Given as 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25
The given equation can be written in matrix form
= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)
= 18 – 12 – 4
= 2
Therefore, the above system has a unique solution, given by
X = A – 1B
Co-factors of A are
C11 = (– 1)1 + 1 (12 – 6) = 6
C21 = (– 1)2 + 1(24 + 4) = – 28
C31 = (– 1)3 + 1(– 12 – 4) = – 16
C12 = (– 1)1 + 2 (0 + 3) = – 3
C22 = (– 1)2 + 1 18 – 2 = 16
C32 = (– 1)3 + 1 – 9 – 0 = 9
C13 = (– 1)1 + 2 (0 – 2) = – 2
C23 = (– 1)2 + 1 (– 6 – 4) = 10
C33 = (– 1)3 + 1 6 – 0 = 6
So, X = -2, Y = 3 and Z = 1
