Ex 3.2
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
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