= Solution
{c}
The four row <Gershgorin discs> have centres $9,3,-4,-5$ and radii $1,2,2,1$, respectively. Their real-axis intersections are
$$
[8,10],\qquad[1,5],\qquad[-6,-2],\qquad[-6,-4].
$$
None of the <Gershgorin discs> contains zero, so the <Gershgorin circle theorem> gives
$$
\boxed{0\notin\operatorname{spec}(A),\quad A\text{ is nonsingular}.}
$$
The discs centred at $9$ and $3$ are each isolated. By <Gershgorin eigenvalue counting>, they contain exactly one <eigenvalue> apiece, and both <eigenvalues> are simple. Each must be real: the <characteristic polynomial> is real, so a nonreal root would bring a distinct conjugate root into the same disc. Denote these <eigenvalues> by $\beta\in[8,10]$ and $\alpha\in[1,5]$. The remaining two <eigenvalues> lie in the union of the negative-centred discs.
For the shift $4$, the distances satisfy
$$
|\alpha-4|\leq3,\qquad |\beta-4|\geq4,
\qquad |\gamma-4|\geq6
$$
for either of the other <eigenvalues> $\gamma$, since its real part is at most $-2$. Thus $\alpha$ is uniquely closest to $4$. Once $4$ is checked not to be an <eigenvalue>, the natural full target-disc <interval enclosure> and a conservative asymptotic direction factor are
$$
\boxed{[a,b]=[1,5],\qquad q_4\leq\frac34.}
$$
As always, <inverse iteration> requires a nonzero initial target component. The following calculation verifies the shift and gives sharper information.
For the <tridiagonal matrix>, let $p_j(t)$ be the <determinant> of the leading $j\times j$ block of $tI-A$. Expansion along the last row gives
$$
\begin{aligned}
p_0(t)&=1,&p_1(t)&=t-9,\\
p_2(t)&=(t-3)(t-9)+1=t^2-12t+28,\\
p_3(t)&=(t+4)p_2(t)-p_1(t)=t^3-8t^2-21t+121,\\
p_4(t)&=(t+5)p_3(t)-p_2(t)
=t^4-3t^3-62t^2+28t+577.
\end{aligned}
$$
In particular, $\det A=577$, $p_4(4)=-239$ and $p_4(5)=-583$. Both shifted <linear systems> are nonsingular. Further exact evaluations are
$$
\begin{array}{c|rrrrrrrr}
t&-6&-5&-4&-3&3&4&8&9\\\hline
p_4(t)&121&-113&-79&97&103&-239&-607&181.
\end{array}
$$
The <intermediate value theorem> gives a real root in each of $(-6,-5)$, $(-4,-3)$, $(3,4)$ and $(8,9)$. As $p_4$ has <polynomial degree> four, these are all its roots and each is simple. In particular, $A$ is a <diagonalizable matrix>, $\alpha\in(3,4)$ and $\beta\in(8,9)$. The closest competitor to $\alpha$ at the shift $4$ is $\beta$, so this refinement yields
$$
\boxed{\alpha\in[3,4],\qquad q_4=\frac{4-\alpha}{\beta-4}<\frac14.}
$$
For the shift $5$, the original <Gershgorin discs> only give $|\alpha-5|\leq4$ and $|\beta-5|\geq3$; those bounds do not decide which is closer. The refined <interval enclosures> do:
$$
1<5-\alpha<2,\qquad3<\beta-5<4,
\qquad|\gamma-5|>8
$$
for the negative <eigenvalues>. Therefore the answer for this particular <matrix> is yes:
$$
\boxed{\mu=5:\quad\lambda_*=\alpha\in[3,4],\qquad
q_5=\frac{5-\alpha}{\beta-5}<\frac23.}
$$
Both shifts select the same <eigenvector> line and alternate signs because $\alpha<\mu$. The factor bounds concern asymptotic directional convergence, rather than a universal error reduction from the first iterate for a <nonsymmetric matrix>. The wider interval $[1,5]$ remains a valid full-disc enclosure for both shifts; $[3,4]$ records the stronger information obtained from the <characteristic polynomial>.
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