Maths · Algebra · chapter 3 of 13

Class 12 Maths Chapter 3: Matrices case study question with answers

One case-based question of the kind the board asks: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
A dairy cooperative near Anand sells milk, curd and ghee at two outlets. Daily sales are recorded in the matrix S, rows for Outlet 1 and Outlet 2, columns for milk (litres), curd (kg) and ghee (kg). The selling prices per unit form the column matrix P and the cost prices per unit the column matrix C, in rupees, in the same order.
MatrixEntries
S[[120, 40, 10], [90, 60, 15]]
P[[56], [80], [600]]
C[[48], [65], [520]]
Part 1 · 1 mark
The order of the revenue matrix SP is
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Answer: (a) order 2 × 1

S is 2 × 3 and P is 3 × 1; the columns of S equal the rows of P, and SP is 2 × 1.

Part 2 · 1 mark
The daily revenue of Outlet 2 is
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Answer: (a) ₹18,840

Row 2 of S times P: 90(56) + 60(80) + 15(600) = 5040 + 4800 + 9000 = ₹18,840.

Part 3 · 2 marks
Find the profit matrix S(P − C) and the total daily profit of the cooperative from both outlets.
Answer and marking scheme

P − C = [[8], [15], [80]]. S(P − C) = [[120(8) + 40(15) + 10(80)], [90(8) + 60(15) + 15(80)]] = [[960 + 600 + 800], [720 + 900 + 1200]] = [[2360], [2820]]. Total daily profit = 2360 + 2820 = ₹5,180.

  • P − C = [[8], [15], [80]] and S(P − C) = [[2360], [2820]] (1)
  • total ₹5,180 (1)
Or
Find the daily cost matrix SC, and use it with SP to find the profit of each outlet. Which law of matrix multiplication shows this equals S(P − C)?
Answer and marking scheme

SC = [[120(48) + 40(65) + 10(520)], [90(48) + 60(65) + 15(520)]] = [[5760 + 2600 + 5200], [4320 + 3900 + 7800]] = [[13560], [16020]]. SP = [[15920], [18840]], so SP − SC = [[2360], [2820]]: ₹2,360 and ₹2,820. By the distributive law, SP − SC = S(P − C).

  • SC = [[13560], [16020]] (1)
  • profits ₹2,360 and ₹2,820, distributive law named (1)
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