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Questions:

Case Study Questions

5 x 4 = 20

Gaurav purchased 5 pens, 3 bags and 1 instrument box and pays Rs. 16. From the same shop, Dheeraj purchased 2 pens, 1 bag and 3 instrument boxes and pays Rs. 19, while Ankur purchased 1 pen, 2 bags and 4 instrument boxes and pays Rs. 25.

Using the concept of matrices and determinants, answer the following questions.
(i) The cost of one pen is

(a) Rs. 2

(b) Rs. 5

(c) Rs. 1

(d) Rs. 3

(ii) What is the cost of one pen and one bag?

(a) Rs. 3

(b) Rs. 5

(c) Rs. 7

(d) Rs. 8

(iii) What is the cost of one pen and one instrument box?

(a) Rs. 7

(b) Rs. 6

(c) Rs. 8

(d) Rs. 9

(iv) Which of the following is correct?

(a) Determinant is a square matrix.	(b) Determinant is a number associated to a matrix
(c) Determinant is a number associated to a square matrix	(d) All of the above

(v) From the matrix equation AB = AC, it can be concluded that B = C provided

(a) A is singular

(b) A is non-singular

(c) A is symmetric

(d) A is square

Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and /h column in a_ijlies and is denoted by M_ij.
Cofactor of an element a_ij'denoted by A_ij,is defined by \(\begin{equation} A_{i j}=(-1)^{i+j} M_{i j} \end{equation}\), where M_ijis minor of a_ij.
Also, the determinant of a square matrix A is the sum of the products of the elements of any row (or column with their corresponding cofactors.
For example if \(\begin{equation} A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \end{equation}\) .
Based on the above information, answer the following questions
(i) Find the sum of the cofactors of all the elements of \(\begin{equation} \left|\begin{array}{cc} 1 & -2 \\ 4 & 3 \end{array}\right| \end{equation}\)

(a) 1

(b) -2

(c) 4

(d) 1

(ii) Find the minor of a₂₁ of \(\begin{equation} \left|\begin{array}{ccc} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{array}\right| \end{equation}\)

(a) 3

(b) -3

(c) 39

(d) -39

(iii) In the determinant \(\begin{equation} \left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) find the value of a₃₂·A₃₂

(a) 27

(b) -110

(c) 110

(d) -27

(iv) If \(\begin{equation} \Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \end{equation}\), find the value of a₃₂·A₃₂ .

(a) -10

(b) -7

(c) 10

(d) 7

(v) If \(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\), then find the value of \(\begin{equation} |\Delta| \end{equation}\).

(a) 26

(b)28

(c) 72

(d) 46

Three shopkeepers Salim, Vijay and Venket are using polythene bags, handmade bags (prepared by prisoners) and newspaper's envelope as carry bags. It is found that the shopkeepers Salim, Vijay and Venket are using (20, 30, 40), (30, 40, 20) and (40, 20, 30) polythene bags, handmade bags and newspaper's envelopes respectively. The shopkeepers Salim, Vijay and Venket spent Rs. 250, Rs. 270 and Rs. 200 on these carry bags respectively.

Using the concept of matrices and determinants, answer the following questions.
(i) What is the cost of one polythene bag?

(a) Rs. 1

(b) Rs. 2

(c) Rs. 3

(d) Rs. 5

(ii) What is the cost of one handmade bag?

(a) Rs. 1

(b) Rs. 2

(c) Rs. 3

(d) Rs. 5

(iii) What is the cost of one newspaper envelope

(a) Rs. 1

(b) Rs. 2

(c) Rs. 3

(d) Rs. 5

(iv) Keeping in mind the social conditions, which shopkeeper is better?

(a) Salim

(b) Vijay

(c) Venket

(d) None of these

(v) Keeping in mind the environmental conditions, which shopkeeper is better?

(a) Salim

(b) Vijay

(c) Venket

(d) None of these

Two schools A and B want to award their selected students on the values of Honesty, Hard work and Punctuality. The school A wants to award Rs. x each, Rs. y each and Rs.z each for the three respective values to its 3, 2 and 1 students respectivefy with a total award money of Rs. 2200. School B wants to spend Rs. 3100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school A). The total amount of award for one prize on each value is Rs. 1200.

Using the concept of matrices and determinants, answer the following questions.
(i) What is the award money for Honesty?

(a) Rs.350

(b) Rs.300

(c) Rs.500

(d)Rs.400

(ii) What is the award money for Punctuality?

(a) Rs.300

(b) Rs.280

(c) Rs.450

(d) Rs.500

(iii) What is the award money for Hard work?

