${\displaystyle {\begin{array}{l}K_{11}=k_{11}^{(1)}={\dfrac {9}{16}}\\K_{12}=k_{12}^{(1)}={\dfrac {3{\sqrt {3}}}{16}}\\K_{33}=k_{33}^{(1)}+k_{11}^{(2)}={\dfrac {9}{16}}+{\dfrac {5}{2}}=3.0625\\K_{34}=k_{34}^{(1)}+k_{12}^{(2)}={\dfrac {3{\sqrt {3}}}{16}}+\left(-{\dfrac {5}{2}}\right)=-2.175\\K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.175\\K_{44}=k_{44}^{(1)}+k_{22}^{(2)}={\dfrac {3}{16}}+{\dfrac {5}{2}}=2.6875\end{array}}}$

In your article, replace the image with:

<math>\begin{array}{l}
K_{11}=k_{11}^{(1)}=\dfrac{9}{16}\\
K_{12}=k_{12}^{(1)}=\dfrac{3\sqrt 3}{16}\\
K_{33}=k_{33}^{(1)}+k_{11}^{(2)}=\dfrac{9}{16}+\dfrac{5}{2}=3.0625\\
K_{34}=k_{34}^{(1)}+k_{12}^{(2)}=\dfrac{3\sqrt 3}{16}+\left(-\dfrac{5}{2}\right)=-2.175\\
K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.175\\
K_{44}=k_{44}^{(1)}+k_{22}^{(2)}=\dfrac{3}{16}+\dfrac{5}{2}=2.6875
\end{array}</math>

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