Question

NCERT Solutions Class 12 Maths Chapter 3 Matrices is perfect study material that students must refer to when preparing for their CBSE board exams and also for tough competitive exams. Our NCERT Solutions is designed in such a way that it makes it easy for students to study all different concepts. NCERT Solutions Class 12 have covered all topics and provided detail of various concepts in detail.

• Matrices – the rectangular array of the number symbols or expressions which are arranged in rows and columns and is used to represent a mathematical objective or any property of such object is called a Matrix (Matrices: plural). Without any specification, the matrices are used to represent linear maps which help carry out explicit computations of linear algebra. The matrices are expressed in the form of linear algebra. The matrix multiplication can also be used to represent the simple composition of linear maps.
• Types of Matrices – 

Type of Matrix	Details
Row Matrix	A = [aij]1×n
Column Matrix	A = [aij]m×1
Zero or Null Matrix	A = [aij]mxn where, aij = 0
Singleton Matrix	A = [aij]mxn where, m = n =1
Horizontal Matrix	[aij]mxn where, n > m
Vertical Matrix	[aij]mxn where, m > n
Square Matrix	[aij]mxn where, m = n
Diagonal Matrix	A = [aij] when i ≠ j
Scalar Matrix	A = [aij]mxn where, aij = where k is a constant.
Identity (Unit) Matrix	A = [aij]m×n where,
Equal Matrix	A = [aij]mxn and B = [bij]rxs where, aij = bij, m = r, and n = s
Triangular Matrices	Can be either upper triangular (aij = 0, when i > j) or lower triangular (aij = 0 when i < j)
Singular Matrix	|A| = 0
Non-Singular Matrix	|A| ≠ 0
Symmetric Matrices	A = [aij] where, aij = aji
Hermitian Matrix	A = Aθ
Skew – Hermitian Matrix	Aθ = -A
Orthogonal Matrix	A AT = In = AT A
Idempotent Matrix	A2 = A
Involuntary Matrix	A2 = I, A-1 = A
Nilpotent Matrix	∃ p ∈ N such that AP = 0

• Operations on Matrices – there are three basic operations of matrix and those are addition, subtraction and multiplication.

NCERT Solutions Class 12 Maths is one such study material our experts suggest to go through while their preparation.

Answer 1

NCERT Solutions Class 12 Maths Chapter 3 Matrices

1. In the matrix

 

write:

(i) The order of the matrix 

(ii) The number of elements, 

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

Answer:

(i) In the given matrix, the number of rows is 3 and the number of columns is 4. 

Therefore, the order of the matrix is 3 × 4. 

(ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it. 

(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = \(\frac52\)

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

Answer:

We know that if a matrix is of the order m × 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?

Answer:

We know that if a matrix is of the order m × 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 = [a_ij], whose elements are given by:

(i) \(a_{ij} = \frac{(i+j)^2}{2}\)

(ii) \(a_{ij} = \frac{i}{j}\) 

(iii) \(a_{ij} =\frac{(i+2j)^2}{2}\)

Answer:

(i) Given that \(a_{ij} = \frac{(i+j)^2}{2}\)

In general a 2 × 2 matrix is given by

So,

Therefore, the required matrix is

(ii) Given that  \(a_{ij} = \frac{i}{j}\) 

In general a 2 × 2 matrix is given by 

So,

Therefore, the required matrix is 

(iii) Given that \(a_{ij} =\frac{(i+2j)^2}{2}\)

 In general a 2 × 2 matrix is given by

So,

Therefore, the required matrix is 

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

(i) \(a_{ij} = \frac12|-3i + j|\) 

(ii) \(a_{ij} = 2i - j\) 

Answer:

(i) \(a_{ij} = \frac12|-3i + j|\)  

In general a 3 × 4 matrix is given by

Therefore, the required matrix is 

(ii) \(a_{ij} = 2i - j\) 

In general a 3 × 4 matrix is given by

Therefore, the required matrix is 

Answer 2

6. Find the value of x, y, and z from the following equation:

Answer:

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

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 

Now, 5 + z = 5 

⇒ z = 0 

we know that: 

(x − y)² = (x + y)² − 4xy 

⇒ (x − y)² = 36 − 32 = 4 

⇒ x − y = ±2

Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2 

When x − y = − 2 and x + y = 6, we get x = 2 and y = 4 

∴ x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0

As the two 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 

Then, from (3), we have:

4 + z = 7

⇒ z = 3 

∴ x + z = 5 

⇒ x = 2 

∴ x = 2, y = 4 and z = 3.

