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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.