NCERT Grade 12-Matrices-Answers

NCERT Solutions for Class 12 Maths

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1.    (i) There are 3 horizontal lines (rows) and 4 vertical lines (columns) in the given matrix A.

Therefore, Order of the matrix is 3 x 4.

(ii) The number of elements in the matrix A is 3 x 4 = 12.

(iii) a13 → Elaement in first row and third column = 19

a21 → Element in second row and first column = 35

a33 →Element in third row and third column = -5

a24 → Element in second row and fourth column = 12

a23 →Element in second row and third column = 5/2.

2.    Since, a matrix having element is of order

(i) Therefore, there are 8 possible matrices having 24 elements of orders 1 × 24, 2 × 12, 3 × 8, 4 × 6, 24 × 1, 12 × 2, 8 × 3, 6 × 4.

(ii) Prime number 13 = 1 × 13 and 13 × 1

Therefore, there are 2 possible matrices of order 1 × 13 (Row matrix) and 13 × 1 (Column matrix).

3.    Since, a matrix having element is of order

(i) Therefore, there are 6 possible matrices having 18 elements of orders 1 × 18, 2 × 9, 3 × 6, 18 × 1, 9 × 2, 6 × 3.

(ii) Prime number 5 = 1 × 5 and 5 × 1

Therefore, there are 2 possible matrices of order 1 × 5 (Row matrix) and 5 × 1 (Column matrix).

4.    

5.    

6.    

By definition of Equal matrices, x = 1 y = 4, z = 3

Equating corresponding entries, x + y = 6 …….(i)

5 + z = 5 ⇒ z = 5 – 5 ⇒ z = 0       ……(ii)

And xy = 8 ⇒ x(6 – x) = 8 [From eq.(i), y = 6 – x]

⇒ 6x – x2 = 8

⇒ x2 – 6x + 8 = 0

⇒ (x – 4) (x – 1) = 0

⇒ x = 4 or x = 2

Putting these values of x in eq.(i), we have y = 2 and y = 4

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

Equating corresponding entries, x + y + z = 9        …..(i)

x + z = 5        …(ii)

And y + z = 7        …(iii)

Eq.(i) – Eq.(ii) = y = 9 – 5 = 4

Eq.(i) – Eq.(ii) = x = 9 – 7 = 2

Putting values of x and y in eq.(i),

2 + 4 + z = 9 ⇒ z = 3

∴ x = 2, y = 4, z = 3.

7.    Equating corresponding entries,

a – b = 1         …(i)

2a – b = 0     ….(ii)

2a + c = 5     …(iii)

3c + d = 13   …(iv)

Eq.(i) – Eq.(ii) = -a = -a

⇒ a = 1

Putting a = 1 in eq. (i), 1 – b = -1

⇒ -b = -2 ⇒ b = 2

Putting a = 1 in eq. (iii), 2 + c = 5

⇒ c = 5 – 2 ⇒ c = 3

Putting c = 3 in eq. (iv), 9 + d = 13

⇒ d = 13 – 9 ⇒ d = 4

∴ a = a, b = 2, c = 3, d = 4

8.    By definition of square matrix m = n, option (C) is correct.

9.    Equating corresponding sides,

Since, values of x are not equal, therefore, no values of x and y exist to make the two matrices equal.

Therefore, option (B) is correct.

10.  Since, general matrix of order 3 x 3 is 

This matrix has 9 elements.

The number of choices for a11 is 2 (as 0 or 1 can be used)

Similarly, the number of choices for each other element is 2.

Therefore, total possible arrangements (matrices) = 2 × 2 × 2 ×….9 times = 29 = 512

Therefore, option (D) is correct.

11.   

12.  

13.  

14.  

15.  

16.  

17.   

Adding eq.(i) and (ii), we get

Subtracting eq.(i) and (ii), we get

18.  

19.  

Equating corresponding entries, we have

2 + y = 5 and 2x + 2 = 8

⇒ y = 5 – 2 and 2(x + 1) = 8

⇒ y = 3 and x + 1 = 4

⇒ y = 3 and x = 3.

20.  