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

Answer:

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), we have: 

b = 2a

Then, from (1), we have:

a − 2a = −1 

⇒ a = 1 

⇒ b = 2 

Now, from (3), we have:

2 ×1 + c = 5 

⇒ c = 3 

From (4) we have:

3 × 3 + d = 13 

⇒ 9 + d = 13 

⇒ d = 4 

∴ a = 1, b = 2, c = 3 and d = 4.

8. \(A = \left[a_{ij}\right]_{m\times n}\) is a square matrix, if 

(A) m < n 

(B) m > n 

(C) m = n 

(D) None of these

Answer:

The correct answer is C. 

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, \(A = \left[a_{ij}\right]_{m\times n}\) is a square matrix, if m = n.

9. Which of the given values of x and y make the following pair of matrices equal

(A) x = \(\frac{-1}3\), y = 7

(B) Not possible to find

(C) y = 7, x = \(\frac{-2}3\) 

(D) x = \(\frac{-1}3\), y = \(\frac{-2}3\)

Answer:

The correct answer is B. 

It is given that

Equating the corresponding elements, we get:

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.

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

Answer:

The correct answer is D. 

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. 

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

11. Let

Find each of the following

(i) A + B

(ii) A - B

(iii) 3A - C

(iv) AB

(v) BA

Answer:

(i)

(ii)

(iii)

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. 

Therefore, AB is defined as:

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. 

Therefore, BA is defined as:

Answer 3

12. Compute the following:

(i)

(ii)

(iii)

(iv)

Answer:

(i)

(ii)

(iii)

(iv)

13. Compute the indicated products

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

14. If, 

then compute (A + B) and (B - C). Also, verify that A + (B - C) = (A  + B) - C

Answer:

Hence,  we have verified that A + (B - C) = (A + B) - C.

15. If 

and

then compute 3A - 5B.

Answer:

16. Simplify

Answer:

17. Find X and Y, if

Answer:

(i)

Adding equations (1) and (2), we get:

(ii)

Multiplying equation (3) with (2), we get:

Multiplying equation (4) with (3), we get:

From (5) and (6), we have:

Now,

18. Find X, if 

Answer:

19. Find x and y, if

Answer:

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

2 + y = 5

⇒ y = 3

2x + 2 = 8

⇒ x = 3

\(\therefore\) x = 3 and y = 3

Answer 4

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

Answer:

Comparing the corresponding elements of these two matrices, we get:

2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

2y = 12

⇒ y = 6

2z - 3 = 15

⇒ 2z = 18

⇒ z = 9

2t + 6 = 18

⇒ 2r = 12

⇒ t =6

\(\therefore\) x = 3, y = 6,z = 9 and t = 6

21. If

,

find values of x and y.

Answer:

Comparing the corresponding elements of these two matrices, we get: 

2x − y = 10 and 3x + y = 5 

Adding these two equations, we have: 

5x = 15

⇒ x = 3 

Now, 

3x + y = 5 

⇒ y = 5 − 3x 

⇒ y = 5 − 9 = −4 

∴ x = 3 and y = −4

22. Given 

,

find the values of x, y, z and w.

Answer:

Comparing the corresponding elements of these two matrices, we get:

3x = x + 4

⇒ 2x = 4

⇒ x =2

3y = 6 + x + y

⇒ 2y = 6 + x = 6 + 2 = 8

⇒ y = 4

3w = 2w + 3

⇒ w = 3

3z = -1 + z + w

⇒ 2z = -1 + w = -1 + 3 = 2

⇒ z = 1

∴ x = 2, y = 4, z = 1 and w = 3

23. If

,

show that F(x) F(y) = F(x + y).

