Question 1 OR (b)

Daily output of three products from machine hours (matrix inverse)

6 marks

Time to write: About 12 minutes
Topic: Matrices and determinants

Question

Question 1 OR (b) · 6 marks

A firm makes three products X, Y and Z, each processed by three machines M₁, M₂ and M₃ until complete. M₁ can process 30 units of X or 60 units of Y or 90 units of Z per hour. M₂ can process 60 units of any one product per hour. M3 can process, per hour, 60 or 30 or 120 units of products X, Y and Z respectively. Machines M₁, M₂ and M₃ have 12,12 and 13 processing hours available respectively. Use the matrix inverse or Cramer's rule to calculate how many units of the three products can be produced per day?

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

Before you write

Read off the question and its marks

12 minfor 6 marks

Read from the question

calculateFormula first, then the working, then the answer with its unit (₹, %, units).

What the paper leaves open

As printed, the third machine is written "M3" once without a subscript, and the hours are written "12,12" without a space; both are kept.

Write it in this order

Read the data; note exactly what is asked 1 min
1 Equations from machine hours 2 marks 3 min
2 Determinant, adjoint and inverse 3 marks 5 min
3 Units of X, Y, Z with check 1 mark 2 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

123

Short on time? Step 2 carries 3 of the 6 marks: guard its minutes.

Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.

Solution

Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.

Let x, y and z be the units of X, Y and Z produced a day. Each product passes through all three machines. The time a unit takes on a machine is the reciprocal of the machine's hourly rate for it, so the hours used on each machine must equal the hours available (12, 12 and 13):

M₁: x/30 + y/60 + z/90 = 12 M₂: x/60 + y/60 + z/60 = 12 M₃ (M3 in the question): x/60 + y/30 + z/120 = 13

Multiplying by 180, 60 and 120 respectively to clear fractions:

6x + 3y + 2z = 2,160; x + y + z = 720; 2x + 4y + z = 1,560.

Matrix inverse method. A = [[6, 3, 2], [1, 1, 1], [2, 4, 1]], B = [2160, 720, 1560]ᵀ.

|A| = 6(1 − 4) − 3(1 − 2) + 2(4 − 2) = −18 + 3 + 4 = −11 ≠ 0, so a unique solution exists.

Cofactors: A₁₁ = −3, A₁₂ = 1, A₁₃ = 2; A₂₁ = 5, A₂₂ = 2, A₂₃ = −18; A₃₁ = 1, A₃₂ = −4, A₃₃ = 3.

adj A = [[−3, 5, 1], [1, 2, −4], [2, −18, 3]]; A⁻¹ = −(1/11) adj A.

X = A⁻¹B = −(1/11) × [−3(2160) + 5(720) + 1(1560), 1(2160) + 2(720) − 4(1560), 2(2160) − 18(720) + 3(1560)]ᵀ = −(1/11) × [−1,320, −2,640, −3,960]ᵀ = [120, 240, 360]ᵀ.

Daily production: X = 120 units, Y = 240 units, Z = 360 units.

Machine	Hours on X	Hours on Y	Hours on Z	Hours used
M₁	4	4	4	12
M₂	2	4	6	12
M₃	2	8	3	13

Check: every machine's hours are used exactly (e.g. M₁: 120/30 + 240/60 + 360/90 = 4 + 4 + 4 = 12).

Where the 6 marks sit

#	Component	Marks
1	Equations from machine hours	2
2	Determinant, adjoint and inverse	3
3	Units of X, Y, Z with check	1
	Total	6

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