CBSE Class 12 Mathematics Matrices Case Studies

Read the following and answer any four questions from (i) to (v). A manufacturer produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below:

If the unit sale price of Pencil, Eraser and Sharpener are Rs 2.50, Rs 1.50 and Rs 1.00 respectively, and unit cost of the above three commodities are Rs 2.00, Rs 1.00 and Rs 0.50 respectively, then,

Question. Total revenue of market A is
(a) Rs 64,000
(b) Rs 60,400
(c) Rs 46,000
(d) Rs 40,600
Answer: (c) Total revenue for market \( A = \begin{bmatrix} 10,000 & 2,000 & 18,000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = 10,000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00 = Rs 46,000 \)

Question. Total revenue of market B is
(a) Rs 35,000
(b) Rs 53,000
(c) Rs 50,300
(d) Rs 30,500
Answer: (b) Total revenue for market \( B = \begin{bmatrix} 6,000 & 20,000 & 8,000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = 6,000 \times 2.50 + 20,000 \times 1.50 + 8,000 \times 1.00 = Rs 53,000 \)

Question. Cost incurred in market A is
(a) Rs 13,000
(b) Rs 30,100
(c) Rs 10,300
(d) Rs 31,000
Answer: (d) Cost incurred in market \( A = \begin{bmatrix} 10,000 & 2,000 & 18,000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} = 10,000 \times 2.00 + 2,000 \times 1.00 + 18,000 \times 0.50 = Rs 31,000 \)

Question. Profit in market A and B respectively are
(a) (Rs 15,000, Rs 17,000)
(b) (Rs 17,000, Rs 15,000)
(c) (Rs 51,000, Rs 71,000)
(d) (Rs 10,000, Rs 20,000)
Answer: (a) Cost incurred in market \( B = \begin{bmatrix} 6,000 & 20,000 & 8,000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} = 12,000 + 20,000 + 4000 = Rs 36,000 \). Profit in market \( A = 46,000 - 31,000 = 15,000 \) and profit in market \( B = 53,000 - 36,000 = 17,000 \).

Question. Gross profit in both market is
(a) Rs 23,000
(b) Rs 20,300
(c) Rs 32,000
(d) Rs 30,200
Answer: (c) Gross profit = Total SP - Total CP for both market A and B. \( = (46,000 + 53,000) - (31,000 + 36,000) = 32,000 \).

Read the following and answer any four questions from (i) to (v). Amit, Biraj and Chirag were given the task of creating a square matrix of order 2. Below are the matrices created by them. A, B, C are the matrices created by Amit, Biraj and Chirag respectively.
\( A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix}, C = \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} \)
If \( a = 4 \) and \( b = -2 \), based on the above information answer the following:

Question. Sum of the matrices A, B and C, \( A + (B + C) \) is
(a) \( \begin{bmatrix} 1 & 6 \\ 2 & 7 \end{bmatrix} \)
(b) \( \begin{bmatrix} 6 & 1 \\ 7 & 2 \end{bmatrix} \)
(c) \( \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} \)
(d) \( \begin{bmatrix} 2 & 1 \\ 7 & 6 \end{bmatrix} \)
Answer: (c) \( A + (B + C) = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} + \left( \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} \right) = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix} \)

Question. \( (A^T)^T \) is equal to
(a) \( \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \)
(b) \( \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \)
(d) \( \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} \)
Answer: (a) \( A^T = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \Rightarrow (A^T)^T = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = A \)

Question. \( (bA)^T \) is equal to
(a) \( \begin{bmatrix} -2 & -4 \\ 2 & -6 \end{bmatrix} \)
(b) \( \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} \)
(c) \( \begin{bmatrix} -2 & 2 \\ -6 & -4 \end{bmatrix} \)
(d) \( \begin{bmatrix} -6 & -2 \\ 2 & 4 \end{bmatrix} \)
Answer: (b) \( bA = -2 \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -2 & -4 \\ 2 & -6 \end{bmatrix} \Rightarrow (bA)^T = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix} \)

Question. \( AC - BC \) is equal to
(a) \( \begin{bmatrix} -4 & -6 \\ -4 & 4 \end{bmatrix} \)
(b) \( \begin{bmatrix} -4 & -4 \\ 4 & -6 \end{bmatrix} \)
(c) \( \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} \)
(d) \( \begin{bmatrix} -6 & 4 \\ -4 & -4 \end{bmatrix} \)
Answer: (c) \( AC - BC = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ 1 & -6 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix} \)

Question. \( (a + b)B \) is equal to
(a) \( \begin{bmatrix} 0 & 8 \\ 10 & 2 \end{bmatrix} \)
(b) \( \begin{bmatrix} 2 & 10 \\ 8 & 0 \end{bmatrix} \)
(c) \( \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} \)
(d) \( \begin{bmatrix} 2 & 0 \\ 8 & 10 \end{bmatrix} \)
Answer: (c) \( (a + b)B = (4 - 2) \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix} = 2 \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix} \)

Read the following and answer any four questions from (i) to (v). Three schools DPS, CVC and KVS decided to organize a fair for collecting money for helping the flood victims. They sold handmade fans, mats and plates from recycled material at a cost of Rs 25, Rs 100 and Rs 50 each respectively. The numbers of articles sold are given as

Question. What is the total money (in Rs ) collected by the school DPS?
(a) 700
(b) 7,000
(c) 6,125
(d) 7875
Answer: (b) Total money collected by school \( DPS = \begin{bmatrix} 40 & 50 & 20 \end{bmatrix} \begin{bmatrix} 25 \\ 100 \\ 50 \end{bmatrix} = 40 \times 25 + 50 \times 100 + 20 \times 50 = 1000 + 5000 + 1000 = Rs 7000 \)

Question. What is the total amount of money (in Rs ) collected by schools CVC and KVS?
(a) 14,000
(b) 15,725
(c) 21,000
(d) 13,125
Answer: (a) Amount of money collected by CVC and KVS \( = \begin{bmatrix} 25 & 40 & 30 \\ 35 & 50 & 40 \end{bmatrix} \begin{bmatrix} 25 \\ 100 \\ 50 \end{bmatrix} = \begin{bmatrix} 25 \times 25 + 40 \times 100 + 30 \times 50 \\ 35 \times 25 + 50 \times 100 + 40 \times 50 \end{bmatrix} = \begin{bmatrix} 6125 \\ 7875 \end{bmatrix} \). Total amount \( = 6125 + 7875 = Rs 14,000 \).

Question. What is the total amount of money collected by all three schools DPS, CVC and KVS?
(a) Rs 15,775
(b) Rs 14,000
(c) Rs 21,000
(d) Rs 17,125
Answer: (c) otal amount of money collected by all three schools \( = 7000 + 6125 + 7875 = Rs 21,000 \).

Question. If the number of handmade fans and plates are interchanged for all the schools, then what is the total money collected by all schools?
(a) Rs 18,000
(b) Rs 6,750
(c) Rs 5,000
(d) Rs 21,250
Answer: (d) After interchanging, number of articles:

Amount of money collected by all schools:
\( \begin{bmatrix} 20 & 50 & 40 \\ 30 & 40 & 25 \\ 40 & 50 & 35 \end{bmatrix} \begin{bmatrix} 25 \\ 100 \\ 50 \end{bmatrix} = \begin{bmatrix} 20 \times 25 + 50 \times 100 + 40 \times 50 \\ 30 \times 25 + 40 \times 100 + 25 \times 50 \\ 40 \times 25 + 50 \times 100 + 35 \times 50 \end{bmatrix} = \begin{bmatrix} 7500 \\ 6000 \\ 7750 \end{bmatrix} \). Total amount \( = 7500 + 6000 + 7750 = Rs 21,250 \).

Question. How many articles (in total) are sold by three schools?
(a) 230
(b) 130
(c) 430
(d) 330
Answer: (d) Total number of articles sold by three schools \( = (40 + 25 + 35) + (50 + 40 + 50) + (20 + 30 + 40) = 100 + 140 + 90 = 330 \).

Read the following and answer any four questions from (i) to (v). On her birthday, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got Rs 10 more. However, if there were 16 children more, everyone would have got Rs 10 less. Let the number of children be \( x \) and the amount distributed by Seema for one child be \( y \) (in Rs ).

Question. The equations in terms of \( x \) and \( y \) are
(a) \( 5x - 4y = 40, 5x - 8y = -80 \)
(b) \( 5x - 4y = 40, 5x - 8y = 80 \)
(c) \( 5x - 4y = 40, 5x + 8y = -80 \)
(d) \( 5x + 4y = 40, 5x - 8y = -80 \)
Answer: (a) Let children be \( x \) and amount per child be \( y \). Total distribution \( = xy \).
Case 1: \( (x - 8)(y + 10) = xy \Rightarrow 10x - 8y = 80 \Rightarrow 5x - 4y = 40 \)
Case 2: \( (x + 16)(y - 10) = xy \Rightarrow -10x + 16y = 160 \Rightarrow 5x - 8y = -80 \)

Question. Which of the following matrix equations represent the information given above?
(a) \( \begin{bmatrix} 5 & -4 \\ 5 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 40 \\ -80 \end{bmatrix} \)
(b) \( \begin{bmatrix} 5 & -4 \\ 5 & -8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 40 \\ 80 \end{bmatrix} \)
(c) \( \begin{bmatrix} 5 & -4 \\ 5 & -8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 40 \\ -80 \end{bmatrix} \)
(d) \( \begin{bmatrix} 5 & 4 \\ 5 & -8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 40 \\ -80 \end{bmatrix} \)
Answer: (c)

Question. The number of children who were given some money by Seema, is
(a) 30
(b) 40
(c) 23
(d) 32
Answer: (d) \( AX = B \Rightarrow X = A^{-1}B \). \( |A| = -40 + 20 = -20 \). adj \( A = \begin{bmatrix} -8 & 4 \\ -5 & 5 \end{bmatrix} \). \( A^{-1} = \frac{1}{-20} \begin{bmatrix} -8 & 4 \\ -5 & 5 \end{bmatrix} \). \( \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-20} \begin{bmatrix} -8 \times 40 + 4 \times (-80) \\ -5 \times 40 + 5 \times (-80) \end{bmatrix} = \frac{1}{-20} \begin{bmatrix} -640 \\ -600 \end{bmatrix} = \begin{bmatrix} 32 \\ 30 \end{bmatrix} \). Number of children \( x = 32 \).

Question. How much amount is given to each child by Seema?
(a) Rs 32
(b) Rs 30
(c) Rs 62
(d) Rs 26
Answer: (b) \( y = 30 \).

Question. How much amount Seema spends in distributing the money to all the students of the orphanage?
(a) Rs 609
(b) Rs 960
(c) Rs 906
(d) Rs 690
Answer: (b) Total amount \( = xy = 32 \times 30 = Rs 960 \).