Miscellaneous

Chapter 3 Class 12 Matrices
Serial order wise

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Misc 6 Find the values of x, y, z if the matrix A = [â 8(0&2đŠ&đ§@đ„&đŠ&âđ§@đ„&âđŠ&đ§)] satisfy the equation AâČA = I. Given, A = [â 8(0&2đŠ&đ§@đ„&đŠ&âđ§@đ„&âđŠ&đ§)] Aâ = [â 8(0&đ„&đ„@2đŠ&đŠ&âđŠ@đ§&âđ§&đ§)] I = [â 8(1&0&[email protected]&1&[email protected]&0&1)] Now, AâA = I Putting values [â 8(0&đ„&đ„@2đŠ&đŠ&âđŠ@đ§&âđ§&đ§)][â 8(0&2đŠ&đ§@đ„&đŠ&âđ§@đ„&âđŠ&đ§)] = [â 8(1&0&[email protected]&1&[email protected]&0&1)] [â 8(0(0)+đ„(đ„)+đ„(đ„)&0(2đŠ)+đ„(đŠ)+đ„(âđŠ)&0(đ§)+đ„(âđ§)+đ„(đ§)@2đŠ(0)+đŠ(đ„)âđŠ(đ„)&2đŠ(2đŠ)+đŠ(đŠ)âđŠ(âđŠ)&2đŠ(đ§)+đŠ(âđ§)âđŠ(đ§)@đ§(0)âđ§(đ„)+đ§(đ„)&đ§(2đŠ)âđ§(đŠ)+đ§(âđŠ)&đ§(đ§)âđ§(âđ§)+đ§(đ§))] = [â 8(1&0&[email protected]&1&[email protected]&0&1)] [â 8(0+đ„^2+đ„^2&0+đ„đŠâđ„đŠ&0âđ„đ§+đ„đ§@0+đ„đŠâđ„đŠ&4đŠ^2+đŠ^2+đŠ^2&2đ§đŠâđ§đŠâđ§đŠ@0âđ„đ§+đ„đ§&2đ§đŠâđ§đŠâđ§đŠ&đ§^2+đ§^2+đ§^2 )]= [â 8(1&0&[email protected]&1&[email protected]&0&1)] [â 8(2đ„^2&0&[email protected]&6đŠ^2&[email protected]&0&3đ§^2 )]= [â 8(1&0&[email protected]&1&[email protected]&0&1)] Since matrices are equal, corresponding elements are equal Thus, x = Â± 1/â2 , y = Â± 1/â6 , z = Â± 1/â3 2x2 = 1 x2 = 1/2 x = Â±â(1/2) x = Â± đ/âđ 6y2 = 1 y2 = 1/6 y = Â±â(1/6) y = Â± đ/âđ 3z2 = 1 z2 = 1/3 z = Â±â(1/3) z = Â± đ/âđ