= Solution
The <bilinear square finite element> contains the terms $1,x,y,xy$. In anticlockwise corner order beginning at $(0,0)$, its <shape functions> are
$$
N_1=(1-x/h)(1-y/h),\quad N_2=(x/h)(1-y/h),\quad
N_3=xy/h^2,\quad N_4=(1-x/h)(y/h).
$$
The <mass matrix> has entries $M_{ij}=\int_0^h\int_0^hN_iN_j\,dx\,dy$. Each product separates into one-dimensional factors, with $\int_0^1(1-s)^2\,ds=\int_0^1s^2\,ds=1/3$ and $\int_0^1s(1-s)\,ds=1/6$. Diagonal, adjacent-corner and opposite-corner entries are respectively $h^2/9$, $h^2/18$, $h^2/36$. Thus
$$
\boxed{M_E=\frac{h^2}{36}\begin{pmatrix}4&2&1&2\\2&4&2&1\\1&2&4&2\\2&1&2&4\end{pmatrix}.}
$$
Number global grid nodes by $k=i+(n+1)j$, with $x_i=ih$, $y_j=jh$ and $0\leq i,j\leq n$. For an interior node, four elements contribute to its diagonal, two to each axial neighbour and one to each diagonal neighbour. Its global row therefore has the nine-point stencil
$$
\boxed{\frac{h^2}{36}\begin{pmatrix}1&4&1\\4&16&4\\1&4&1\end{pmatrix}.}
$$
In the stated ordering the axial entries occur at columns $k\pm1$ and $k\pm(n+1)$; the four diagonal ones occur at $k\pm(n+1)\pm1$. No boundary degrees of freedom have been removed.
For general valid quadrilaterals, use an <isoparametric finite element>. Map the unit square to an element by $F(\xi,\eta)=\sum_{i=1}^4\mathbf x_iN_i(\xi,\eta)$ and define physical <shape functions> $N_i\circ F^{-1}$. Require a one-to-one nondegenerate map and compatible common edges. On a shared edge the map and the trace of each <shape function> are linear between the same endpoints; the neighbouring traces therefore agree. The assembled functions are continuous and lie in $H^1$, giving a <conforming finite element space>. On a non-parallelogram the pulled-back functions are generally not bilinear polynomials in the physical coordinates themselves.
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