= Solution
There is a defect in the stated geometric interpretation of the data. The supplied numbers cannot be <Euclidean distances>, because
$$
d(B,D)=7.0>4.2+1.2=d(B,A)+d(A,D).
$$
They violate the triangle inequality; independently, $d(C,D)=10.3>5.9+1.2$. Nevertheless, both clustering algorithms are defined for this symmetric <dissimilarity matrix>. The calculations below retain the supplied numbers and use <agglomerative clustering of nonmetric dissimilarities>, rather than silently inventing Euclidean replacement data.
In <single-linkage clustering>, the intercluster dissimilarity is
$$
d_{\min}(G,H)=\min_{r\in G,\,t\in H}d(r,t).
$$
The closest pair is $A,D$, merging at $1.2$. Among the remaining clusters the next closest pair is $B,E$, merging at $2.6$. The three intercluster dissimilarities are now
$$
\begin{aligned}
d_{\min}(AD,BE)&=\min(4.2,6.1,7.0,7.8)=4.2,\\
d_{\min}(AD,C)&=\min(5.9,10.3)=5.9,\\
d_{\min}(BE,C)&=\min(7.6,5.4)=5.4.
\end{aligned}
$$
Consequently $AD$ merges with $BE$ at $4.2$, using the cross-pair $A,B$. The remaining point $C$ joins at $\min(5.9,7.6,10.3,5.4)=5.4$, using $C,E$. The <dendrogram> thus has
$$
\boxed{AD\text{ at }1.2;\quad BE\text{ at }2.6;\quad AD+BE\text{ at }4.2;\quad ADBE+C\text{ at }5.4.}
$$
The close pairs $AD$ and $BE$ are the clearest early groupings. A cut between $2.6$ and $4.2$ produces these two pairs and the singleton $C$; a cut between $4.2$ and $5.4$ produces $ADBE$ and $C$. <Single-linkage clustering> builds chains through close cross-pairs, so the merge at $4.2$ does not mean every pair in $ADBE$ is that close: $d(D,E)=7.8$. The data alone do not prescribe a unique number of clusters.
For <complete-linkage clustering>, replace the minimum by the maximum:
$$
d_{\max}(G,H)=\max_{r\in G,\,t\in H}d(r,t).
$$
The first two merges remain $AD$ at $1.2$ and $BE$ at $2.6$. Subsequently
$$
\begin{aligned}
d_{\max}(AD,BE)&=7.8,\\
d_{\max}(AD,C)&=10.3,\\
d_{\max}(BE,C)&=7.6.
\end{aligned}
$$
Thus $C$ joins $BE$ at $7.6$. The final maximum between $AD$ and $BCE$ is $d(D,C)=10.3$, giving
$$
\boxed{AD\text{ at }1.2;\quad BE\text{ at }2.6;\quad BE+C\text{ at }7.6;\quad AD+BCE\text{ at }10.3.}
$$
Both the topology and the later heights of the <dendrogram> change: <complete-linkage clustering> attaches $C$ to $BE$ before joining that group to $AD$, rather than merging the two pairs first. Its control of the largest within-cluster separation makes it less susceptible to the chaining seen with <single-linkage clustering>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-46-dendrograms.png]
{title=Single-linkage and complete-linkage dendrograms for the supplied five-subject dissimilarities}
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