Slide76.JPG

Slide77.JPG

  Slide78.JPG

Slide79.JPG

Go Ad-free

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo