(i) Given as x + y + z = 6
x + 2z = 7
3x + y + z = 12
The given equation can be written in matrix form
= 1(– 2) – 1(1 – 6) + 1(1)
= – 2 + 5 + 1
= 4
Therefore, the above system has a unique solution, given by
X = A – 1B
Co-factors of A are
C11 = (– 1)1 + 1 0 – 2 = – 2
C21 = (– 1)2 + 1 1 – 1 = 0
C31 = (– 1)3 + 1 2 – 0 = 2
C12 = (– 1)1 + 2 1 – 6 = 5
C22 = (– 1)2 + 1 1 – 3 = – 2
C32 = (– 1)3 + 1 2 – 1 = – 1
C13 = (– 1)1 + 2 1 – 0 = 1
C23 = (– 1)2 + 1 1 – 3 = 2
C33 = (– 1)3 + 1 0 – 1 = – 1
So, X = 3, Y = 1 and Z = 2
(ii)Given as (2/x) + (3/y) + (10/z) = 4,
(4/x) – (6/y) + (5/z) = 1,
(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0
The given equation can be written in matrix form
AX = B
Now,
|A| = 2(75) – 3(– 110) + 10(72)
= 150 + 330 + 720
= 1200
Therefore, the above system has a unique solution, given by
X = A – 1B
Co-factors of A are
C11 = (– 1)1 + 1 120 – 45 = 75
C21 = (– 1)2 + 1 – 60 – 90 = 150
C31 = (– 1)3 + 1 15 + 60 = 75
C12 = (– 1)1 + 2 – 80 – 30 = 110
C22 = (– 1)2 + 1 – 40 – 60 = – 100
C32 = (– 1)3 + 1 10 – 40 = 30
C13 = (– 1)1 + 2 36 + 36 = 72
C23 = (– 1)2 + 1 18 – 18 = 0
C33 = (– 1)3 + 1 – 12 – 12 = – 24
So, X = 2, Y = 3 and Z = 5
