Q: By using matrices determine current I1 , I2 & I3
Solution: We have to calculate the value of currents I1 ,I2 and I3. By using matrices proceed like
By applying KCL eq. at node 1
-1m + (1/6k)(V1 - V2) + (1/12k) (V1 - 0) + =0
(1/12k + 1/6k)V1 - (1/6k) V2 =1mA ....................(i)
By applying KCL eq. for node 2
4m - (1/6k)(V1 - V2) + (1/6k)(V2 -0 )=0
Which may be expressed like
(1/6k + 1/6k) V2 -(1/6k)V1 = - 4mA .....................(ii)
By simplifying these two equations (i) & (ii)
V1/4k - V2/6k = 1mA
-V1/6k + V2/3k = -4mA
In matrix form
The circuit equations may be solve utilizing matrix analysis. The common form of matrix equation will be
A X = Y or GV=I we have to determine X or V
where in this X or V = 
The solution of the matrix equation is
X = A-1 Y OR V = G-1 I ............... (iii)
To determine A-1or G-1 we know that G-1 = Adj(G) / |G| or A-1 = Adj(A) / |A|
Ad joint of the coefficient matrix A will be
Adj (A) =
And the determinant shall be
| A| = (1/3k)(1/4k) - (-1/6k)(-1/6k)
= 1/18k2
A-1 = Adj(A) / |A|
Put the value of A-1 in equation (iii) we get
By solving out the above equation we get
Therefore
V1 = - 6Volts
V2 =- 15 Volts
By knowing the voltages we may find all currents utilizing OHM's LAW
I1 = V1/12k
=-6/12k =-1/2mA
I1 =-0.5mA
Now I2 may be determine like
I2 = (V1 - V2)/6k
= -6 - (-15)/6k
I2= 3/2mA
Now for calculating value of I3
I3 = V2/6k= -15/6k
I3= - 5/2mA