= Solution
{c}
<Gershgorin circle theorem> places every complex <eigenvalue> of $A$ in at least one disk $D_i=\{z:|z-a_{ii}|\leq\sum_{j\ne i}|a_{ij}|\}$. To prove it, take a nonzero <eigenvector> $v$ and choose $i$ with $|v_i|=\max_j|v_j|>0$. The $i$th <eigenvector> equation gives
$$
|\lambda-a_{ii}|\,|v_i|=\left|\sum_{j\ne i}a_{ij}v_j\right|
\leq\sum_{j\ne i}|a_{ij}|\,|v_i|.
$$
Division by $|v_i|$ proves the disk bound.
Here $C=I$, and the three disks for $B+\mu I$ have real centres $2+\mu$, $3+\mu$, $4+\mu$ and radii $2,3,4$ respectively. Every point in every disk has real part at least $\mu>0$. Hence zero cannot be an <eigenvalue>, proving \b[$B+\mu C$ is nonsingular for every $\mu>0$].
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