Answer:

24. Show that

Answer:

(i)

(ii)

25. if

Find A² - 5A + 6I 

Answer:

We have A² = A × A

26. If

,

prove that A³ - 6A² + 7A + 2I = O

Answer:

∴ A³ - 6A² + 7A + 2I = O

27. If 

,

find k so that A² = kA - 2I

Answer:

Comparing the corresponding elements, we have:

3k - 2 = 1

⇒ 3k = 3

⇒ k = 1

Thus, the value of k is 1.

28. If 

and I is the identity matrix of order 2, show that

Answer:

On the L.H.S.

On the R.H.S.

Thus, from (1) and (2), we get L.H.S. = R.H.S.

Answer 5

29. A trust fund has Rs 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 Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: 

(a) Rs 1,800

(b) Rs 2,000

Answer:

(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (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 Rs 1800, we have:

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

(b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x). 

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

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

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

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. 

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40. 

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:

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

31. Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. 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

Answer:

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 p × k 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.

32. Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. If n = p, then the order of the matrix 7X - 5Z is

A) p × 2 

B) 2 × n 

C) n × 3 

D) p × n

Answer:

The correct answer is B. 

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.

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

Answer:

(i)

(ii)

(iii)

34. If,

then verify that 

(i) (A + B)' = A' + B'

(ii) (A - B)' = A' - B'

Answer:

We have:

(i)

Hence, we have verified that (A + B)' = A' + B'

(ii)

Hence, we have verified that (A - B)' = A' - B'.

35. If,

then verify that 

(i) (A + B)' = A' + B'

(ii) (A - B)' = A' - B'

Answer:

(i) It is known that A = (A')'

Therefore, we have:

Thus, we have verified that (A + B)' = A' + B'

(ii)

Thus, we have verified that (A - B)' = A' - B'

36. If

,

then find (A + 2B)'

Answer:

We know that A = (A')'

37. For the matrices A and B, verify that (AB)′ = B'A' where

Answer:

(i)

Hence, we have verified that (AB)' = B'A'.

(ii)

Hence, we have verified that (AB)' = B'A'.

Answer 6

38. If 

,

then verify that A'A = I

Answer:

Hence, we have verified that A'A = I.

39. If

then verify that A'A = I

Answer:

Hence, we have verified that A'A = I.

40. Show that the

matrix is a symmetric matrix.

Answer:

We have:

\(\therefore A' = A\)

Hence, A is a symmetric matrix.

41. Show that the 

matrix is a skew symmetric matrix.

Answer:

We have:

\(\therefore A' = A\)

Hence, A is a skew-symmetric matrix.

42. For the matrix

,

verify that

(i) (A + A') is a symmetric matrix

(ii) (A - A') is a skew symmetric matrix

Answer:

(i)

Hence, (A + A') is a symmetric matrix.

(ii)

Hence, (A - A') is a skew-symmetric matrix.

43. Find \(\frac12(A +A')\) and \(\frac12(A -A')\), when

Answer:

The given matrix is

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

Answer:

(i)

Thus, \(P =\frac12(A +A')\) is a symmetric matrix.

Thus, \(Q =\frac12(A -A')\) is a symmetric matrix.

Representing A as the sum of P and Q:

(ii)

Thus, \(P =\frac12(A +A')\) is a symmetric matrix.

Thus, \(Q =\frac12(A -A')\) is a symmetric matrix.

Representing A as the sum of P and Q:

(iii)

Thus, \(P =\frac12(A +A')\) is a symmetric matrix.

Thus, \(Q =\frac12(A -A')\) is a symmetric matrix.

Representing A as the sum of P and Q:

(iv)

Thus, \(P =\frac12(A +A')\) is a symmetric matrix.

Thus, \(Q =\frac12(A -A')\) is a symmetric matrix.

