= Solution
{c}
Write $u_h=U_B\phi_B$. On $ABD,BCE,BDE$, the <gradients> are $(0,-1)$, $(-1,0)$ and $(-1,-1)$, respectively. Thus
$$
\int_\Omega|\nabla\phi_B|^2=\tfrac12+\tfrac12+1=2.
$$
The integral of an affine nodal function over a triangle equals its area times the mean of its three vertex values. Each supported triangle therefore contributes $1/6$ to $\int\phi_B$, giving total $1/2$. The reduced system is $[2][U_B]=[1/2]$, and the complete nodal answer is
$$
\boxed{U_B=\tfrac14,\qquad U_A=U_C=U_D=U_E=U_F=0.}
$$
Only one algebraic unknown remains despite the two-dimensional domain.
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