Ex 4.2

Chapter 4 Class 12 Determinants
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Ex 4.2, 13 By using properties of determinants, show that: |■8(1+a2−b2&2ab&−[email protected]&1−a2+b2&[email protected]&−2a&1−a2−b2)| = (1 + a2+b2)3 Solving L.H.S |■8(1+a2−b2&2ab&−[email protected]&1−a2+b2&[email protected]&−2a&1−a2−b2)| Applying R1 → R1 + bR3 = |■8(1+a2−b2+𝑏(2𝑏)&2ab+b(−2a)&−2b+𝑏(1−𝑎2−𝑏2)@2ab&1−a2−b2&[email protected]&−2a&1−a2−b2)| = |■8(1+a2−b2+2𝑏^2&2ab−2ab&−2b+𝑏−𝑏𝑎^2−𝑏^[email protected]&1−a2−b2&[email protected]&−2a&1−a2−b2)| = |■8(𝟏+𝐚𝟐+𝐛𝟐&0&−𝑏(𝟏+𝐚𝟐+𝐛𝟐)@2ab&1−a2+b2&[email protected]&−2a&1−a2−b2)| Taking Common (1+𝑎2+𝑏2) from R1 = (1+𝑎2+𝑏2) |■8(1&0&−[email protected]&1−a2+b2&[email protected]&−2a&1−a2−b2)| Applying R2 → R2 − aR3 = (1+𝑎2+𝑏2) |■8(1&0&−[email protected]−𝑎(2𝑏)&1−a2+b2−𝑎(−2𝑎) &2a−a(1−a2−b2) @2b&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−[email protected]−2𝑎𝑏&1−a2+b2+2𝑎2&2a−[email protected]&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−[email protected]&1+b2+𝑎2&[email protected]&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−[email protected]&𝟏+𝒂𝟐+𝒃𝟐&a(𝟏+𝐚𝟐+𝒃𝟐) @2b&−2a&1−a2−b2)| Taking Common (1+𝑎2+𝑏2) from R2 = (1+𝑎2+𝑏2)2 |■8(1&0&−[email protected]&1&[email protected]&−2a&1−a2−b2)| = (1+𝑎2+𝑏2)2 ( 1|■8(1&𝑎@−2𝑎&1−a2−b2)|−0|■8(0&[email protected]−2𝑎&1−a2−b2)|−𝑏|■8(0&[email protected]&−2𝑏)|) = (1+𝑎2+𝑏2)2 ( 1|■8(1&𝑎@−2𝑎&1−a2−b2)|−0−𝑏|■8(0&[email protected]&−2𝑏)|) = (1+𝑎2+𝑏2)2 (1(1 – a2 – b2) + 2a2) – b (0 – 2b)) = (1+𝑎2+𝑏2)2 (1 – a2 – b2 + 2a2 + 2b2) = (1+𝑎2+𝑏2)2 (1 + a2 + b2) = (1+𝑎2+𝑏2)3 = R.H.S Hence proved