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Misc 8 Find the matrix X so that X [■8(1&2&3@4&5&6)] =[■8(−7&−8&−9@2&4&6)] X [■8(1&2&3@4&5&6)] = [■8(−7&−8&−9@2&4&6)] X [■8(𝟏&𝟐&𝟑@𝟒&𝟓&𝟔)]_(𝟐 × 𝟑) = [■8(−𝟕&−𝟖&−𝟗@𝟐&𝟒&𝟔)]_(𝟐 × 𝟑) So X will be a × matrix Let X =[■8(𝑢&𝑤@𝑣&𝑥)]_(2 × 2) So, our equation becomes [■8(𝑢&𝑤@𝑣&𝑥)]_(2 × 2) [■8(1&2&3@4&5&6)]_(2 × 3) = [■8(−7&−8&−9@2&4&6)] [■8(𝑢(1)+𝑤(4)&𝑢(2)+𝑤(5)&𝑢(3)+𝑤(6)@𝑣(1)+𝑥(4)&𝑣(2)+𝑥(5)&𝑣(3)+𝑥(6))] = [■8(−7&−8&−9@2&4&6)] [■8(𝒖+𝟒𝒘&𝟐𝒖+𝟓𝒘&𝟑𝒖+𝟔𝒘@𝒗+𝟒𝒙&𝟐𝒗+𝟓𝒙&𝟑𝒗+𝟔𝒙)]_(𝟐×𝟑) = [■8(−𝟕&−𝟖&−𝟗@𝟐&𝟒&𝟔)]_(𝟐×𝟑) Since the matrices are equal Corresponding elements are equal u + 4w = - 7 2u + 5w = - 8 3u + 6w = - 9 v + 4x = 2 2v + 5x = 4 3v + 6x = 6 Solving (1) u + 4w = −7 u = −7 – 4w Putting value of u in (2) 2u + 5w = - 8 2(−7 – 4w) + 5w = - 8 −14 – 8w + 5w = - 8 −14 – 3w = - 8 −3w = - 8 + 14 −3w = 6 w = 6/(−3) w = –2 Now, u = – 7 – 4w Putting w = −2 u = – 7 – 4 (-2) u = – 7 + 8 u = 1 Solving (4) v + 4x = 2 v = 2 – 4x Putting value of v in (5) 2v + 5x = 4 2 (2 – 4x) + 5x = 4 4 – 8x + 5x = 4 4 – 3x = 4 −3x = 4 – 4 −3x = 0 x = 0 Putting value of x = 0 in (4) v + 4x = 2 v + 4(0) =2 v + 0 = 2 v = 2 Hence, u = 1 , v = 2 , w = − 2 & x = 0 Hence, matrix X = [■8(u&w@v&x)] = [■8(𝟏&−𝟐@𝟐&𝟎)]

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.