_3^2+b_4^2 & -b_2b_3 & -b_2b_4 \\ b_1b_3 & -b_2b_3 & -b_1^2+b_2^2+b_4^2 & -b_3b_4\\ b_1b_4 & -b_2b_4 & -b_3b_4 & -b_1^2+b_2^2+b_3^2 \end{matrix} \right) $$ Where $$C=(-b_1^3+b_1b_2^2+b_1b_3^2+b_1b_4^2)^{-1}$$
I didn't do anything fancy, just a straight-forward calculation: