Solution Step 1 The situation is easily identifiable Step 2 Suppose the firm produces x units of P1, y units of P2 and z units of P3 Since
product P1 requires 3 units of R1, P2 requires 7 units of R1 and P3 requires 5 units of R1
(observe matrix B) and the total number of units, of R1, available is 330, we have
3x + 7y + 5z = 330 (for raw material R1)
Similarly, we have
4x + 9y + 12z = 455 (for raw material R2)
and
3y + 7z = 140 (for raw material R3)
This system of equations can be expressed in matrix form as
3
7
5
4
9
12
0
3
7
330
455
140















=








x
y
z
Rationalised 2023-24
202
MATHEMATICS
Step 3 Using elementary row operations, we obtain
1
0
0
0
1
0
0
0
1
20
35
5















=








x
y
z
This gives x = 20, y = 35 and z = 5 Thus, the firm can produce 20 units of P1, 35
units of P2 and 5 units of P3 to make full use of its available raw material