Equating corresponding entries, we have

2x + 3 = 9 ⇒ 2x = 9 – 3 ⇒ 2x = 6 ⇒ x = 3

And 2z – 3 = 15 ⇒ 2z = 15 + 3 ⇒ 2z = 18 ⇒ z = 9

And 2y = 12 ⇒ y = 6

And 2t + 6 = 18 ⇒ 2t = 18 – 6 ⇒ 2t = 12 ⇒ t = 6

∴ x = 3, y = 6, z = 9, t = 6.

21.  

Equating corresponding entries, we have

2x – y = 10   ……….(i) and 3x + y = 5    ……….(ii)

Adding eq. (i) and (ii), we have 5x = 15 ⇒ x = 3

Putting x = 3 in eq. (ii), 9 + y = 5 ⇒ y = -4.

22.  

Equating corresponding entries, we have

3x = x + 4 ⇒ 2x = 4 ⇒ x = 2

And 3y = 6 + x + y

⇒ 2y = 6 + 2

⇒ 2y = 8

⇒ y = 4

And 3z = -1 + z + w ⇒ 2z – w = -1    …(i)

And 3w = 2w + 3 ⇒ w = 3

Putting w = 3 in eq.(i), 2z – 3 = -1

⇒ 2z = 2 ⇒ z = 1

∴ x = 2, y = 4, w = 3.

23.  

24.  

25.  

26.  L.H.S = A3 – 6A2 + 7A + 2I

27.  

Equating corresponding entries, we have

3k – 2 = 1 ⇒ 3k = 3 ⇒ k = 1

And 4k = k ⇒ k = 1 and -4 = -2k – 2

⇒ 2k = 2

⇒ k = 1

∴ k = 1.

28.  

∴ L.H.S. = R.H.S. Proved.

29.  Let the investment in first bond be Rs x, then the investment in the second bond = Rs (3000 – x)

Interest paid by first bond  per rupee and interest paid by second bond = 5% = per rupee.

Matrix of investment is A = [x   3000 – x]1×2

Matrix of annual interest per rupee 

⇒ 210000 – 2x = 180000

⇒ x = 15000

Therefore, Investment in first bond = Rs. 15000

And investment in second bond = Rs (30000 – 15000) = Rs 15000

(b) According to question, 

⇒ 210000 – 2x = 200000

⇒ x = 5000

Therefore, Investment in first bond = Rs. 5000

And investment in second bond = Rs (30000 – 15000) = Rs 25000

30.  Let the number of books as a 1 x 3 matrix 

Let the selling prices of each book as a 3 x 1 matrix 

∴ Total amount received by selling all books 

= [120(80) + 96(60) + 120(40)] = [9600 + 5760 + 4800] = [20160]

Therefore, Total amount received by selling all the books = Rs 20160.

31.  Given: x2×n, y3×k, z2×p, wn×3, pp×k

Now, py + wy = pp×k × y3×k + wn×3 × y3×k

On comparing, k = 3 and p = n

Therefore, option (A) is correct.

32.  Here n = p (given), the order of matrices X and Z are equal.

∴ 7X – 5Z is well defined and the order of 7X – 5Z is same as the order of X and Z.

∴ The order of 7X – 5Z is either equal to 2 × n or 2 × p

But it is given that n = p

Therefore, the option (B) is correct.

33.  

34.  

35.  

36.  

37.  

38.  

39.  

Changing rows of matrix A as the columns of new matrix 

∴ A’ = A

Therefore, by definitions of symmetric matrix, A is a symmetric matrix.

Therefore, by definition matrix A is a skew-symmetric matrix.

40.  

41.  

42.  

43.  Given: A and B are symmetric matrices ⇒ A = A’ and B = B’

Now, (AB – BA)’ = (AB)’ – (BA)’

⇒ (AB – BA)’ = B’A’ – A’B’ [Reversal law]

⇒ (AB – BA)’ = BA – AB [From eq. (i)]

⇒ (AB – BA)’ = – (AB – BA)

∴ (AB – BA) is a skew matrix.

Therefore, option (A) is correct.

44.  

Equating corresponding entries, we have

Therefore, option (B) is correct.

45.  

46.  

47.  

48.  

49.  

50.  

51.  

52.  

53.  

54.  

55.  

56.  

Here, all entries in second row of left side are zero.

∴ A-1 does not exist.

57.  

58.  

Here, all entries in second row of left side are zero.

∴ A-1 does not exist.

59.  

60.  

61.  

62.  By definition of inverse of square matrix,

Option (A) is correct.

