NCERT Class 12 Mathematics Chapter 3 Matrices

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NCERT Class 12 Mathematics Chapter 3 Matrices

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Matrices

Chapter – 3

Exercise 3.1

(i) The order of the matrix.

Ans: Since, in the given matrix, the number of rows is 3 and the number of columns is 4, the order of the matrix is 3 × 4.

(ii) The number of elements.

Ans: Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements.

(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Ans: Here,

a₁₃ = 19 

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a₂₁ = 35

a₃₃ = -5

a₂₄ = 12

a₂₃ = 5/2

2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Ans: We know that if a matrix is of the order m x n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24,1), (2,12), (12,2), (3,8), (8,3), (4,6) and (6,4).

Hence, the possible orders of a matrix having 24 elements are: (1×24), (24×1), (2×12), (12×2), (3×8), (8×3), (4×6) and (6×4). 

(1,13) and (13,1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are (1×13) and (13×1).

3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Ans: We know that if a matrix is of the order m x n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.

The ordered pairs are: (1,18), (18,1), (2,9), (9,2), (3,6) and (6,3). 

Hence, the possible orders of a matrix having 18 elements are: (1×18), (18×1), (2×9), (9×2), (3×6) and (6×3). 

(1×5) and (5×1) are the ordered pairs of natural numbers whose product is 5. 

Hence, the possible orders of a matrix having 5 elements are (1×5) and (5×1).

4. Construct a 2 × 2 matrix, A = [aᵢⱼ] whose elements are given by:

Ans: 

Ans: 

Ans: 

5. Construct a 3 × 4 matrix, whose elements are given by:

(i) aᵢⱼ = 1/ 2 |- 3i + j|.

Ans: 

Thus, the required matrix is

(ii) aᵢⱼ = 2i – j.

Ans: 

6. Find the values of x, y and z from the following equations:

Ans: 

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get: x = 1, Y = 4 and z = 3.

(ii)

Ans: 

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

x + y = 6

xy = 8

5 + z = 5

Hence,

⇒ 5 + z = 5

⇒ z = 0

We know that (a – b)² = (a + b)² – 4ab 

⇒ (x – y)² = (6)² – 8 × 4 

⇒ (x – y)² = 36 – 32 

⇒ (x – y)² = 4 

⇒ (x – y) = ± 2 

Equating x – y = 2 and x + y = 6, we get x = 4, y = 2 

Similarly, Equating x – y = – 2 and x + y = 6 , we get x = 2, y = 4 

Thus, x = 4, y = 2, z = 0 or x = 2, y = 4, z = 0

(iii)

Ans: 

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

x + y + z = 9    …(1)

x + z = 5    …(2)

y + z = 7   …(3)

From (1) and (2), we have 

⇒ y + 5 = 9 

⇒ y = 4

From (3), we have 

⇒ 4 + z = 7 

⇒ z = 3 

Therefore, 

⇒ x + z = 5 

⇒ x + 3 = 5 

⇒ x = 2 

Thus, x = 2 y = 4 z = 3

7. Find the value of a,b,c and d from the equation:

Ans: 

As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

a – b = – 1     …(1)

2a – b = 0     …(2)

2a + c = 5       …(3)

3c + d = 13     …(4)

From (2),

b = 2a

Putting this value in (1),

⇒ a – 2a = – 1

⇒ a = 1

Hence,

⇒ b = 2

Putting a = 1 in (3), 

⇒ 2(1) + c = 5 

⇒ c = 3

Putting c = 3 in (4), 

⇒ 3(3) + d = 13 

⇒ d = 4 

Thus, a = 1, b = 2, c = 3 and d = 4

8. A = [aᵢⱼ]ₘ ₓ ₙ is a square matrix, if:

(A) m < n.

(B) m > n.

(C) m = n.

(D) None of these.

Ans: It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, A = [aᵢⱼ]ₘ ₓ ₙ is a square matrix, if m = n

Thus, the correct option is C.

Ans: 

Equating the corresponding elements, we get:

3x + 7 = 0 ⇒ x = -7/3

y – 2 = 5 ⇒ y = 7

y + 1 = 8 ⇒ y = 7

2 – 3x = 4 ⇒ x = -2/3

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.

Hence, it is not possible to find the values of x and y for which the given matrices are equal.

Thus, the correct option is B.

10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512

Ans: The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1

Now, each of the 9 elements can be filled in two possible ways.