(a) Rs 500

(b) Rs.400

(c) 0

(d) none of these

(iv) If a matrix P is both symmetric and skew-symmetric, then IPI is equal to

(a) 1

(b) -1

(c) 0

(d) none of these

(v) If P and Q are two matrices such that PQ = Q and QP = P, then IQ²1 is equal to

(a) IQI

(b) IPI

(c) 1

(d) 0

Let \(\begin{equation} A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \end{equation}\) , and U₁ U₂ are first and second columns respectively of a 2 x 2 matrix U.
Also, let the column matrices U_I and U₂ satisfying \(\begin{equation} A U_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \end{equation}\) and \(\begin{equation} A U_{2}=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{equation}\)
Based on the above information, answer the following questions
(i) The matrix U₁ + U₂ is equal to

(a) \(\begin{equation} \left[\begin{array}{c} 1 \\ -1 \end{array}\right] \end{equation}\)

(b) \(\begin{equation} \left[\begin{array}{c} 2 \\ -2 \end{array}\right] \end{equation}\)

(c) \(\begin{equation} \left[\begin{array}{c} 3 \\ -3 \end{array}\right] \end{equation}\)

(d) \(\begin{equation} \left[\begin{array}{c} 4 \\ -4 \end{array}\right] \end{equation}\)

(ii) The value of IUI is

(a) 2

(b) -2

(c) 3

(d) -3

(iii) If \(\begin{equation} X=\left[\begin{array}{ll} 3 & 2 \end{array}\right] U\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{equation}\),then the value of IXI =

(a) 3

(b) -3

(c) -5

(d) 5

(iv) The minor of element at the position a₂₂in U is

(a) 1

(b) 2

(c) -2

(d) -1

(v) If \(\begin{equation} U=\left[a_{i j}\right]_{2 \times 2} \end{equation}\) , then the value of a₁₁A₁₁ + a₁₂A_l2 where A_ij denotes the cofactor of a_ijis

(a) 1

(b) 2

(c) -3

(d) 3

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Answers Key:

Case Study Questions

5 x 4 = 20

Let the cost of 1 pen = Rs. x, the cost of 1 bag = Rs. y, and the cost of 1 instrument box = Rs. z
According to the question, we have
5x + 3y + z = 16, 2x + Y + 3z = 19, x + 2y + 4z = 25
This system of equation can be written as AX = B,
where \(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \) and \(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\)
IAI = 5(4 - 6) - 3(8 - 3) + 1(4 - 1)
\(=-10-3(5)+3=-22 \neq 0\)
\(\therefore\) A^-1 exists.
Now, X = A^-1B, where \(A^{-1}=\frac{1}{|A|} \operatorname{adj} A\)
Here, \(\operatorname{adj} A=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{\prime}=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\)
\(\therefore \quad A^{-1}=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\)
\(\therefore \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{c} 16 \\ 19 \\ 25 \end{array}\right]\)
\(=\frac{1}{-22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right]=\frac{-1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]\)
\(\therefore\) x = 1, y = 2, z = 5
Hence, cost of one pen, one bag and an instrument box isRs. 1, Rs. 2 and Rs. 5 respectively.
(i) (c) : Cost of one pen is Rs. 1.
(ii) (a) : Cost of one pen. and one bag = Rs. (1 + 2) = Rs. 3
(iii) (b) : Cost of one pen and one instrument box
= Rs. (1 + 5) = Rs. 6
(iv) (c) : According to the definition of determinant, determinartt is a number associated to a square matrix.
(v) (b) : Given matrix equation is AB = AC Pre-multiplying by A^-I on both sides, we get
\(A^{-1} A B=A^{-1} A C \Rightarrow\left(A^{-1} A\right) B=\left(A^{-1} A\right) C\)
\(\Rightarrow \quad I B=I C \quad\left(\because A A^{-1}=A^{-1} A=I\right)\)
\(\Rightarrow B=C\)
Since A^-1exists only if A i's non-singular
\(\therefore\) For B = C, A should be non-singular

