Maths · Algebra · chapter 4 of 13

Class 12 Maths Chapter 4: Determinants case study question with answers

One case-based question of the kind the board asks: a passage, 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 snack stall in Indore sells plates of poha, jalebi and samosa at fixed prices of ₹x, ₹y and ₹z a plate. On Monday it sold 2 poha, 1 jalebi and 1 samosa plate for ₹120; on Tuesday 1 poha, 3 jalebi and 2 samosa plates for ₹190; on Wednesday 3 poha, 2 jalebi and 4 samosa plates for ₹250. The owner writes the three days as AX = B, with A = [[2, 1, 1], [1, 3, 2], [3, 2, 4]], X = [[x], [y], [z]] and B = [[120], [190], [250]].
Part 1 · 1 mark
The value of |A| is
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Answer: (d) 11

|A| = 2(12 − 4) − 1(4 − 6) + 1(2 − 9) = 16 + 2 − 7 = 11.

Part 2 · 1 mark
The cofactor A₂₁ of the element in row 2, column 1 of A is
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Answer: (d) −2

M₂₁ = |[[1, 1], [2, 4]]| = 4 − 2 = 2, so A₂₁ = (−1)3(2) = −2.

Part 3 · 2 marks
Given adj A = [[8, −2, −1], [2, 5, −3], [−7, −1, 5]], find the price of a plate of each item.
Answer and marking scheme

X = A⁻¹B = 111(adj A)B = 111[[960 − 380 − 250], [240 + 950 − 750], [−840 − 190 + 1250]] = 111[[330], [440], [220]] = [[30], [40], [20]]. Poha ₹30, jalebi ₹40, samosa ₹20.

  • X = \frac{1}{11}(adj A)B set up (1)
  • ₹30, ₹40, ₹20 (1)
Or
Given adj A = [[8, −2, −1], [2, 5, −3], [−7, −1, 5]], verify that A (adj A) = |A| I and write A⁻¹.
Answer and marking scheme

Row 1 of A times the columns of adj A: 2(8) + 1(2) + 1(−7) = 11, 2(−2) + 1(5) + 1(−1) = 0, 2(−1) + 1(−3) + 1(5) = 0; similarly rows 2 and 3 give (0, 11, 0) and (0, 0, 11). So A (adj A) = 11I = |A| I, and A⁻¹ = 111[[8, −2, −1], [2, 5, −3], [−7, −1, 5]].

  • product shown to be 11I (1)
  • A⁻¹ = \frac{1}{11} adj A (1)
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