Handloom Stock System with a Singular Coefficient Matrix
A handloom cooperative in Sualkuchi produces three items: stoles x, shawls y and scarves z. Each stole uses 2 units of silk, 1 of cotton and 1 of zari; each shawl uses 3, 2 and 1; each scarf uses 1, 1 and 0. The cooperative has 90 units of silk, 55 of cotton and 35 of zari. The manager writes the stock equations as 2x+3y+z=90, x+2y+z=55 and x+y=35. An order requires at least 5 scarves and at most 12 shawls. The accountant computes the determinant of the coefficient matrix, finds it zero, and claims the order cannot be completed. The production supervisor disagrees, pointing out that the stock figures may still be consistent and that several production plans might use all materials. They ask for the general solution, the feasible range of shawls under the order, and a plan that maximises stoles without violating the order.
1. Write the system as AX=B, compute det A, and classify the solutions. Does det A=0 mean no production plan exists?
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Here A = [[2,3,1],[1,2,1],[1,1,0]], X = [x,y,z]^T and B = [90,55,35]^T. Expanding along the third row: det A = 1(3·1−1·2) − 1(2·1−1·1) + 0 = (3−2) − (2−1) = 1−1 = 0. The rows are dependent: R2 − R1 = (−1,−1,0) = −R3, so R3 = R1 − R2. Checking the constants: 90 − 55 = 35, so the augmented system is consistent. Therefore rank A = rank[A|B] = 2 < 3, and there are infinitely many solutions. A zero determinant means no unique solution, not necessarily no solution.
2. Find the general solution and apply the order constraints. Give two valid production plans.
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From x+y=35, x=35−y. Substitute into the silk equation: 2(35−y)+3y+z=90, so 70−2y+3y+z=90, giving y+z=20, hence z=20−y. The cotton equation then gives (35−y)+2y+(20−y)=55, which is identically true. Thus (x,y,z)=(35−y, y, 20−y). For non-negative outputs, 0≤y≤20. The order requires z≥5, so 20−y≥5, i.e. y≤15; and y≤12 because at most 12 shawls are allowed. Hence 0≤y≤12. Two plans are y=0: (35,0,20), and y=12: (23,12,8). Both use all three materials exactly.
3. Which feasible plan maximises stoles? If the order also insists on at least 5 shawls, what is the new maximum and plan?
Read the worked answer
Since x=35−y, the number of stoles decreases as y increases. Without a shawl minimum, the largest x occurs at the smallest feasible y, namely y=0. This gives (x,y,z)=(35,0,20), with 20 scarves, satisfying z≥5 and y≤12. If the order additionally requires y≥5, then y must lie in [5,12]. The largest x is again at the smallest y, so y=5 gives x=30 and z=15. The plan is (30,5,15). In both cases the maximum is determined by the constraint boundary, not by a unique solution of the linear system.