(i) (a) : Let \(\begin{equation} \Delta=\left|\begin{array}{cc} 1 & -2 \\ 4 & 3 \end{array}\right| \end{equation}\)
Cofactor of 1 = 3, cofactor of -2 =-4
Cofactor of 4 = 2, cofactor of 3 = 1
\(\therefore\) Required sum = 3 - 4 + 2 + 1 = 2
(ii) (b) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{array}\right| \end{equation}\)
Minor of \(\begin{equation} a_{21}=\left|\begin{array}{cc} 6 & -3 \\ -7 & 3 \end{array}\right|=18-21=-3 \end{equation}\)
(iii) (c) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\)
Clearly, a₃₂ = 5
and A₃₂ = cofactor of a₃₂ in \(\begin{equation} \Delta=(-1)^{3+2}\left|\begin{array}{ll} 2 & 5 \\ 6 & 4 \end{array}\right| \end{equation}\)
= (-1)(8-30) = 22
\(\begin{equation} \therefore \end{equation}\) a₃₂·A₃₂ = 5 x 22 = 110
(iv) (d) : Here, \(\begin{equation} \Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \end{equation}\)
\(\therefore\) Minor of \(\begin{equation} a_{23}=\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=10-3=7 \end{equation}\)
(v) (b) : Here,\(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\)
\(\begin{equation} A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & 4 \\ 5 & -7 \end{array}\right|=1(0-20)=-20 \end{equation}\)
\(\begin{equation} A_{12}=(-1)^{1+2}\left|\begin{array}{cc} 6 & 4 \\ 1 & -7 \end{array}\right|=-1(-42-4)=46 \end{equation}\)
\(\begin{equation} A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 6 & 0 \\ 1 & 5 \end{array}\right|=1(30-0)=30 \end{equation}\)
\(\begin{equation} \therefore \quad \Delta=a_{11} A_{11}+a_{12} a_{12}+a_{13} A_{13} \end{equation}\)
= 2(-20) -3(46) + 5(30) = -28
\(\begin{equation} \Rightarrow|\Delta|=28 \end{equation}\)

Let the cost of a polythene bag = Rs. x,
the cost of a handmade bag = Rs. y
and the cost of a newspaper bag = Rs. z
According to question,
20x + 30y + 40z = 250, 30x + 40y + 20z = 270
40x + 20y + 30z .=. 200 This system can be written as AX = B, where
\(\begin{equation} A=\left[\begin{array}{ccc} 20 & 30 & 40 \\ 30 & 40 & 20 \\ 40 & 20 & 30 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ \frac{1}{4} \end{array}\right] \end{equation}\) and \(\begin{equation} B=\left[\begin{array}{c} 250 \\ 270 \\ 200 \end{array}\right] \end{equation}\)
\(\begin{equation} |A|=\left|\begin{array}{ccc} 20 & 30 & 40 \\ 30 & 40 & 20 \\ 40 & 20 & 30 \end{array}\right| \end{equation}\)
= 20(1200 - 400) - 30(900 - 800) + 40(600 - 1600)
= 20(800) - 30(100) + 40(-1000)
= 16000 - 3000 - 40000 = -27000≠ 0
So, A^-1 exists and system has a solution given by
X = A^-1B.
Now, \(\begin{equation} \operatorname{adj} A=\left[\begin{array}{ccc} 800 & -100 & -1000 \\ -100 & -1000 & 800 \\ -1000 & 800 & -100 \end{array}\right] \end{equation}\)
\(\begin{equation} =\left[\begin{array}{ccc} 800 & -100 & -1000 \\ -100 & -1000 & 800 \\ -1000 & 800 & -100 \end{array}\right] \end{equation}\)
\(\begin{equation} \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-27000}\left[\begin{array}{ccc} 800 & -100 & -1000 \\ -100 & -1000 & 800 \\ -1000 & 800 & -100 \end{array}\right] \end{equation}\)
Now \(\begin{equation} X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{27000}\left[\begin{array}{ccc} -800 & 100 & 1000 \\ 100 & 1000 & -800 \\ 1000 & -800 & 100 \end{array}\right]\left[\begin{array}{c} 250 \\ 270 \\ 200 \end{array}\right] \end{equation}\)

\(\begin{equation} X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{27000}\left[\begin{array}{ccc} -800 & 100 & 1000 \\ 100 & 1000 & -800 \\ 1000 & -800 & 100 \end{array}\right]\left[\begin{array}{c} 250 \\ 270 \\ 200 \end{array}\right] \end{equation}\)
Hence, cost of a polythene bag, a handmade bag and a newspaper envelope is Rs. 1, Rs. 5 and Rs. 2 respectively.
(i) (a)
(ii) (d)
(iii) (b)
(iv) (b) : Vijay investing most of the money on hand-rnade bags.
(v) (a) : Salim investing less amount of money on polythene bags.