Representing A as the sum of P and Q:

Answer 7

45. 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

Answer:

The correct answer is A. 

A and B are symmetric matrices, therefore, we have:

A' = A and B' = B       .......(i)

Consider

\(\therefore\) (AB - BA)' = - (AB - BA)

Thus, (AB − BA) is a skew-symmetric matrix.

46. If 

then A + A' = I, if the value of α is

(A) \(\frac{\pi}6\)

(B) \(\frac{\pi}3\)

(C) n

(D) \(\frac{3\pi}2\) 

Answer:

The correct answer is B.

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

2cos α = 1

⇒ \(cos\,\alpha = \frac12=cos\frac{\pi}{3}\)

\(\therefore \alpha = \frac{\pi}{3}\)

47. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

48. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

49. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

50. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

51. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

52. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

53. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = AI

54. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

55. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

56. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = AI

Answer 8

57. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = AI

58. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

Now, in the above equation, we can see all the zeros in the second row of the matrix on the L.H.S. 

Therefore, A −1 does not exist.

59. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

60. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

Applying \(R_1 \rightarrow R_1 - \frac12R_2\), we have:

Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S. 

Therefore, A −1 does not exist.

61. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

Applying R₂ → R₂ + 3R₁ and R₃ → R₃ − 2R₁, we have:

Applying R₁ → R₁ + 3R₃ and R₂ → R₂ − 8R₃, we have:

Applying R₃ → R₃ + 3R₂, we have:

Applying R₃ → \(\frac1{25}\) R₃, we have:

Applying R₁ → R₁ + 10R₃ and R₂ → R₂ − 21R₃, we have:

62. Find the inverse of each of the matrices, if it exists.

Answer:

Let A =

We know that A = IA

Applying R₁ → \(\frac12\) R₁, we have:

Applying R₂ → R₂ - 5R₁, we have:

Applying R₃ → R₃ - R₂, we have:

Applying R₃ → 2R₃, we have:

Applying R₁ → R₁ + \(\frac12\) R₃ and R₂ → R₂ - \(\frac52\) R₃, we have:

63. 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

Answer:

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.

64. Let 

,

show that (aI + bA)ⁿ = aⁿI + na^n-1bA, where I is the identity matrix of order 2 and n \(\in\) N

Answer:

It is given that 

To show: P(n) : (aI + bA)ⁿ = aⁿI + na^n-1bA, n \(\in\) N

We shall prove the result by using the principle of mathematical induction. 

For n = 1, we have:

P(1) : (aI + bA) = aI + ba⁰A = aI + bA

Therefore, the result is true for n = 1. 

Let the result be true for n = k. 

That is,

P(n) : (aI + bA)^k = a^kI + na^k-1bA

Now, we prove that the result is true for n = k + 1. 

Consider

From (1), we have:

Therefore, the result is true for n = k + 1. 

Thus, by the principle of mathematical induction, we have:

(aI + bA)ⁿ = aⁿI + na^n-1bA where 

,

n \(\in\) N

Answer 9

65. If 

,

prove that

,

 n \(\in\) N

Answer:

It is given that

To show:

We shall prove the result by using the principle of mathematical induction. 

For n = 1, we have:

Therefore, the result is true for n = 1. 

Let the result be true for n = k.

That is

Now, we prove that the result is true for n = k + 1.

Now, A^k+1 = A . A^k

Therefore, the result is true for n = k + 1. 

Thus by the principle of mathematical induction, we have:

66. If

,

then prove

where n is any positive integer.

Answer:

It is given that 

To prove:

We shall prove the result by using the principle of mathematical induction. 

For n = 1, we have:

Therefore, the result is true for n = 1.

Let the result be true for n = k. 

That is,

Now, we prove that the result is true for n = k + 1.

Consider

Therefore, the result is true for n = k + 1. 

Thus, by the principle of mathematical induction, we have:

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

Answer:

It is given that A and B are symmetric matrices. Therefore, we have:

A' = A and B' = B        ......(1)

Now,

\(\therefore\) (AB - BA)' = - (AB - BA)

Thus, (AB − BA) is a skew-symmetric matrix.

