Since L.H.S. = R.H.S.

Hence, proved.

Since L.H.S. = R.H.S.

Hence, proved.

**5.** Evaluate the determinants:

= 3(0 – 5) + 1{0 – (-3)} – 2(0 – 0)

= -15 + 3 – 0 = -12

= 3(1 + 6) + 4{1 – (-4)} + 5(3 – 2)

= 3 × 7 + 4 × 5 + 5 × 1

= 21 + 20 + 5 = 46

= 0(0 + 9) – (0 – 6) + 2 (-3 – 0)

= 0 + 6 – 6 = 0

= 2(0 – 5) + (0 + 3) – 2(0 – 6)

= -10 + 3 + 12 = 5

= {-9 -(-12)} – {-18 -(-15)}-2 (8 – 5)

= -9 + 12 -(-18 + 15) -2 (3)

= 3 – (-3) – 6

= 3 + 3 – 6 = 0.

⇒ 2 – 20 = 2x^{2} – 24

⇒ -18 = 2x^{2} – 24

⇒ 2x^{2} = -18 + 24

⇒ 2x^{2} = 6

⇒ x^{2} = 3

⇒ x = ±√3.

⇒ 10 – 12 = 5x – 6x

⇒ -2 = -x

⇒ x = 2

⇒ x^{2} – 36 = 36 – 36

⇒ x^{2} – 36 = 0

⇒ x^{2} = 36

⇒ x = ±6

Therefore, option (B) is correct.

[∵ All entries of one column here first are zero]

= 9 × 0 = 0 [∵ two columns are identical] Proved.

[Taking (-1) common from each row]

Interchanging rows and columns in the determinants on R.H.S.,

Taking common a, b, c from R_{1}, R_{2}, R_{3} respectively.

**24. **Since, Determinant is a number associated to a square matrix.

Therefore, option (C) is correct.

Taking positive sign, -k + 4 = 4

⇒ k = 0

Taking negative sign, -k + 4 = -4

⇒ k = 8

Taking positive sign, -k + 4 = 4

⇒ k = 0

Taking negative sign, -k + 4 = -4

⇒ k = 8

**28. **(i) Let P be any point on the line joining the points (1, 2) and (3, 6).

Then, Area of triangle that could be formed by these points is zero.

⇒ y = 2x which is required line.

(ii) Let P(x, y) be any point on the line joining the points (3, 1) and (9, 3).

Then, Area of triangle that could be formed by these points is zero.

⇒ x – 3y = 0 which is required line.

Taking positive sign, 25 – 5k = 35

⇒ k = -2

Taking negative sign, 25 – 5k = -35

⇒ k = 12

Therefore, option (D) is correct.

M_{11} = Minor of a_{11} = |3| = 3 and A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} (3) = 3

M_{12} = Minor of a_{12} = |0| = 0 and A_{12} = (-1)^{1+2} M_{12 }= (-1)^{3} (0) = 0

M_{21} = Minor of a_{21} = |-4| = 4 and A_{21} = (-1)^{2+1} M_{21} = (-1)^{3} (-4) = 4

M_{22} = Minor of a_{22} = |2| = 2 and A_{22} = (-1)^{2+2} M_{22} = (-1)^{4} (2) = 2

M_{11} = Minor of a_{11} = |d| = d and A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} (d) = d

M_{12} = Minor of a_{12} = |b| = b and A_{12} = (-1)^{1+2} M_{12} = (-1)^{3} (b) = b

M_{21} = Minor of a_{21} = |c| = c and A_{21} = (-1)^{2+1} M_{21} = (-1)^{3} (c) = c

M_{22} = Minor of a_{22} = |a| = a and A_{22} = (-1)^{2+2} M_{22} = (-1)^{4} (a) = a

**34.** Option (D) is correct.

∴ A_{11} = Cofactor of a_{11} = (-1)^{2} (4) = 4

A_{12} = Cofactor of a_{12} = (-1)^{3} (4) = 4

A_{21} = Cofactor of a_{21} = (-1)^{3} (4) = 4

A_{22} = Cofactor of a_{22} = (-1)^{4} (4) = 4

∴ From eq.(i), (ii) and (iii) A.(adj.A) = (adj.A).A = |A|I

∴ We have 11 + 3a + b = 0 ……(i)

8 + 2a + 0 = 0 ….(ii)

⇒ 2a = -8

⇒ a = -4

Here a = -4 satisfies 4 + a + 0 = 0 also, therefore a= -4

Putting a = -4 in eq.(i), 11 – 12 + b = 0 ⇒ b – 1 = 0 ⇒ b = 1

Here also b = 1 satisfies 3 + a + b = 0, therefore b = 1

Therefore, a = -4 and b = 1

**51. ** If A is a non-singular matrix of order n x n then |adj.A| = |A|^{n-1}

∴ Putting n = 3, |adj.A| = |A|^{2}

Therefore, option (B) is correct.

**52.** Since AA^{-1} = I

Therefore, option (B) is correct.

**53.** Matrix form of given equations is AX = B

Therefore, Unique solution and hence equations are consistent.

**54. **Matrix form of given equations is AX = B

Therefore, Unique solution and hence equations are consistent.

**55.** Matrix form of given equations is AX = B

Therefore, given equations are inconsistent, i.e., have no common solution.

**56.** Matrix form of given equations is AX = B

Therefore, Unique solution and hence equations are consistent.

**57.** Matrix form of given equations is AX = B

Therefore, given equations are inconsistent.

**58.** Matrix form of given equations is AX = B

Therefore, Unique solution and hence equations are consistent.

**59.** Matrix form of given equations is AX = B

**60.** Matrix form of given equations is AX = B

**61.** Matrix form of given equations is AX = B

**62.** Matrix form of given equations is AX = B

Therefore, x = -1 and y = 4.

**63.** Matrix form of given equations is AX = B

**64.** Matrix form of given equations is AX = B

Therefore, x = 2, y = -1 and z = 1.

**65. ** Matrix form of given equations is AX = B

Therefore, x = 1, y = 2 and z = -1.

**66.** Matrix form of given equations is AX = B

Therefore, x = 2, y = 1 and z = 3.

Therefore, x = 1, y = 2 and z = 3.

**68. **Let Rs x, Rs y, Rs z per kg be the prices of onion, wheat and rice respectively.

∴ According to given data, we have three equations,

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

Matrix form of given equations is AX = B

Therefore, x = 5, y = 8 and z = 8

Hence, the cost of onion, wheat and rice are Rs. 5, Rs 8 per kg.

Therefore, from eq.(i) and (ii),

either a + b + c = 0 or a = b = c.

But this is contrary as given that a ≠ 0.

Therefore, from eq.(i) is only the solution.

**85.** According to question, b – a = c – b ……….(i)

[From eq. (i)] = 0 [∵ R_{2} and R_{3} have become identical]

Therefore, option (A) is correct.