**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) a_{13 }→ Elaement in first row and third column = 19

a_{21 }→ Element in second row and first column = 35

a_{33 }→Element in third row and third column = -5

a_{24 }→ Element in second row and fourth column = 12

a_{23 }→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).

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 – x^{2} = 8

⇒ x^{2} – 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 a_{11} 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 = 2^{9} = 512

Therefore, option (D) is correct.

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

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

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.

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.

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.

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.

**26.** L.H.S = A^{3} – 6A^{2} + 7A + 2I

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.

∴ 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

⇒ 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: x_{2×n}, y_{3×k}, z_{2×p}, w_{n×3}, p_{p×k}

Now, py + wy = p_{p×k} × y_{3×k} + w_{n×3} × y_{3×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.

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.

**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.

Equating corresponding entries, we have

Therefore, option (B) is correct.

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

∴ A^{-1} does not exist.

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

∴ A^{-1} does not exist.

**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} = a^{n}I + na^{n-1}bA

**Given:** p(k) is true, i.e. (aI + bA)^{k} = a^{n}I + ka^{k-1}bA

**To prove:** (aI + bA)^{k+1} = a^{k+1}I +(k + 1)a^{k}bA

**Proof:** L.H.S. = (aI + bA)^{k+1} = (aI + bA)^{k} (aI + bA) = (a^{k}I + ka^{k-1}bA) (aI + bA)

= a^{k+1}I × I + ka^{k}bAI + a^{k}bAI + ka^{k-1}b^{2}A.A

= a^{k+1}I × I + ka^{k}bAI + a^{k}bAI + ka^{k-1}b^{2}.0

= a^{k+1}I + (k + 1) a^{k}bA = R.H.S.

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

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.

Equating corresponding entries, we have

**72.** According to question, the matrix

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

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.

∴ 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.

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

**74.** Given: AB = BA ….(i)

Let p(n) : AB^{n} = BA^{n} …(ii)

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

∴ p(1) is true for n = 1.

For n = k, p(k) : AB^{k} = B^{k}A

Multiplying both sides by B, AB^{k}B = B^{k}AB ⇒ AB^{k+1} = B^{k}AB

⇒ AB^{k+1} = BA^{k+1} [From eq.(i)

∴ p(k + 1) is also true.

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

Equating corresponding entries, we have

α^{2} + βγ = 1

⇒ 1 – α^{2} – 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: A^{2} = A …..(i)

Multiplying both sides by A, A^{3} = A^{2} = A [From eq. (i)] ……(ii)

Also given (I + A)^{3} – 7A = I^{3} + A^{3} + 3I^{2}A + 3IA^{2} – 7A

Putting A^{2} = A [from eq. (i)] and A^{3} = 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.