Three equations are formed from the given statements :
3x + 2y + z = 2200
4x + y + 3z = 3100
and x + y + z = 1200
Converting the system of equations in matrix form, we get
\(\begin{equation} \left[\begin{array}{lll} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2200 \\ 3100 \\ 1200 \end{array}\right] \end{equation}\)
i.e., PX= Q,
where \(\begin{equation} P=\left[\begin{array}{lll} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \end{equation}\) and \(\begin{equation} Q=\left[\begin{array}{l} 2200 \\ 3100 \\ 1200 \end{array}\right] \end{equation}\)
\(\begin{equation} |P|=3(1-3)-2(4-3)+1(4-1)=-6-2+3=-5 \neq 0 \end{equation}\)
\(\begin{equation} \Rightarrow X=P^{-1} \end{equation}\) provided p^-lexists
\(\begin{equation} \therefore \text { adj } P=\left[\begin{array}{ccc} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{array}\right] \end{equation}\)
\(\begin{equation} \therefore \quad P^{-1}=\frac{1}{|P|}(\operatorname{adj} P) \end{equation}\)
\(\begin{equation} =\frac{1}{-5}\left[\begin{array}{ccc} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc} 2 & 1 & -5 \\ 1 & -2 & 5 \\ -3 & 1 & 5 \end{array}\right] \end{equation}\)
\(\begin{equation} \therefore \quad X=\frac{1}{5}\left[\begin{array}{ccc} 2 & 1 & -5 \\ 1 & -2 & 5 \\ -3 & 1 & 5 \end{array}\right]\left[\begin{array}{c} 2200 \\ 3100 \\ 1200 \end{array}\right] \end{equation}\)
\(\begin{equation} =\frac{1}{5}\left[\begin{array}{c} 4400+3100-6000 \\ 2200-6200+6000 \\ -6600+3100+6000 \end{array}\right]=\left[\begin{array}{l} 300 \\ 400 \\ 500 \end{array}\right] \end{equation}\)
\(\begin{equation} \Rightarrow x=300, y=400 \text { and } z=500 \end{equation}\)
Hence the money awarded for Honesty, Hardwork and Punctuality are Rs.300, Rs. 400 and Rs.500 respectively
(i) (b)
(ii) (d) .
(iii) (b)
(iv) (c) : If a matrix P is both symmetric and skew skewsymmetric then it will be a zero matrix. So, IPI = 0.
(v) (a): We have Q² = QQ = Q(PQ)
= (QP) Q = PQ = Q
\(\begin{equation} \therefore \end{equation}\) IQ²| = IQI

(i) (c) : we have \(\begin{equation} A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \end{equation}\)
Let \(\begin{equation} U_{1}=\left[\begin{array}{l} a \\ b \end{array}\right] \end{equation}\) then \(\begin{equation} A U_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \end{equation}\)
\(\begin{equation} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \Rightarrow\left[\begin{array}{c} a \\ 2 a+b \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \end{equation}\)
\(\begin{equation} \Rightarrow \end{equation}\) a = 1 and 2a + b = 0 \(\begin{equation} \Rightarrow \end{equation}\) a = 1 and b = -2
Let \(\begin{equation} U_{2}=\left[\begin{array}{l} c \\ d \end{array}\right] \end{equation}\) then \(\begin{equation} A U_{2}=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{equation}\)
\(\begin{equation} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} c \\ d \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \Rightarrow\left[\begin{array}{c} c \\ 2 c+d \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{equation}\)
\(\begin{equation} \Rightarrow \end{equation}\) c = 2 and 2c + d = 3
\(\begin{equation} \Rightarrow \end{equation}\) c = 2 and d = 3 - 4 = -1
Thus,\(\begin{equation} U_{1}+U_{2}=\left[\begin{array}{c} 1 \\ -2 \end{array}\right]+\left[\begin{array}{c} 2 \\ -1 \end{array}\right]=\left[\begin{array}{c} 3 \\ -3 \end{array}\right] \end{equation}\)
(ii) (c) : Clearly \(\begin{equation} U=\left[\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right] \end{equation}\)
\(\begin{equation} \therefore|U|=\left|\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right|=-1+4=3 \end{equation}\)
(iii) (d) : We have \(\begin{equation} X=\left[\begin{array}{ll} 3 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{equation}\)
\(\begin{equation} =\left[\begin{array}{ll} 3 & 2 \end{array}\right]\left[\begin{array}{c} 7 \\ -8 \end{array}\right]=[21-16]=[5] \end{equation}\)
\(\therefore\) IX|=5.
(iv) (a) : a₂₂ in U is -1 and its minor is 1.
(v) (d) : Since, the sum of products of elements of any row (or column) with their corresponding cofactors is equal to the value of determinant
\(\begin{equation} \therefore \quad a_{11} A_{11}+a_{12} A_{12}=|U|=3 \end{equation}\)