Hence, by the multiplication principle, the required number of possible matrices is 2⁹ = 512

Thus, the correct option is D.

EXERCISE 3.2

Find each of the following:

(i) A + B.

Ans: 

(ii) A – B.

Ans: 

(iii) 3A – C.

Ans:

(iv) AB.

Ans: 

(v) BA.

Ans: 

2. Compute the following:

Ans: 

Ans: 

Ans: 

Ans: 

3. Compute the indicated products:

Ans:

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

(A + B) and (B – C) Also, verify that A + (B – C) = (A + B) – C

Ans: 

5. 

Ans: 

Ans: 

7. Find X and Y, if:

Ans: 

Ans: 

Ans: 

Ans: 

Comparing the corresponding elements of these two matrices,

2 + y = 5 

⇒ y = 3

2x + 2 = 8 

⇒ x = 3

Therefore, x = 3 and y = 3

10. Solve the equation for x, y, z and t, if 2

Ans: 

Comparing the corresponding elements of these two matrices,

2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

2y = 12 

⇒ y = 6

2z – 3 = 15

⇒ 2z = 18

⇒ z = 9

2t + 6 = 18

⇒ 2t = 12

⇒ t = 6

Therefore, x = 3, y = 6, z = 9 and t = 6

Ans: 

Comparing the corresponding elements of these two matrices, 

2x – y = 10   …(1)

3x + y = 5   …(2) 

By adding these two equations, we get 

5x = 15 

⇒ x = 3 

Now, putting this value in (2) 

⇒ 3x + y = 5 

⇒ y = 5 – 3x 

⇒ y = 5 – 3(3) 

⇒ y = 5 – 9 

⇒ y = – 4 

Therefore, x = 3 and y = 4

Ans: 

Comparing the corresponding elements of these two matrices,

⇒ 3x = x + 4 

⇒ 2x = 4 

⇒ x = 2

⇒ 3y = 6 + x + y 

⇒ 2y = 6 + x 

⇒ 2y = 6 + 2 

⇒ 2y = 8 

⇒ y = 4

⇒ 3w = 2w + 3 

⇒ w = 3

⇒ 3z = – 1 + z + w 

⇒ 2z = w – 1 

⇒ 2z = 3 – 1 

⇒ 2z ==2 

⇒ z = 1

Therefore, x = 2, y = 4, z = 1 and w = 3

Ans: 

14. Show that:

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

19. A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) 1800

Ans: Let ₹ x be invested in the first bond. Then, the sum of money invested in the second bond will be ₹ (30000 – x)

It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.

Therefore, in order to obtain an annual total interest of ₹ 1800, we have:

⇒ 5x + 210000 – 7x = 180000

⇒ 210000 – 2x = 180000 

⇒ 2x = 210000 – 180000

⇒ 2x = 30000 

⇒ x = 15000

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.

(b) ₹2000

Ans: Let ₹ x be invested in the first bond. Then, the sum of money invested in the second bond will be ₹(30000 – x)

Therefore, in order to obtain an annual total interest of ₹2000, we have:

⇒ 5x + 210000 – 7x = 200000 

⇒ 210000 – 2x = 200000 

⇒ 2x = 210000 – 200000 

⇒ 2x = 10000

⇒ x = 5000

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹5000 in the first bond and the remaining ₹25000 in the second bond.

20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k respectively. Choose the correct answer in Exercises 21 and 22.

Ans: The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

= 12(800 + 480 + 400)

=12(1680)

=20160

Thus, the bookshop will receive ₹ 20160 from the sale of all these books.

21. The restriction on n, k and p so that PY + WY will be defined are:

(A) k = 3, p = n.

(B) k is arbitrary, p = 2

(C) p is arbitrary, k = 3

(D) k = 2 p = 3

Ans: Matrices P and Y are of the orders p × k and 3 × k respectively. 

Therefore, matrix PY will be defined if k = 3 

Consequently, PY will be of the order p × k 

Matrices W and Y are of the orders n × 3 and 3 × k respectively.

Since the number of columns in W is equal to the number of rows in Y, matrix WY is well- defined and is of the order n × k.

Matrices PY and WY can be added only when their orders are the same.

However, PY is of the order pk and WY is of the order n × k Therefore, we must have p = n.

Thus, k = 3 and p = n are the restrictions on n, k and P so that PY + WY will be defined.

