Question 2.14: Write the mesh equations directly in matrix form for the cir...
Write the mesh equations directly in matrix form for the circuit of Figure 2.36.

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The matrix equation is:
\left[\begin{matrix} (R_2+R_4+R_5)& -R_2 & -R_5 \\ -R_2& (R_1+R_2+R_3)& -R_3\\- R_5 & -R_3 & (R_3+R_5+R_6) \end{matrix} \right] \left[\begin{matrix} i_1 \\ i_2 \\ i_3 \end{matrix} \right] = \left[\begin{matrix} -v_A+v_B \\ v_A \\ -v_B \end{matrix} \right]
Notice that mesh 1 includes R_2, R_4 , and R_5, so the r_{11} element of R is the sum of these resistances. Similarly, mesh 2 contains R_1, R_2, and R_3, so r_{22} is the sum of these resistances. Because R_2 is common to meshes 1 and 2, we have r_{12} = r_{21} = −R_2.
Similar observations can be made for the other elements of R.
As we go around mesh 1 clockwise, we encounter the positive reference for v_A first and the negative reference for v_B first, so we have v_1 = −v_A + v_B, and so forth.
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