= Solution
Fix the <quantum enveloping algebra of sl2> convention
$$
KEK^{-1}=q^2E,\qquad KFK^{-1}=q^{-2}F,\qquad
EF-FE=\frac{K-K^{-1}}{q-q^{-1}},\qquad
[j]=\frac{q^j-q^{-j}}{q-q^{-1}}.
$$
The printed pair of matrices contains an error. On its first <basis> vector, the printed matrices give $(EF-FE)v_0=\epsilon[2]^2v_0$, whereas the defining commutator gives $\epsilon[2]v_0$. Since $q$ is not a <root of unity>, $[2]\ne0,1$: $[2]=0$ would give $q^2=-1$, and $[2]=1$ would give $q^2-q+1=0$. Thus \b[the two printed matrices cannot represent this algebra]. Keeping the printed $F$ matrix requires exchanging the two nonzero entries of the printed $E$ matrix. We derive that corrected pair rather than assume a classification theorem.
As $q$ is not a <root of unity>, successive applications of $E$ multiply $K$ <eigenvalues> by $q^2$. Finiteness of the <spectrum> therefore makes $E$ a <nilpotent operator>. Start from a $K$ <eigenvector> and apply $E$ until the last nonzero vector $v$; then $Ev=0$ and $Kv=\lambda v$, with $\lambda\ne0$. The defining commutator gives the <highest-weight string for quantum sl2>
$$
EF^jv=[j]\frac{\lambda q^{1-j}-\lambda^{-1}q^{j-1}}{q-q^{-1}}F^{j-1}v.
$$
For completeness, at $j=1$ this is the commutator applied to $v$. To advance from $j$ to $j+1$, write $EF^{j+1}v=F(EF^jv)+(K-K^{-1})F^jv/(q-q^{-1})$, use $KF^jv=\lambda q^{-2j}F^jv$, and combine the two coefficients; their sum is the displayed coefficient at $j+1$.
Let $N$ be the first index with $F^Nv=0$. Such an index exists because nonzero $F^jv$ have distinct $K$ <eigenvalues> $\lambda q^{-2j}$. The <linear span> of $v,Fv,\ldots,F^{N-1}v$ is stable under $K,E,F$, so simplicity makes it all of $V$. The distinct <eigenvalues> make these vectors independent; hence $N=3$. Applying the string formula to $F^3v=0$ gives
$$
0=[3]\frac{\lambda q^{-2}-\lambda^{-1}q^2}{q-q^{-1}}F^2v.
$$
As $[3]\ne0$ and $F^2v\ne0$, $\lambda^2=q^4$, so $\lambda=\epsilon q^2$ with $\epsilon=\pm1$.
Choose the <basis> $v_0=v$, $v_1=Fv/[2]$, $v_2=F^2v/[2]$. Then $Fv_0=[2]v_1$, $Fv_1=v_2$, $Ev_1=\epsilon v_0$ and $Ev_2=\epsilon[2]v_1$. The correct matrices in this <basis> are
$$
\boxed{E=\epsilon\begin{pmatrix}0&1&0\\0&0&[2]\\0&0&0\end{pmatrix},\quad
F=\begin{pmatrix}0&0&0\\{[2]}&0&0\\0&1&0\end{pmatrix},\quad
K=\epsilon\begin{pmatrix}q^2&0&0\\0&1&0\\0&0&q^{-2}\end{pmatrix}.}
$$
Alternatively, preserving the printed $E$ instead requires $F_{21}=1$, $F_{32}=[2]$. Both corrected pairs are related by a diagonal change of <basis>.
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