KVL arround abca loop:

${\displaystyle I_{1}*R_{1}+(I_{1}-I_{3})*R_{2}+(I_{1}-I_{2})*R_{3}=-V_{s}}$


Therefore

${\displaystyle I_{1}(R_{1}+R_{2}+R_{3})-I_{2}(R_{3})-I_{3}(R_{2})=-V_{s}}$   ...............   (1)


KVL arround acda loop:

 ${\displaystyle (I_{2}-I_{1})*R_{3}+(I_{2}-I_{3})*R_{4}+I_{2}*R_{5}=V_{s}}$


Therefore

${\displaystyle -I_{1}(R_{3})+I_{2}(R_{3}+R_{4}+R_{5})-I_{3}(R_{4})=V_{s}}$   ...............   (2)


KVL arround bcdb loop:

 ${\displaystyle I_{3}*R_{6}+(I_{3}-I_{2})*R_{4}+(I_{3}-I_{1})*R_{2}=0}$


Therefore

${\displaystyle -I_{1}(R_{2})-I_{2}(R_{4})+I_{3}(R_{2}+R_{4}+R_{6})=0}$   ...............   (3)


Now we can create a matrix with the above equations as follows:

${\displaystyle {\begin{bmatrix}(R_{1}+R_{2}+R_{3})&(-R_{3})&(-R_{2})\\(-R_{3})&(R_{3}+R_{4}+R_{5})&(-R_{4})\\(-R_{2})&(-R_{4})&(R_{2}+R_{4}+R_{6})\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-V_{s}\\V_{s}\\0\end{bmatrix}}}$


The following matrix is the above with values substituted:

${\displaystyle A.{\vec {X}}={\vec {Y}}}$${\displaystyle {\begin{bmatrix}10220&-10000&-20\\-10000&30000&-5000\\-20&-5000&6020\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-9\\9\\0\end{bmatrix}}}$

Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:

Solving determinants of:

• Matrix A  : General matrix A from KVL equations
• Matrix A1 : Genral Matrix A with Column 1 substituted by ${\displaystyle {\vec {Y}}}$.
• Matrix A2 : Genral Matrix A with Column 2 substituted by ${\displaystyle {\vec {Y}}}$.
• Matrix A3 : Genral Matrix A with Column 3 substituted by ${\displaystyle {\vec {Y}}}$.

As follows:

${\displaystyle detA=9.86\times 10^{11}}$

${\displaystyle detA1=-857700000}$

${\displaystyle detA2=11016000}$

${\displaystyle detA3=6300000}$


Now we can use the solved determinants to arrive at solutions for Mesh Currents ${\displaystyle I_{1};I_{2}andI_{3}}$ as follows:

1. ${\displaystyle I_{1}={\frac {detA1}{detA}}=-0.00086968A}$

2. ${\displaystyle I_{2}={\frac {detA2}{detA}}=0.00001117A}$

3. ${\displaystyle I_{3}={\frac {detA3}{detA}}=0.000006388A}$

Now we can solve for the current through ${\displaystyle R_{3}}$ as follows:

${\displaystyle I_{R_{3}}=I_{1}-I_{2}=-0.881mA}$


The negative sign means that ${\displaystyle I_{R_{3}}}$ is flowing in the direction of ${\displaystyle I_{2}}$.