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Questions:

Case Study Questions

10 x 4 = 40

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

Area of a triangle whose vertices are (x₁, y₁), (x_2' y₂) and (x_3, y₃) is given by thedeterminant
\(\begin{equation} \Delta=\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right| \end{equation}\)
Since, area is a positive quantity, so we always take the absolute value of the determinant Δ. Also, the area of the triangle formed by three collinear points is zero.
Based on the above information, answer the following questions
(i) Find the area of the triangle whose vertices are (-2, 6), (3, -6) and (1, 5).

(a) 30 sq. units

(b) 35 sq. units

(c) 40 sq. units

(d) 15.5 sq. units

(ii) If the points (2, -3), (k, -1) and (0, 4) are collinear, then find the value of 4k

(a) 4

(b) \(\begin{equation} \frac{7}{140} \end{equation}\)

(c) 4

(d) \(\begin{equation} \frac{40}{7} \end{equation}\)

(iii) If the area of a triangle ABC, with vertices A(1, 3), B(O, 0) and C(k, 0) is 3 sq. units, then a value of k is

(a) 2

(b) 3

(c) 4

(d) 5

(iv) Using determinants, find the equation of the line joining the points A(1, 2) and B(3, 6).

(a) y = 2x

(b) x = 3y

(c) y = x

(d) 4x-y = 5

(v) If A = (11, 7), B = (5, 5) and C = (-1, 3), then

(a) \(\begin{equation} \Delta A B C \end{equation}\) is scalene triangle	(b) \(\begin{equation} \Delta A B C \end{equation}\) is equilateral triangle
(c) A, B and C are collinear	(d) None of these

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

The upward speed v(t) of a rocket at time t is approximated by \(\begin{equation} v(t)=a t^{2}+b t+c, 0 \leq t \leq 100 \end{equation}\) ,where a, band c are constants. It has been found that the speed at times t = 3, t = 6 and t = 9 seconds are respectively 64, 133 and 208 miles per second.

If \(\begin{equation} \left(\begin{array}{ccc} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{array}\right)^{-1}=\frac{1}{18}\left(\begin{array}{ccc} 1 & -2 & 1 \\ -15 & 24 & -9 \\ 54 & -54 & 18 \end{array}\right) \end{equation}\) ,then answer the following questions
(i) The value of b + c is

(a) 20

(b) 21

(c) 3/4

(d) 4/3

(ii) The value of a + c is

(a) 1

(b) 20

(c) 4/3

(d) none of these

(iii) v(t) is given by

(a) t²+20t+l

(b) \(\begin{equation} \frac{1}{3} t^{2}+20 t+1 \end{equation}\)

(c) \(\begin{equation} t^{2}+\frac{1}{3} t+20 \end{equation}\)

(d) P + t + 1

(iv) The speed at time t = 15 seconds is

(a) 346 miles/see

(b) 356 miles/see

(c) 366 miles/see

(d) 376 miles/see

(v) The time at which the speed of rocket is 784 miles/see is

(a) 20 seconds

(b) 30 seconds

(c) 25 seconds

(d) 27 seconds

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

Each triangular face of the Pyramid of Peace in Kazakhstan is made up of 25 smaller equilateral triangles as shown in the figure.

Using the above information and concept of determinants, answer the following questions
(i) If the vertices of one of the smaller equilateral triangle are (0, 0), \(\begin{equation} (3, \sqrt{3}) \end{equation}\) and \(\begin{equation} (3,-\sqrt{3}) \end{equation}\) then the area of such triangle is

(a) \(\begin{equation} \sqrt{3} \text { sq. units } \end{equation}\)

(b) \(\begin{equation} 2 \sqrt{3} \text { sq. units } \end{equation}\)

(c) \(\begin{equation} 3 \sqrt{3} \text { sq. units } \end{equation}\)

(d) none of these

(ii) The area of a face of the Pyramid is

(a) \(\begin{equation} 25 \sqrt{3} \text { sq. units } \end{equation}\)

