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Question 8 (i) - Whole row/column one - Chapter 4 Class 12 Determinants
Last updated at Dec. 16, 2024 by Teachoo
Chapter 4 Class 12 Determinants
Concept wise
- Finding determinant of a 2x2 matrix
- Evalute determinant of a 3x3 matrix
- Area of triangle
- Equation of line using determinant
- Finding Minors and cofactors
- Evaluating determinant using minor and co-factor
- Find adjoint of a matrix
- Finding Inverse of a matrix
- Inverse of two matrices and verifying properties
- Finding inverse when Equation of matrice given
- Checking consistency of equations
- Find solution of equations- Equations given
- Find solution of equations- Statement given
- Verifying properties of a determinant
- Two rows or columns same
- Whole row/column zero
-
Whole row/column one
- Making whole row/column one and simplifying
- Proving Determinant 1 = Determinant 2
- Solving by simplifying det.
- Using Property 5 (Determinant as sum of two or more determinants)
Transcript
Question 8 By using properties of determinants, show that: (i) |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| = (a - b) (b - c)(c – a) Solving L.H.S |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| Applying R1 → R1 − R2 = |■8(𝟏−𝟏&𝑎−𝑏&𝑎^2−𝑏^2@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(𝟎&(𝑎−𝑏)&(𝑎−𝑏)(𝑎+𝑏)@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(0(𝐚−𝐛)&(𝐚−𝐛)&(𝐚−𝒃)(a+b)@1&b&b2@1&c&c2 ) | Taking Common (a – b) from R1 = (𝐚−𝒃) |■8(0&1&a+b@1&b&b2@1&c&c2 ) | Applying R2 → R2 − R3 = (a−b) |■8(0&1&a+b@𝟏−𝟏&b−c&b2−c2@1&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&a+b@0&1&b+c@1&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
