Given as 8x + 4y + 3z = 18
2x + y + z = 5
x + 2y + z = 5
The given equation can be written in matrix form
= 8(– 1) – 4(1) + 3(3)
= – 8 – 4 + 9
= – 3
Therefore, the above system has a unique solution, given
X = A – 1B
Co-factors of A are
C11 = (– 1)1 + 1 1 – 2 = – 1
C21 = (– 1)2 + 1 4 – 6 = 2
C31 = (– 1)3 + 1 4 – 3 = 1
C12 = (– 1)1 + 2 2 – 1 = – 1
C22 = (– 1)2 + 1 8 – 3 = 5
C32 = (– 1)3 + 1 8 – 6 = – 2
C13 = (– 1)1 + 2 4 – 1 = 3
C23 = (– 1)2 + 1 16 – 4 = – 12
C33 = (– 1)3 + 1 8 – 8 = 0
So, X = 1, Y = 1 and Z = 2