The correct option is A.

22. If n = p, then the order of the matrix 7X – 5Z is:

(A) p × 2

(B) 2 × n.

(C) n × 3

(D) p × n.

Ans: Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 × p , i.e., 2 × n [Since n = p] 

Therefore, matrix 5Z is also of the same order. 

Now, both the matrices 7X and 5Z are of the order 2 × n. 

Thus, matrix 7X – 5Z is well-defined and is of the order 2 × n.

The correct option is B.

EXERCISE 3.3

1. Find the transpose of each of the following matrices:

Ans: 

Ans: 

Ans: 

(i) (A + B)’ = A’ + B’,

Ans: 

(ii) (A – B)’ = A’ – B’

Ans: 

(i) (A + B)’ = A’ + B’,

Ans: 

(ii) (A – B)’ = A’ – B’

Ans: 

Ans: It is known that A = (A’)’. 

Therefore,

5. For the matrices A and B, verify that (AB)’ = B’A’ where:

Ans: 

Ans: 

Thus, (AB)’ = B’A’.

Ans: 

Ans: 

Ans: 

Ans: 

(i) (A + A’) is a symmetric matrix.

Ans: 

(ii) (A – A’) is a skew symmetric matrix.

Ans: 

Thus, (A-A’) is a skew symmetric matrix.

Ans: 

10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Ans: 

Ans: 

Ans: Let

Hence,

Ans: 

Choose the correct answer in the Exercises 11 and 12.

11. If A, B are symmetric matrices of same order, then AB – BA is a:

(A) Skew symmetric matrix.

(B) Symmetric matrix.

(C) Zero matrix.

(D) Identity matrix.

Ans: If A and B are symmetric matrices of the same order, then 

A’ = A and B’ = B     …….(1)

Now consider,

(AB – BA)’ = (AB)’ -(BA)’    [∵(A-B)’ = A’-Β’]

=  B’A’ – A’B     [∵(AB)’=B’A’]

= BA-AB    [from (1)]

=-(AB-BA)

Therefore,

(AB-BA)’ = -(AB-BA)

Thus, AB-BA is a skew symmetric matrix.

The Correct option is A.

Ans: 

Comparing the corresponding elements of the two matrices, we have:

⇒ 2 cos α = 1 

⇒ cos α = 1/2 

⇒ α = cos⁻¹ 1/2

⇒ α = π/3

Thus, the correct option is B

EXERCISE 3.4

1. Matrices A and B will be inverse of each other only if:

(A) AB = BA.

(B) AB = BA = 0.

(C) AB = 0, BA = I.

(D) AB = BA = I.

Ans: We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A.

In this case, it is clear that A is the inverse of B.

Thus, matrices A and B will be inverses of each other only if AB = BA = I.

The correct option is D.

Miscellaneous Exercise on Chapter 3

1. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Ans: It is given that A and B are symmetric matrices. 

Therefore, we have:

A’ = A and B’ = B            …(1)

Now,

(AB – BA)’ = (AB)’ – (BA)’     [(A – B)’ = A’ – B’]

 = B’ A’ – A’ B’      [(AB)’ = B’A’]

= BA – AB     [Using (1)]

= – (AB – BA) 

Hence,

(AB – BA)’ = – (AB – BA) 

Thus, AB – BA is a skew symmetric matrix.

2. Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Ans: We suppose that A is a symmetric matrix, then 

A’ = A     …(1)

Consider, 

(B’ AB)’ = {B’ (AB)}’ 

= (AB)’ (B’)’       [∵ (AB)’=B’A] 

= B’A’ (B)       [∵ (B’)’ = B]        

= B’ (A’ B) 

= B’ (AB)     [Using (1)]

 Therefore,

(B’ AB)’ = B’ AB 

Thus, if A is symmetric matrix, then B’AB is a symmetric matrix. 

Now, we suppose that A is a skew symmetric matrix, then 

A’ = – A      …(2)

Consider,

(B’ AB)’ = {B’ (AB)}’

= (AB)’ (B’)’

= (B’ A’) B 

= B’ (- A) B     [Using (2)] 

= – B’ AB 

Therefore,

(B’ AB)’ = – B’ AB

Thus, if A is a skew symmetric matrix, then B’AB is a skew symmetric matrix.

Hence, if A is symmetric or skew symmetric matrix, then B’ AB is symmetric or skew symmetric accordingly.

