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Misc 11 - Find matrix X so that X [1 2 3 4 5 6] - Miscellaneous

Misc. 11 - Chapter 3 Class 12 Matrices - Part 2
Misc. 11 - Chapter 3 Class 12 Matrices - Part 3
Misc. 11 - Chapter 3 Class 12 Matrices - Part 4
Misc. 11 - Chapter 3 Class 12 Matrices - Part 5
Misc. 11 - Chapter 3 Class 12 Matrices - Part 6

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Misc 11 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 These equations have u & w These equations have u & w These equations have u & w 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 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.