(b) \(\begin{equation} 50 \sqrt{3} \text { sq. units } \end{equation}\)

(c) \(\begin{equation} 75 \sqrt{3} \text { sq. units } \end{equation}\)

(d) \(\begin{equation} 35 \sqrt{3} \text { sq. units } \end{equation}\)

(iii) The length of a altitude of a smaller equilateral triangle is

(a) 2 units

(b) 3 units

(c) \(\begin{equation} \sqrt{3} \text { units } \end{equation}\)

(d) 4 units

(iv) If (2, 4), (2, 6) are two vertices of a smaller equilateral triangle, then the third vertex will lie on the line represented by

(a) x +y = 5

(b) \(\begin{equation} x=1+\sqrt{3} \end{equation}\)

(c) \(\begin{equation} x=2+\sqrt{3} \end{equation}\)

(d) 2x + y = 5

(v) Let A(a, 0), B(O, b) and C(1, 1) be three points. If \(\begin{equation} \frac{1}{a}+\frac{1}{b}=1 \end{equation}\) ,then the three points are

(a) vertices of an equilateral triangle

(b) vertices of a right angled triangle

(c)collinear

(d) vertices of an isosceles triangle

If there is a statement involving the natural number n such that
(i) The statement is true for n = 1
(ii) When the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1.
Then, the statement is true for all natural numbers n.
Also, if A is a square matrix of order n, then A² is defined as AA. In general, A^m = AA .... A (m times), where m is any positive integer.
Based on the above information, answer the following questions.
(i) If \(A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\),then for any positive integer n,

(a) \(A^{n}=\left[\begin{array}{cc} 3 n & -4 n \\ n & -n \end{array}\right]\)

(b) \(A^{n}=\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\)

(c) \(A^{n}=\left[\begin{array}{cc} 3 n & -8 n \\ 1 & -n \end{array}\right]\)

(d) \(A^{n}=\left[\begin{array}{cc} 1+3 n & -4 n \\ n & 1-3 n \end{array}\right]\)

(ii) If \(A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\),then |Aⁿ| where \(n \in N\), is equal to

(a) 2ⁿ

(b) 3ⁿ

(c) n

(d) 1

(iii) If \(A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\), and \(I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) then which of the following holds for all natural numbers \(n \geq 1\) ?

(a) Aⁿ= nA-(n-1}I

(b) Aⁿ= 2^n-1A-(n-1}I

(c) An = nA+(n-1)I

(d) Aⁿ= 2^n-1A+(n-1}I

(iv) Let \(A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\) , and \(A^{n}=\left[a_{i j}\right]_{3 \times 3}\) for some positive integer n, then the cofactor of a₁₃ is

(a) aⁿ

(b) -aⁿ

(c) 2aⁿ

(d) 0

(v) If A is a square matrix such that IAI = 2, then for any positive integer n, IAⁿl is equal to

(a) 0

(b) 2n

(c) 2n

(d) n²

10.

A company produces three products every day. Their production on certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product.

Using the concepts of matrices and determinants, answer the following questions.
(i) If x, y and z respectively denotes the quantity (in tons) of first, second and third product produced, then which of the following is true?

(a) x + y + z = 45

(b) x + 8 = z

(c) -2y+z=0

(d) all of these

(ii) If \(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -2 \\ 1 & -1 & 1 \end{array}\right)^{-1}=\frac{1}{6}\left(\begin{array}{ccc} 2 & 2 & 2 \\ 3 & 0 & -3 \\ 1 & -2 & 1 \end{array}\right)\) , then the inverse of \(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \end{array}\right)\) is

(a) \(\left(\begin{array}{lll} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{2} & 0 & \frac{-1}{2} \\ \frac{1}{6} & \frac{-1}{3} & \frac{1}{6} \end{array}\right)\)

(b) \(\left(\begin{array}{ccc} \frac{1}{2} & 0 & -\frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{6} & \frac{-1}{3} & \frac{1}{6} \end{array}\right)\)

(c) \(\left(\begin{array}{ccc} \frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \frac{1}{3} & 0 & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{2} & \frac{1}{6} \end{array}\right)\)

(d) none of these

(iii) x :y : z is equal to

(a) 12: 13: 20

(b) 11:15:19

(c) 15: 19: 11

(d) 13: 12: 20

(iv) Which of the following is not true?

