Ex 4.2

Chapter 4 Class 12 Determinants
Serial order wise

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### Transcript

Ex 4.2, 8 By using properties of determinants, show that: (i) |■8(1&𝑎&𝑎[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2)| = (a - b) (b - c)(c – a) Solving L.H.S |■8(1&𝑎&𝑎[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2)| Applying R1 → R1 − R2 = |■8(𝟏−𝟏&𝑎−𝑏&𝑎^2−𝑏^[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2 ) | = |■8(𝟎&(𝑎−𝑏)&(𝑎−𝑏)(𝑎+𝑏)@1&𝑏&𝑏[email protected]&𝑐&𝑐2 ) | = |■8(0(𝐚−𝐛)&(𝐚−𝐛)&(𝐚−𝒃)(a+b)@1&b&[email protected]&c&c2 ) | Taking Common (a – b) from R1 = (𝐚−𝒃) |■8(0&1&[email protected]&b&[email protected]&c&c2 ) | Applying R2 → R2 − R3 = (a−b) |■8(0&1&[email protected]𝟏−𝟏&b−c&b2−[email protected]&c&c2 ) | = (a – b) |■8(0&1&a+𝑏@𝟎&b−c&(b−c)(b+c)@1&c&c2 ) | Taking common (b – c) from R2 = (a – b) (b – c) |■8(0&1&[email protected]&1&[email protected]&c&c2 ) | Expanding Determinant along C1 = (a – b) (b – c) ( 0|■8(1&𝑏+𝑐@𝑐&𝑐2)|−0|■8(1&𝑎+𝑏@𝑐&𝑐2)|+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) ( 0−0+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) (1(b + c) – 1(a + b) ) = (a – b) (b – c) (b + c – a – b) = (a – b) (b – c)(c – a) = R.H.S Hence Proved