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  1. Chapter 3 Class 12 Matrices
  2. Serial order wise

Transcript

Misc 6 Find the values of x, y, z if the matrix A = [โ– 8(0&2๐‘ฆ&๐‘ง@๐‘ฅ&๐‘ฆ&โˆ’๐‘ง@๐‘ฅ&โˆ’๐‘ฆ&๐‘ง)] satisfy the equation Aโ€ฒA = I. Given, A = [โ– 8(0&2๐‘ฆ&๐‘ง@๐‘ฅ&๐‘ฆ&โˆ’๐‘ง@๐‘ฅ&โˆ’๐‘ฆ&๐‘ง)] Aโ€™ = [โ– 8(0&๐‘ฅ&๐‘ฅ@2๐‘ฆ&๐‘ฆ&โˆ’๐‘ฆ@๐‘ง&โˆ’๐‘ง&๐‘ง)] I = [โ– 8(1&0&0@0&1&0@0&0&1)] Now, Aโ€™A = I Putting values [โ– 8(0&๐‘ฅ&๐‘ฅ@2๐‘ฆ&๐‘ฆ&โˆ’๐‘ฆ@๐‘ง&โˆ’๐‘ง&๐‘ง)][โ– 8(0&2๐‘ฆ&๐‘ง@๐‘ฅ&๐‘ฆ&โˆ’๐‘ง@๐‘ฅ&โˆ’๐‘ฆ&๐‘ง)] = [โ– 8(1&0&0@0&1&0@0&0&1)] [โ– 8(0(0)+๐‘ฅ(๐‘ฅ)+๐‘ฅ(๐‘ฅ)&0(2๐‘ฆ)+๐‘ฅ(๐‘ฆ)+๐‘ฅ(โˆ’๐‘ฆ)&0(๐‘ง)+๐‘ฅ(โˆ’๐‘ง)+๐‘ฅ(๐‘ง)@2๐‘ฆ(0)+๐‘ฆ(๐‘ฅ)โˆ’๐‘ฆ(๐‘ฅ)&2๐‘ฆ(2๐‘ฆ)+๐‘ฆ(๐‘ฆ)โˆ’๐‘ฆ(โˆ’๐‘ฆ)&2๐‘ฆ(๐‘ง)+๐‘ฆ(โˆ’๐‘ง)โˆ’๐‘ฆ(๐‘ง)@๐‘ง(0)โˆ’๐‘ง(๐‘ฅ)+๐‘ง(๐‘ฅ)&๐‘ง(2๐‘ฆ)โˆ’๐‘ง(๐‘ฆ)+๐‘ง(โˆ’๐‘ฆ)&๐‘ง(๐‘ง)โˆ’๐‘ง(โˆ’๐‘ง)+๐‘ง(๐‘ง))] = [โ– 8(1&0&0@0&1&0@0&0&1)] [โ– 8(0+๐‘ฅ^2+๐‘ฅ^2&0+๐‘ฅ๐‘ฆโˆ’๐‘ฅ๐‘ฆ&0โˆ’๐‘ฅ๐‘ง+๐‘ฅ๐‘ง@0+๐‘ฅ๐‘ฆโˆ’๐‘ฅ๐‘ฆ&4๐‘ฆ^2+๐‘ฆ^2+๐‘ฆ^2&2๐‘ง๐‘ฆโˆ’๐‘ง๐‘ฆโˆ’๐‘ง๐‘ฆ@0โˆ’๐‘ฅ๐‘ง+๐‘ฅ๐‘ง&2๐‘ง๐‘ฆโˆ’๐‘ง๐‘ฆโˆ’๐‘ง๐‘ฆ&๐‘ง^2+๐‘ง^2+๐‘ง^2 )]= [โ– 8(1&0&0@0&1&0@0&0&1)] [โ– 8(2๐‘ฅ^2&0&0@0&6๐‘ฆ^2&0@0&0&3๐‘ง^2 )]= [โ– 8(1&0&0@0&1&0@0&0&1)] Since matrices are equal, corresponding elements are equal Thus, x = ยฑ 1/โˆš2 , y = ยฑ 1/โˆš6 , z = ยฑ 1/โˆš3 2x2 = 1 x2 = 1/2 x = ยฑโˆš(1/2) x = ยฑ ๐Ÿ/โˆš๐Ÿ 6y2 = 1 y2 = 1/6 y = ยฑโˆš(1/6) y = ยฑ ๐Ÿ/โˆš๐Ÿ” 3z2 = 1 z2 = 1/3 z = ยฑโˆš(1/3) z = ยฑ ๐Ÿ/โˆš๐Ÿ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.