(a) IAI = IA'I

(b) (A'rl = (A-I),

(c) A is skew symmetric-matrix of odd order, then IAI = 0

(d) IABI = IAI + IBI

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Answers Key:

Case Study Questions

10 x 4 = 40

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
\(\begin{equation} \Rightarrow|\Delta|=28 \end{equation}\)

(i) (d) : Let be the area of the triangle then,

\(\begin{equation} =\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)| \end{equation}\) [Expanding along R_1]
\(\begin{equation} \Rightarrow \quad \Delta=\frac{1}{2}|43-12|=15.5 \mathrm{sq} . \text { units } \end{equation}\)
(ii) (d) : The given points are collinear.
\(\begin{equation} \therefore \frac{1}{2}\left|\begin{array}{ccc} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0 \end{equation}\)
Expanding along R₁',we get
\(\begin{equation} 2(-1-4)+3(k)+1(4 k)=0 \end{equation}\)
\(\begin{equation} \Rightarrow 7 k-10=0 \Rightarrow k=\frac{10}{7} \Rightarrow 4 k=\frac{40}{7} \end{equation}\)
(iii) (a) : Area of \(\begin{equation} \Delta A B C=3 \text { sq. units } \end{equation}\) [Given]
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 3 \Rightarrow\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 6 \end{equation}\)
\(\begin{equation} \Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6 \end{equation}\)
\(\begin{equation} \Rightarrow 3 k=\pm 6 \Rightarrow k=\pm 2 . \end{equation}\)
(iv) (a) : Let Q(x, y) be any point on the line joining
A(1, 2) and B(3, 6). Then, area of \(\begin{equation} \Delta A B Q=0 \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{array}\right|=0 \end{equation}\)
\(\begin{equation} \Rightarrow 1(6-y)-2(3-x)+1(3 y-6 x)=0 \end{equation}\)
\(\begin{equation} \Rightarrow 6-y-6+2 x+3 y-6 x=0 \end{equation}\)
\(\begin{equation} \Rightarrow-4 x=-2 y \Rightarrow 2 x=y \end{equation}\)
(v) (c) : Area of ΔABC is given by
\(\begin{equation} \frac{1}{2}\left|\begin{array}{rrr} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{array}\right|=\frac{1}{2}[11(5-3)-7(5+1)+1(15+5)] \end{equation}\)
\(\begin{equation} =\frac{1}{2}[22-42+20]=0 \end{equation}\)
\(\therefore\) Points are collinear

