(i) Given as 6x + 4y = 2
9x + 6y = 3
The given equation can be written in matrix form

Here

|A| = 36 – 36 = 0
Therefore, A is singular, Now X will be consistence if (Adj A) x B = 0
C11 = (– 1)1 + 1 6 = 6
C12 = (– 1)1 + 2 9 = – 9
C21 = (– 1)2 + 1 4 = – 4
C22 = (– 1)2 + 2 6 = 6

So, AX = B will be infinite solution.
Suppose y = k
So,
6x = 2 - 4k or 9x = 3 - 6k
X = (1 - 2k)/3
So, X = (1 - 2k)/3 and Y = k.
(ii) Given as 2x + 3y = 5
6x + 9y = 15

Here

|A| = 18 – 18 = 0
Therefore, A is singular,
Then X will be consistence if (Adj A) x B = 0
C11 = (– 1)1 + 1 9 = 9
C12 = (– 1)1 + 2 6 = – 6
C21 = (– 1)2 + 1 3 = – 3
C22 = (– 1)2 + 2 2 = 2

So, AX = B will be infinite solution.
Suppose y = k
So,

Therefore, X = (5 - 3k)/2, Y = k