A’A = I.

Ans: 

Ans: we have:

Hence,

⇒ [6(0) + 2(2) + 4(x)] = 0

⇒ [4 + 4x] = 0 

⇒ 4x = – 4 

⇒ x = – 1

Thus, the required value of x = – 1

Ans: 

Ans: 

⇒ [x(x – 2) – 40 + 2x – 8] = 0

⇒ [x² – 2x – 40 + 2x – 8] = 0

⇒ [x² – 48] = 0

⇒ x² – 48 = 0

⇒ x² = 48

⇒ x = ± 4√3

Thus, x = ± 4√3

7. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

MarketProducts
I10,0002,00018,000
II6,00020,0008,000

(a) If unit sale prices of x, y and z are ₹2.50, ₹1.50 and ₹1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Ans: The unit sale prices of x, y and z are ₹2.50, ₹1.50 and ₹1.00 respectively.

Consequently, the total revenue in market I can be represented in the form of a matrix as:

= 25000 + 3000 + 18000

= 46000

The total revenue in market II can be represented in the form of a matrix as:

= 15000 + 30000 + 8000 

= 53000

Thus, the total revenue in market I is ₹46000 and the total revenue in market II is ₹ 53000

(b) If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively. Find the gross profit.

Ans: The unit costs of x, y and z are ₹2.00, ₹1.00 and 50 paise respectively.

Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:

= 20000 + 2000 + 9000

=31000

Since the total revenue in market I is ₹ 46000, the gross profit in this market in ₹ is 

46000-31000 = 15000

The total cost prices of all the products in market II can be represented in the form of a matrix as:

= 12000 + 20000 + 4000 

= 36000

Since the total revenue in market I is ₹ 53000, the gross profit in this market in ₹ is 

53000-36000 = 17000

Thus, the gross profit in market I is ₹15000 and in market II is ₹17000

Ans: It is given that 

The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix.

Therefore, X has to be a 2 × 2 matrix.

Equating the corresponding elements of the two matrices, we have:

a + 4c = 72a + 5c = – 83a + 6c = – 9a
b + 4d = 22b + 5d = 43b + 6d = 6

Now, 

a + 4c = – 7 

⇒ a = – 7 – 4c

Therefore, 

2a + 5c = – 8 

⇒ 2(- 7 – 4c) + 5c = – 8 

⇒ – 14 – 8c + 5c = – 8 

⇒ – 3c = 6 

⇒ c = – 2

Hence,

⇒ a = – 7 – 4(- 2) 

⇒ a = – 7 + 8 

⇒ a = 1

Now,

b + 4d = 2

⇒ b = 2 – 4d

Therefore,

2b + 5d = 4

⇒ 2(2 – 4d) + 5d = 4

⇒ 4 – 8d + 5d = 4

⇒ – 3d = 0

⇒ d = 0

Hence,

b = 2 – 4d

⇒ b = 2

Thus, a = 1, b = 2, c = – 2 and d = 0

Hence, the required matrix

(Α) 1 + α² + βγ = 0

(Β) 1 – α² + βγ = 0

(C) 1 – α² – βγ = 0

(D) 1 + α² – βγ = 0

Ans: 

On comparing the corresponding elements, we have:

α² + βγ = 1 

⇒ α² + βγ – 1 = 0 

⇒ 1 – α² – βγ = 0 

Thus, the correct option is C.

10. If the matrix A is both symmetric and skew symmetric, then

(A) A is a diagonal matrix.

(B) A is a zero matrix.

(C) A is a square matrix.

(D) None of these.

Ans: If the matrix A is both symmetric and skew symmetric, then 

A’ = A and A’ = – A

Hence,

⇒ A = – A

⇒ A + A = 0

⇒ 2A = 0

⇒ A = 0

Therefore, A is a zero matrix.

Thus, the correct option is B.

11. If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to

(A) A.

(B) I – A.

(C) I.

(D) 3A.

Ans: It is given that A is a square matrix such that A² = A.

Now,

(I + A)³ – 7A = I³ + A³ + 3I² A + 3A² I – 7A

= I+A² A + 3A + 3A² – 7A 

= I+A. A + 3A + 3A – 7A     [∵ A² = A]

= I + A² – A 

= I + A – A     [∵A² = A]

= I 

Hence,

(I + A)³ – 7A = I

Thus, the correct option is C.

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