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

Since v(3) = 64, v(6) = 133 and v(9) = 208, we get the following system of linear equations
9a + 3b + c = 64
36a + 6b + c = 133
81a + 9b + c = 208
This can be written in matrix form as
\(\begin{equation} \left[\begin{array}{ccc} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=\left[\begin{array}{c} 64 \\ 133 \\ 208 \end{array}\right] \end{equation}\)
or AX = B
Since,\(\begin{equation} A^{-1}=\frac{1}{18}\left(\begin{array}{ccc} 1 & -2 & 1 \\ -15 & 24 & -9 \\ 54 & -54 & 18 \end{array}\right) \end{equation}\)
\(\begin{equation} \therefore \quad X=\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=A^{-1} B=\frac{1}{18}\left(\begin{array}{ccc} 1 & -2 & 1 \\ -15 & 24 & -9 \\ 54 & -54 & 18 \end{array}\right)\left(\begin{array}{c} 64 \\ 133 \\ 208 \end{array}\right) \end{equation}\)
\(\begin{equation} =\frac{1}{18}\left(\begin{array}{c} 64-266+208 \\ -960+3192-1872 \\ 3456-7182+3744 \end{array}\right)=\frac{1}{18}\left(\begin{array}{c} 6 \\ 360 \\ 18 \end{array}\right)=\left(\begin{array}{c} 1 / 3 \\ 20 \\ 1 \end{array}\right) \end{equation}\)
Thus,\(\begin{equation} a=\frac{1}{3} \end{equation}\) ,b = 20 and c = 1
(i) (b)
(ii) (c)
(iii) (b) : \(\begin{equation} v(t)=\frac{1}{3} t^{2}+20 t+1 \end{equation}\)
(iv) (d) : Clearly, required speed = v(15)
\(\begin{equation} =\frac{1}{3} \times 225+20 \times 15+1 \end{equation}\)
= 75 + 300 + 1 = 376 miles per second
(v) (d) : Consider, v(t) = 784
\(\begin{equation} \Rightarrow \frac{1}{3} t^{2}+20 t+1=784 \end{equation}\)
\(\begin{equation} \Rightarrow \end{equation}\) t² + 60t = 2349
\(\begin{equation} \Rightarrow \end{equation}\) t²+ (87 - 27)t - 2349 = 0
\(\begin{equation} \Rightarrow \end{equation}\) t(t + 87) - 27(t + 87) = 0
\(\begin{equation} \Rightarrow \end{equation}\) (t - 27)(t + 87) = 0
\(\begin{equation} \Rightarrow \end{equation}\) = 27 seconds [\(\therefore\) Time can't be negative]

(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
\(\begin{equation} \therefore \quad a_{11} A_{11}+a_{12} A_{12}=|U|=3 \end{equation}\)

(i) (c) : Area of triangle is given by

\(\begin{equation} =\left|\frac{1}{2}[1(-3 \sqrt{3}-3 \sqrt{3})]\right| \end{equation}\) [Expanding along R_I]
\(\begin{equation} =3 \sqrt{3} \text { sq. units } \end{equation}\)
(ii) (c) : Since a face of the Pyramid consist of 25 smaller equilateral triangles
\(\therefore\) Required area = \(\begin{equation} 25 \times 3 \sqrt{3}=75 \sqrt{3} \text { sq. units } \end{equation}\)
(iii) (b) : Area of equilateral triangle \(\begin{equation} =\frac{\sqrt{3}}{4} a^{2} \end{equation}\)
\(\begin{equation} \therefore 3 \sqrt{3}=\frac{\sqrt{3}}{4} a^{2} \Rightarrow a^{2}=12 \Rightarrow a=2 \sqrt{3} \end{equation}\)
Let h be the length of altitude of a smaller equilateral triangle. Then
\(\begin{equation} \frac{1}{2} \times \text { base } \times \text { height }=3 \sqrt{3} \end{equation}\)
\(\begin{equation} \Rightarrow \quad \frac{1}{2} \times 2 \sqrt{3} \times h=3 \sqrt{3} \Rightarrow h=3 \text { units } \end{equation}\)
(iv) (c) : Let the third vertex be (x, y), then we get
\(\begin{equation} \frac{1}{2}\left|\begin{array}{lll} 2 & 4 & 1 \\ 2 & 6 & 1 \\ x & y & 1 \end{array}\right|=\pm 3 \sqrt{3} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{2}[2(6-y)-4(2-x)+1(2 y-6 x)]=\pm 3 \sqrt{3} \end{equation}\)
\(\begin{equation} \Rightarrow \quad 12-2 y-8+4 x+2 y-6 x=\pm 6 \sqrt{3} \end{equation}\)
\(\begin{equation} \Rightarrow \quad 4-2 x=\pm 6 \sqrt{3} \end{equation}\)
\(\begin{equation} \Rightarrow 2-x=\pm 3 \sqrt{3} \Rightarrow x=2 \pm 3 \sqrt{3} \end{equation}\)
(v) (c) : Area of \(\begin{equation} \Delta A B C=\frac{1}{2}\left|\begin{array}{lll} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{array}\right| \end{equation}\)
\(\begin{equation} =\frac{1}{2}[a(b-1)-0+1(0-b)]=\frac{1}{2}(a b-a-b)=0 \end{equation}\)
\(\begin{equation} \left[\because \frac{1}{a}+\frac{1}{b}=1 \Rightarrow b+a=a b\right] \end{equation}\)
\(\therefore\) Points A, Band C are collinear.

