Given as 2x + 6y = 2
3x – z = -8
2x – y + z = -3
The given equation can be written in matrix form
= 2(0 – 1) – 6(3 + 2)
= – 2 – 30
= – 32
So, the above system has a unique solution, given by
X = A – 1B
Co-factors of A are
C11 = (– 1)1 + 1 0 – 1 = – 1
C21 = (– 1)2 + 16 + 0 = – 6
C31 = (– 1)3 + 1 – 6 = – 6
C12 = (– 1)1 + 2 3 + 2 = 5
C22 = (– 1)2 + 1 2 – 0 = 2
C32 = (– 1)3 + 1 – 2 – 0 = 2
C13 = (– 1)1 + 2 – 3 – 0 = – 3
C23 = (– 1)2 + 1 – 2 – 12 = 14
C33 = (– 1)3 + 1 0 – 18 = – 18
So, X = -2, Y = 1 and Z = 2