68. Show that the matrix B'AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

We suppose that A is a symmetric matrix, then A' = A … (1) 

Consider

\(\therefore\) (B'AB)' = B'AB

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

Now, we suppose that A is a skew-symmetric matrix.

Then, A' = - A

Consider

 \(\therefore\) (B'AB)' = - B'AB

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

Hence, if A is a symmetric or skew-symmetric matrix, then B'AB is a symmetric or skew-symmetric matrix accordingly.

69. Solve system of linear equations, using matrix method.

2x - y = -2

3x + 4y = 3

Answer:

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 8 +3 = 11 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

Hence, x = \(\frac{-5}{11}\) and y = \(\frac{12}{11}\).

70. For what values of x,

 ?

Answer:

We have:

∴ 4 + 4x = 0 

⇒ x = −1 

Thus, the required value of x is −1.

71. If

, 

show that A² - 5A + 7I = O

Answer:

It is given that 

∴ L.H.S. = A² - 5A + 7I

= O = R.H.S.

∴ A² - 5A + 7I = O

72. Find x, if

Answer:

We have:

Answer 10

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

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

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

Answer:

(a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00. 

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

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

Therefore, the total revenue in market I isRs 46000 and the same in market II isRs 53000. 

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise. 

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

Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs 46000 − Rs 31000) Rs 15000. 

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

Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000.

74. Find the matrix X so that

Answer:

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.

Now, let

Therefore, we have:

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

\(\therefore\) b = 2 - 4(0) = 2

Thus, a = 1, b = 2, c = −2, d = 0 

Hence, the required matrix X is 

.

75. If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n \(\in\) N

Answer:

A and B are square matrices of the same order such that AB = BA.

To prove: P(n): ABⁿ = BⁿA,  n \(\in\) N

For n = 1, we have:

P(1): AB = BA        [Given]

⇒ AB¹ = B¹A

Therefore, the result is true for n = 1. 

Let the result be true for n = k.

P(k): AB^k = B^kA        .........(1)

Now, we prove that the result is true for n = k + 1.

Therefore, the result is true for n = k + 1. 

Thus, by the principle of mathematical induction, we have  ABⁿ = BⁿA,  n \(\in\) N

Now, we prove that (AB)ⁿ = Aⁿbⁿ for all n N 

For n = 1, we have:

(AB)¹ = A¹B¹ = AB

Therefore, the result is true for n = 1. 

Let the result be true for n = k.

(AB)^k = A^kB^k       ......(2)

Now, we prove that the result is true for n = k + 1.

Therefore, the result is true for n = k + 1. 

Thus, by the principle of mathematical induction, we have (AB)ⁿ = AⁿBⁿ, for all natural numbers.

76. Choose the correct answer in the following questions:

If

is such that A² = I then

(A) \(1 + \alpha^2 +\beta\gamma = 0\)

(B) \(1 - \alpha^2 +\beta\gamma = 0\)

(C) \(1 - \alpha^2 -\beta\gamma = 0\)

(D) \(1 + \alpha^2 -\beta\gamma = 0\)

Answer:

Correct option is (C) \(1 - \alpha^2 -\beta\gamma = 0\) 

On comparing the corresponding elements, we have:

\(\alpha^2 + \beta\gamma = 1\)

⇒ \(\alpha^2 + \beta\gamma -1=0 \)

⇒ \(1 - \alpha^2 - \beta\gamma=0\)

77. 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

Answer:

Correct option is B. A is a zero matrix 

If A is both symmetric and skew-symmetric matrix, then we should have 

A' = A and A' = - A

⇒ A = -A

⇒ A + A = 0

⇒ 2A = 0

⇒ A = 0

Therefore, A is a zero matrix.

78. If A is square matrix such that A² = A then (I + A)³ - 7A is equal to

A. A 

B. I − A 

C. I 

D. 3A

Answer:

Correct option is C. I