= Solution
The <Gershgorin circle theorem> states that every complex <eigenvalue> $\lambda$ of $A=(a_{ij})$ belongs to at least one closed disc
$$
D_i=\left\{z\in\mathbb C:|z-a_{ii}|\leq\sum_{j\ne i}|a_{ij}|\right\}.
$$
To prove it, take an <eigenvector> $v\ne0$ and choose $i$ with $|v_i|=\max_j|v_j|>0$. The $i$th row of $Av=\lambda v$ gives
$$
|\lambda-a_{ii}|\,|v_i|
=\left|\sum_{j\ne i}a_{ij}v_j\right|
\leq\sum_{j\ne i}|a_{ij}|\,|v_j|
\leq|v_i|\sum_{j\ne i}|a_{ij}|.
$$
Division by $|v_i|$ proves the assertion. The proof works for real matrices even when their eigenvalues or eigenvectors are complex.
For $B+\mu I$, the three discs have centres $2+\mu,3+\mu,4+\mu$ and radii $2,3,4$. Every point in their <union> has real part at least $\mu$. Thus for $\mu>0$ no disc contains zero, and zero cannot be an <eigenvalue>. Hence \b[$B+\mu C$ is nonsingular].
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