= Solution
Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero <quadratic residue> modulo $11$. The square subgroup is
$$
Q=\langle4\rangle=\{1,4,5,9,3\}\cong C_5.
$$
Matrix multiplication becomes
$$
(r,b)(s,c)=(rs,b+rc),
\qquad
(r,b)^{-1}=(r^{-1},-r^{-1}b).
$$
Thus $H$ is the <affine semidirect product of cyclic groups of orders eleven and five>
$$
H\cong(\mathbb F_{11},+)\rtimes Q\cong C_{11}\rtimes C_5.
$$
There are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since
$$
(4,0)(1,1)=(4,4)\ne(4,1)=(1,1)(4,0).
$$
Let
$$
N=\{(1,b):b\in\mathbb F_{11}\}\cong C_{11}.
$$
Conjugation by the complement gives
$$
(r,0)(1,b)(r,0)^{-1}=(1,rb).
$$
Hence the ten nonidentity translations split into the two orbits
$$
T_Q=\{(1,b):b\in Q\},
\qquad
T_N=\{(1,b):b\in2Q\},
$$
each of size five. For $r\ne1$,
$$
(1,t)(r,b)(1,t)^{-1}
=(r,b+(1-r)t).
$$
Since $1-r\ne0$, varying $t$ gives every $b\in\mathbb F_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, with
$$
C_k=\{(4^k,b):b\in\mathbb F_{11}\},
\qquad 1\leq k\leq4,
$$
the seven conjugacy classes are
$$
\{1\},\quad T_Q,\quad T_N,\quad C_1,\quad C_2,\quad C_3,\quad C_4,
$$
of sizes $1,5,5,11,11,11,11$, agreeing with the <conjugacy classes in the affine semidirect product of orders eleven and five>.
The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in $\mathbb F_{11}$. Hence the <commutator subgroup> is $N$, and the <abelianization> is
$$
H^{\mathrm{ab}}\cong H/N\cong C_5.
$$
Put $\zeta=e^{2\pi i/5}$. The <one-dimensional characters factor through the abelianization>, giving five characters
$$
\chi_j(4^k,b)=\zeta^{jk},
\qquad 0\leq j\leq4.
$$
For the remaining characters, put $\omega=e^{2\pi i/11}$ and define a character of $N$ by
$$
\theta_m(1,b)=\omega^{mb}.
$$
The complement has two free orbits on the nontrivial $\theta_m$, indexed by $m\in Q$ and $m\in2Q$. By <induction from an abelian normal subgroup with a free character orbit>, the characters
$$
\psi_Q=\operatorname{Ind}_N^H\theta_1,
\qquad
\psi_N=\operatorname{Ind}_N^H\theta_2
$$
are irreducible of degree five and vanish outside $N$.
Set
$$
\eta=\sum_{q\in Q}\omega^q
=\frac{-1+i\sqrt{11}}2.
$$
The <quadratic periods modulo eleven> give the nonsquare sum $\overline\eta=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete <character table> is
$$
\begin{array}{c|ccc|cccc}
&1&T_Q&T_N&C_1&C_2&C_3&C_4\\
\text{class size}&1&5&5&11&11&11&11\\ \hline
\chi_0&1&1&1&1&1&1&1\\
\chi_1&1&1&1&\zeta&\zeta^2&\zeta^3&\zeta^4\\
\chi_2&1&1&1&\zeta^2&\zeta^4&\zeta&\zeta^3\\
\chi_3&1&1&1&\zeta^3&\zeta&\zeta^4&\zeta^2\\
\chi_4&1&1&1&\zeta^4&\zeta^3&\zeta^2&\zeta\\
\psi_Q&5&\eta&\overline\eta&0&0&0&0\\
\psi_N&5&\overline\eta&\eta&0&0&0&0
\end{array}.
$$
Finally,
$$
5\cdot1^2+2\cdot5^2=55=|H|,
$$
and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the <irreducible characters of the affine semidirect product of orders eleven and five>.
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