Ex 3.2

Ex 3.2, 1

Ex 3.2, 2 (i)

Ex 3.2, 2 (ii) Important

Ex 3.2, 2 (iii)

Ex 3.2, 2 (iv)

Ex 3.2, 3 (i)

Ex 3.2, 3 (ii) Important

Ex 3.2, 3 (iii)

Ex 3.2, 3 (iv) Important

Ex 3.2, 3 (v)

Ex 3.2, 3 (vi) Important

Ex 3.2, 4

Ex 3.2, 5

Ex 3.2, 6

Ex 3.2, 7 (i)

Ex 3.2, 7 (ii) Important

Ex 3.2, 8

Ex 3.2, 9

Ex 3.2, 10

Ex 3.2, 11

Ex 3.2, 12 Important

Ex 3.2, 13 Important

Ex 3.2, 14

Ex 3.2, 15

Ex 3.2, 16 Important

Ex 3.2, 17 Important

Ex 3.2, 18

Ex 3.2, 19 Important

Ex 3.2, 20 Important

Ex 3.2, 21 (MCQ) Important You are here

Ex 3.2, 22 (MCQ) Important

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 3.2, 21 (Introduction) Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 , and p × 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 PY + WY =[■8(1&0@5&6@3&0)]_(3 × 2) + [■8(10&15@3&0@9&3)]_(3 × 2)= [■8(11&15@8&6@12&3)]_(3 × 2) Thus, PY is defined as 3 × 2 & 2 × 2 is 3 × 2 WY is defined as 3 × 2 & 2 × 2 is 3 × 2 & Order of PY + WY = Order of PY = Order of WY Ex 3.2, 21 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 , and p × 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 Order of P is p × k Order of Y is 3 × k PY = [P]_(p × k) [Y]_(3 × 𝑘) This is possible only if k = 3 So, 〖𝑷𝒀〗_(𝒑 × 𝒌) Order of W is n × 3 Order of Y is 3 × k WY = [W]_(𝑛 ×3) [Y]_(3 × 𝑘) Since 3 = 3 it is defined So, 〖𝑾𝒀〗_(𝒏 × 𝒌) Now, PY_(𝑝 × 𝑘) + WY_(𝑛 × 𝑘) is possible if p × k = n × k p = n Thus p = n and k = 3 Hence, correct answer is A