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+// Exa 1.8
+clc;
+clear;
+close;
+format('v',7)
+// Given data
+R1= 2;// in ohm
+R2= 4;// in ohm
+R3= 1;// in ohm
+R4= 6;// in ohm
+R5= 4;// in ohm
+V1= 10;// in V
+V2= 20;// in V
+//Applying KVL in ABGHA :  I1*(R1+R2) - R2*I2 = V1                   (i)
+//Applying KVL in BCFGB :  I1*R5-I2*(R3+R4+R5)+I3*R4 = 0      (ii)
+//Applying KVL in CDEFC:  R4*I2-I3*(R2+R4)=V2                          (iii)
+A= [(R1+R2) R5 0; -R2 -(R3+R4+R5) R4; 0 R4 -(R2+R4)];
+B= [V1 0 V2];
+I= B*A^-1;// Solving eq(i), (ii) and (iii) by Matrix method
+I1= I(1);// in A
+I2= I(2);// in A
+I3= I(3);// in A
+I6_ohm_resistor= I2-I3;//The current through 6 ohm resistance  in A
+disp(I6_ohm_resistor,"The current through 6 ohm resistance in A is : ")