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Misc 3 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(๐ŸŽ&๐’™&๐’™@๐Ÿ๐’š&๐’š&โˆ’๐’š@๐’›&โˆ’๐’›&๐’›)] I = [โ– 8(1&0&0@0&1&0@0&0&1)] Now, Aโ€™A = I Putting values [โ– 8(๐ŸŽ&๐’™&๐’™@๐Ÿ๐’š&๐’š&โˆ’๐’š@๐’›&โˆ’๐’›&๐’›)][โ– 8(๐ŸŽ&๐Ÿ๐’š&๐’›@๐’™&๐’š&โˆ’๐’›@๐’™&โˆ’๐’š&๐’›)] = [โ– 8(๐Ÿ&๐ŸŽ&๐ŸŽ@๐ŸŽ&๐Ÿ&๐ŸŽ@๐ŸŽ&๐ŸŽ&๐Ÿ)] [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(๐Ÿ๐’™^๐Ÿ&๐ŸŽ&๐ŸŽ@๐ŸŽ&๐Ÿ”๐’š^๐Ÿ&๐ŸŽ@๐ŸŽ&๐ŸŽ&๐Ÿ‘๐’›^๐Ÿ )]= [โ– 8(๐Ÿ&๐ŸŽ&๐ŸŽ@๐ŸŽ&๐Ÿ&๐ŸŽ@๐ŸŽ&๐ŸŽ&๐Ÿ)] 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‘

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

About the Author

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