Question 1 (a)

Selling prices from commission records (matrix inverse)

6 marks

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

Question

Question 1 (a) · 6 marks

During three weeks of a certain month, a salesperson has the following record of sales of items P, Q and R. Her sales commission is at the rate of 2%, 2.5% and 1% respectively. The sales and commission details are given below:

Week	Sales in units			Total Commission earned (₹)
	P	Q	R
First	400	200	200	5,400
Second	500	400	700	10,500
Third	400	600	400	10,400

Using matrices, you are required to:

Find the selling price per unit for items P, Q and R.
The total commission received by the salesperson in the fourth week if her sales are 720, 180 and 400 units of products P, Q and R respectively.

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

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

Write it in this order

Read the data; note exactly what is asked 1 min
1 Equations in matrix form 1 mark 2 min
2 Determinant, adjoint and inverse 2 marks 3 min
3 Selling prices, with check 1 mark 2 min
4 Fourth-week commission 2 marks 3 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

1234

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.

Setting up the system

Let the selling prices per unit of P, Q and R be ₹p, ₹q and ₹r. Commission on a week's sales is 2% of the value of P sold, 2.5% of the value of Q sold and 1% of the value of R sold. Each week therefore gives one linear equation:

Week	Commission on P	Commission on Q	Commission on R	Commission earned (₹)
First	0.02 × 400p = 8p	0.025 × 200q = 5q	0.01 × 200r = 2r	5,400
Second	0.02 × 500p = 10p	0.025 × 400q = 10q	0.01 × 700r = 7r	10,500
Third	0.02 × 400p = 8p	0.025 × 600q = 15q	0.01 × 400r = 4r	10,400

In matrix form AX = B:

A = [[8, 5, 2], [10, 10, 7], [8, 15, 4]], X = [p, q, r]ᵀ, B = [5400, 10500, 10400]ᵀ.

(i) Selling prices by the inverse method

|A| = 8(10 × 4 − 7 × 15) − 5(10 × 4 − 7 × 8) + 2(10 × 15 − 10 × 8) = 8(−65) − 5(−16) + 2(70) = −520 + 80 + 140 = −300.

Since |A| ≠ 0, A is non-singular and the system has the unique solution X = A⁻¹B, with A⁻¹ = (adj A)/|A|.

The cofactors are A₁₁ = −65, A₁₂ = 16, A₁₃ = 70; A₂₁ = 10, A₂₂ = 16, A₂₃ = −80; A₃₁ = 15, A₃₂ = −36, A₃₃ = 30, so

adj A = [[−65, 10, 15], [16, 16, −36], [70, −80, 30]] and A⁻¹ = −(1/300) × adj A.

X = A⁻¹B = −(1/300) × [−65(5400) + 10(10500) + 15(10400), 16(5400) + 16(10500) − 36(10400), 70(5400) − 80(10500) + 30(10400)]ᵀ

= −(1/300) × [−90,000, −1,20,000, −1,50,000]ᵀ = [300, 400, 500]ᵀ.

Selling price per unit: P ₹300, Q ₹400, R ₹500.

Check by substitution: 8(300) + 5(400) + 2(500) = 5,400; 10(300) + 10(400) + 7(500) = 10,500; 8(300) + 15(400) + 4(500) = 10,400.

(ii) Commission in the fourth week

The fourth week's commission is the product of the row of commission per unit sold and the column of units, a row-by-column product of matrices:

Commission per unit = [0.02 × 300, 0.025 × 400, 0.01 × 500] = [6, 10, 5] (₹).

Item	Units sold	Commission per unit (₹)	Commission (₹)
P	720	6	4,320
Q	180	10	1,800
R	400	5	2,000
Total			8,120

[6 10 5] × [720, 180, 400]ᵀ = 4,320 + 1,800 + 2,000 = ₹8,120.

Where the 6 marks sit

#	Component	Marks
1	Equations in matrix form	1
2	Determinant, adjoint and inverse	2
3	Selling prices, with check	1
4	Fourth-week commission	2
	Total	6

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