Transpose of a matrix

Chapter 3 Class 12 Matrices (Term 1)
Concept wise

### Transcript

Misc 6 Find the values of x, y, z if the matrix A = [β 8(0&2π¦&π§@π₯&π¦&βπ§@π₯&βπ¦&π§)] satisfy the equation Aβ²A = I. Given, A = [β 8(0&2π¦&π§@π₯&π¦&βπ§@π₯&βπ¦&π§)] Aβ = [β 8(0&π₯&π₯@2π¦&π¦&βπ¦@π§&βπ§&π§)] I = [β 8(1&0&0@0&1&0@0&0&1)] Now, AβA = I Putting values [β 8(0&π₯&π₯@2π¦&π¦&βπ¦@π§&βπ§&π§)][β 8(0&2π¦&π§@π₯&π¦&βπ§@π₯&βπ¦&π§)] = [β 8(1&0&0@0&1&0@0&0&1)] [β 8(0(0)+π₯(π₯)+π₯(π₯)&0(2π¦)+π₯(π¦)+π₯(βπ¦)&0(π§)+π₯(βπ§)+π₯(π§)@2π¦(0)+π¦(π₯)βπ¦(π₯)&2π¦(2π¦)+π¦(π¦)βπ¦(βπ¦)&2π¦(π§)+π¦(βπ§)βπ¦(π§)@π§(0)βπ§(π₯)+π§(π₯)&π§(2π¦)βπ§(π¦)+π§(βπ¦)&π§(π§)βπ§(βπ§)+π§(π§))] = [β 8(1&0&0@0&1&0@0&0&1)] [β 8(0+π₯^2+π₯^2&0+π₯π¦βπ₯π¦&0βπ₯π§+π₯π§@0+π₯π¦βπ₯π¦&4π¦^2+π¦^2+π¦^2&2π§π¦βπ§π¦βπ§π¦@0βπ₯π§+π₯π§&2π§π¦βπ§π¦βπ§π¦&π§^2+π§^2+π§^2 )]= [β 8(1&0&0@0&1&0@0&0&1)] [β 8(2π₯^2&0&0@0&6π¦^2&0@0&0&3π§^2 )]= [β 8(1&0&0@0&1&0@0&0&1)] Since matrices are equal, corresponding elements are equal Thus, x = Β± 1/β2 , y = Β± 1/β6 , z = Β± 1/β3 2x2 = 1 x2 = 1/2 x = Β±β(1/2) x = Β± π/βπ 6y2 = 1 y2 = 1/6 y = Β±β(1/6) y = Β± π/βπ 3z2 = 1 z2 = 1/3 z = Β±β(1/3) z = Β± π/βπ

#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.