Solve the nine internal equations from (4 to 12) to find the values of the nine internal voltages by creating a code in the Matlab program

Solve the nine internal equations from 4 to 12 to find the values of the nine internal voltages by creating a code in the Matlab program (V4, 126) (V44-6) (14,3 =6) V474 V3,3 - (V2,4 =o) vi 1,3 Van (Vo, eso) (120) V. 3.) 0.4 --Ec Veri), CV, edic V4:ss=6 { dou), Wwe), (V.1) =0 , , , Vies. C.Vejo),V5,01$ = 0 Vi-19 - 4V +V;-1; + Vi, j+1 + V1-1 = 0 We substitute the points into equation (3); ( (3); At point V., (i=1, j =1) in eq(3). V2,1 - 4V 1,1 + 10,1 + V 1.2 + V 1,0 = 0 ----(4") At point V1,2 (i=1, j= 2 ) in eq(3"). V212 - 4V 1,2 +V0.2 + V 13 + V1.1 = 0 ----(5") At point V1,3 (i=1, j= 3 ) in eq(3). V2.3 - 4V 1,3 +V03 + V 1.4 + V 12 = 0 -- At point V2,1 (i=2, j=1 ) in eq(3'). V3,1 - 4V 2,1 + V1,1 + V2,2 + V 2,0 = 0 --- (7) At point V2,2 (i=2, j= 2 ) in eq(3). V3,2 - 4V 2.2 + V1.2 + V2.3 + V2.1 = 0 ----(8) At point V2,3 (i=2, j = 3 ) in eq(3). V3,3 - 4V 2,3 + V1,3 + V 2.4 + V2.2 = 0 ----(9) At point V3,1 (i = 3, j = 1) in eq(3"). V4.1 - 4V 3,1 + V2,1 + V3,2 + V 3,0 = 0 ----(10") At point V3,2 (i= 3, j = 2 ) in eq(3"). V4.2 2 - 4V3.2 +12,2 + V3,3 + V3,1 = 0 ----(11") At point V3,3 (i= 3, j= 3 ) in eq(3) V4,3 - 4V 3,3 + V2,3 + V3,4 + V3,2 = 0 ---- (12")