= Solution
For a complex <matrix> $C=(c_{ij})$, Gershgorin's theorem puts every <eigenvalue> in the union of the row discs $D_i=\{z:|z-c_{ii}|\leq R_i\}$, where $R_i=\sum_{j\ne i}|c_{ij}|$. To prove it, choose an <eigenvector> $v$ for an <eigenvalue> $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation gives
$$
|\lambda-c_{ii}|\,|v_i|\leq\sum_{j\ne i}|c_{ij}|\,|v_j|\leq R_i|v_i|.
$$
Division proves $\lambda\in D_i$.
The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ <eigenvalues>, with <algebraic multiplicity>. For completeness, deform the <matrix> from its diagonal part to the original by multiplying all off-diagonal entries by $t\in[0,1]$. Its discs have radii $tR_i$ and stay inside the original discs. Continuous variation of the roots of the <characteristic polynomial> keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.
For the application put $s=-\mu>0$. The <matrix> is
$$
A+sB=\begin{pmatrix}2+2s&-2&0\\-3&5+2s&s\\1&2+s&3+2s\end{pmatrix}.
$$
Its three row discs have positive real centres and leftmost real coordinates
$$
(2+2s)-2=2s,\qquad(5+2s)-(3+s)=2+s,\qquad(3+2s)-(3+s)=s.
$$
All are strictly positive, so zero lies in none of the discs. It cannot be an <eigenvalue>. Thus \b[$A-\mu B$ is nonsingular for every real $\mu<0$].
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