63.  Using Mathematical Induction, we see the result is true for n = 1, for

(aI + bA)n = anI + nan-1bA

Given: p(k) is true, i.e. (aI + bA)k = anI + kak-1bA

To prove: (aI + bA)k+1 = ak+1I +(k + 1)akbA

Proof: L.H.S. = (aI + bA)k+1 = (aI + bA)k (aI + bA) = (akI + kak-1bA) (aI + bA)

= ak+1I × I + kakbAI + akbAI + kak-1b2A.A

= ak+1I × I + kakbAI + akbAI + kak-1b2.0

= ak+1I + (k + 1) akbA = R.H.S.

Thus, p(k + 1) is true, therefore, p(n) is true.

64.  

65.  

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

Hence, by the principal of mathematical induction, the result is true for all positive integers n.

66.  A and B are symmetric matrices. ⇒ A’ = A and B’ = B ……….(i)

Now, (AB – BA)’ = (AB)’ – (BA)’ ⇒ (AB – BA)’ = B’A’ – A’B’ [Reversal law]

⇒ (AB – BA)’ = BA – AB [Using eq. (i)]

⇒ (AB – BA)’ = – (AB – BA)

Therefore, (AB – BA) is a skew symmetric.

67.  (B’AB)’ = [B’(AB]’ = (AB)’ (B’)’ [∵ (CD)’ = D’C’]

⇒ (B’AB)’ = B’A’B ……….(i)

Case I: A is a symmetric matrix, then ⇒ A’ = A

⇒ From eq. (i) (B’AB)’ = B’AB

⇒ B’AB is a symmetric matrix.

Case II: A is a skew symmetric matrix. ⇒ A’ = – A

Putting A’ = – A in eq. (i), (B’AB)’ = B’(– A)B = – B’AB

⇒ B’AB is a skew symmetric matrix.

68.  

Equating corresponding entries, we have

69.  

70.  

71.  

72.  According to question, the matrix 

(a) Let B be the column matrix representing sale price of each unit of products x, y, z.

Then 

Now Revenue = Sale price * Number of items sold

Therefore, the revenue collected by sale of all items in Market I = Rsb 46,000 and the revenue collected by sale of all items in Market II = Rs 53,000.

(b) Let C be the column matrix representing cost price of each unit of products x, y, z.

Then

∴ The profit collected in two markets is given in matrix form as

Profit matrix = Revenue matrix – Cost matrix

Therefore, the gross profit in both the markets =Rs 15000 + Rs 17000 = Rs 32,000.

73.  

Equating corresponding entries, we have

a + 4b = 7 …..(ii)

2a + 5b = -8  …..(iii)

3a + 6b = -9 …..(iv)

c + 4d = 2 …..(v)

2c + 5d = 4 …..(vi)

3c + 6d = 6 …..(vi)

Solving eq. (ii) and (iii), we have a = 1 and b = -2

Solving eq. (v) and (vi), we have c = 2 and d = 0

Putting these values in 

74.  Given: AB = BA ….(i)

Let p(n) : ABn = BAn  …(ii)

For n = 1, p(1) : becomes AB = BA

∴ p(1) is true for n = 1.

For n = k, p(k) : ABk = BkA

Multiplying both sides by B, ABkB = BkAB ⇒ ABk+1 = BkAB

⇒ ABk+1 = BAk+1 [From eq.(i)

∴ p(k + 1) is also true.

Therefore, p(n) is true for all n ∈ N by P.M.I

75.  

Equating corresponding entries, we have

α2 + βγ = 1

⇒ 1 – α2 – βγ = 0

Therefore, option (C) is correct.

76.  Since, A is symmetric, therefore, A’ = A ……..(i)

And A is skew-symmetric, therefore, A’ = –A

⇒ A = – A [From eq. (i)]

⇒ A + A = 0 ⇒ 2A = 0 ⇒ A = 0

Therefore, A is zero matrix.

Therefore, option (B) is correct.

77.  Given: A2 = A …..(i)

Multiplying both sides by A, A3 = A2 = A [From eq. (i)] ……(ii)

Also given (I + A)3 – 7A = I3 + A3 + 3I2A + 3IA2 – 7A

Putting A2 = A [from eq. (i)] and A3 = A [from eq. (ii)],

= I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A [∵ IA = A]

= I + 7A – 7A = I

Therefore, option (C) is correct.

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