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+
+
+                             // Examle 3.24
+
+               // From the diagram (3.42) Node voltages are
+               // Have { va-vb+0vc = 6 }.......................(1
+               // Apply KCL at Super node 
+               // will get { 0.33va+0.25vb-0.25vc = 2  }.......(2
+               // Apply KCL at node c
+               // will get { 0va-0.25vb+4.5vc = -7  }..........(3
+ 
+              // By using matrix form will get A*X = B  formate 
+
+delta=[1 -1 0 ; 0.33 0.25 -0.25 ; 0 -0.25 0.45];   // value of A
+d=det(delta);                                      // Determinant of A
+
+delta1=[1 6 0 ; 0.33 2 -0.25 ; 0 -7 0.45];         // value of A1 (when 2nd colomn is replace by B) 
+d1=det(delta1);                                    // Determinant of A1
+
+delta2=[1 -1 6 ; 0.33 0.25 2 ; 0 -0.25 -7];        // value of A2 (when 3rd colomn is replace by B) 
+d2=det(delta2);                                    // Determinant of A2
+
+Vb=d1/d;                                           // Voltage at node-b
+Vc=d2/d;                                           // Voltage at node-c
+
+I=(Vb-Vc)/4;                                       // Current through 4 ohm resistor (I)
+disp(' Current through 4 ohm resistor = '+string(I)+' Amp');
+
+
+
+                     //   p 82        3.24
+\ No newline at end of file