The prices of three commodities P,Q and R and ₹ x,y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. C purchases 1 unit of Q. B purchases of Q and 6 units of R. In the process A, B and C earn ₹6000, ₹5000 and ₹13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using the matrix method.
Let the numbers are x, y, z
3x + 5y – 4 z = 6000 …… (i)
Also,
2x – 3y + z = 5000 …… (ii)
Again,
– x + 4y + 6z = 13000 ……(iii)
A X = B
|A| = 3(– 18 – 4) – 2(30 + 16) – 1(5 – 12)
= 3(– 22) – 2(46) + 7
= – 66 – 92 + 7
= – 151
Hence, the unique solution given by x = A – 1B
C11 = (– 1)1 + 1 (– 18 – 4) = – 22
C12 = (– 1)1 + 2 (12 + 1) = – 13
C13 = (– 1)1 + 3 (8 – 3) = 5
C21 = (– 1)2 + 1 (30 + 16) = – 46
C22 = (– 1)2 + 2 (18 – 4) = 14
C23 = (– 1)2 + 3 (12 + 5 ) = – 17
C31 = (– 1)3 + 1 (5 – 12) = – 7
C32 = (– 1)3 + 2 (3 + 8) = – 11
C33 = (– 1)3 + 3 (– 9 – 10) = – 19
Adj A = 
X = A – 1 B = 
X = 
X = 
= 
Hence, x = 3000, y = 1000 and z = 2000