Ex 4.2, 13

Transcript

Ex 4.2, 13 By using properties of determinants, show that: 1+a2−b2﷮2ab﷮−2b﷮2ab﷮1−a2+b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ = (1 + a2+b2)3 Taking L.H.S 1+a2−b2﷮2ab﷮−2b﷮2ab﷮1−a2+b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ Applying R1 → R1 + bR3 = 1+a2−b2+𝑏(2𝑏)﷮2ab+b(−2a)﷮−2b+𝑏(1−𝑎2−𝑏2)﷮2ab﷮1−a2−b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ = 1+a2−b2+2 𝑏﷮2﷯﷮2ab−2ab﷮−2b+𝑏−𝑏 𝑎﷮2﷯− 𝑏﷮3﷯﷮2ab﷮1−a2−b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ = 𝟏+𝐚𝟐+𝐛𝟐﷮0﷮−𝐛(𝟏+𝐛𝐚𝟐+𝐛𝟐)﷮2ab﷮1−a2+b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ Taking Common (1+𝑎2+𝑏2) from R1 = (1+𝑎2+𝑏2) 1﷮0﷮−b﷮2ab﷮1−a2+b2﷮2a﷮2b﷮−2a﷮1−a2−b2﷯﷯ Applying R2 → R2 − aR3 = (1+𝑎2+𝑏2) 1﷮0﷮−b﷮2ab−𝑎(2𝑏)﷮1−a2+b2−𝑎(−2𝑎) ﷮2a−a(1−a2−b2) ﷮2b﷮−2a﷮1−a2−b2﷯﷯ = (1+𝑎2+𝑏2) 1﷮0﷮−b﷮2ab−2𝑎𝑏﷮1−a2+b2+2𝑎2﷮2a−a+a3+ab2﷮2b﷮−2a﷮1−a2−b2﷯﷯ = (1+𝑎2+𝑏2) 1﷮0﷮−b﷮0﷮1+b2+𝑎2﷮a+a3+ab2﷮2b﷮−2a﷮1−a2−b2﷯﷯ Taking Common (1+𝑎2+𝑏2) from R2 = (1+𝑎2+𝑏2)2 1﷮0﷮−b﷮0﷮1﷮a﷮2b﷮−2a﷮1−a2−b2﷯﷯ = (1+𝑎2+𝑏2)2 1 1﷮𝑎﷮−2𝑎﷮1−a2−b2﷯﷯−0 0﷮x﷮−2𝑎﷮1−a2−b2﷯﷯+𝑏 0﷮1﷮2b﷮−2𝑏﷯﷯﷯ = (1+𝑎2+𝑏2)2 1 1﷮𝑎﷮−2𝑎﷮1−a2−b2﷯﷯−0+𝑏 0﷮1﷮2b﷮−2𝑏﷯﷯﷯ = (1+𝑎2+𝑏2)2 (1(1 – a2 – b2) + 2a2) – b (0 – 2b)) = (1+𝑎2+𝑏2)2 (1 – a2 – b2 + 2a2 + 2a2) = (1+𝑎2+𝑏2)2 (1 + a2 + b2) = 1+𝑎2+𝑏2﷯3 = R.H.S Hence proved