(i) (b) : we have \(A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]\)
\(\therefore A^{2}=\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{cc} 5 & -8 \\ 2 & -3 \end{array}\right]\) ,
which can be obtained from \(A^{n}=\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]\) for n = 2.
(ii) (d) : We have, \(A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\)
\(\therefore|A|=\left|\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right|=1-0=1\)
Also, IAⁿl = IA .A A(n times)| = lAiⁿ = 1ⁿ = 1
(iii) (a) : For n = 1, all options are true
\(A^{2}=A \cdot A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\)
and \(A^{3}=A^{2} \cdot A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]\)
Putting n = 3, in (a), we get A³ = 3A - 2I
\(=3\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{ll} 3 & 0 \\ 3 & 3 \end{array}\right]-\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right]\) ,which is true
All other options are different' from A³ = 3A - 21 for n = 3.
(iv) (d) : we have \(A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\)
\(\therefore A^{2}=A \cdot A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\)
\(=\left[\begin{array}{ccc} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{array}\right]\)
Similarly, \(A^{n}=\left[\begin{array}{ccc} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{array}\right]\)
Now, cofactor of \(a_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & a^{n} \\ 0 & 0 \end{array}\right|=0\)
(v) (c) : We have, IAI = 2
and IAⁿl = IA . A A(n-times)|
= IAIIAI IAI(n-times) = lA|ⁿ = 2n

10.

(i) (d) : According to given condition, we have the following system of linear equations.
x + y + z = 45
x + 8 = z or x + 0·y - z = -8
and x + z = 2y or x - 2y + z = 0
(ii) (c): Let \(A=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -2 \\ 1 & -1 & 1 \end{array}\right)\) then we have,
\(A^{-1}=\frac{1}{6}\left(\begin{array}{ccc} 2 & 2 & 2 \\ 3 & 0 & -3 \\ 1 & -2 & 1 \end{array}\right)\)
Now, (A')^-1 = (A^-1)'
\(=\frac{1}{6}\left(\begin{array}{ccc} 2 & 3 & 1 \\ 2 & 0 & -2 \\ 2 & -3 & 1 \end{array}\right)=\left(\begin{array}{ccc} \frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \frac{1}{3} & 0 & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{2} & \frac{1}{6} \end{array}\right)\)
(iii) (b) : The above system of equations can be written in matrix form as
A'X = B, where
\(A^{\prime}=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \end{array}\right), X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 45 \\ -8 \\ 0 \end{array}\right]\)
\(\Rightarrow X=\left(A^{\prime}\right)^{-1} B=\frac{1}{6}\left[\begin{array}{ccc} 2 & 3 & 1 \\ 2 & 0 & -2 \\ 2 & -3 & 1 \end{array}\right]\left[\begin{array}{c} 45 \\ -8 \\ 0 \end{array}\right]\)
\(=\frac{1}{6}\left[\begin{array}{c} 90-24 \\ 90 \\ 90+24 \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 66 \\ 90 \\ 114 \end{array}\right]=\left[\begin{array}{c} 11 \\ 15 \\ 19 \end{array}\right]\)
Thus, x:y:z = 11: 15: 19
(iv) (d) : Clearly, IABI = IAI ·IBI
(v) (b)