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![Ex 4.1, 3 - Show that |2A| = 4|A|, if A = [1 2 4 2] - Ex 4.1 - Ex 4.1](https://cdn.teachoo.com/8cf4ab8c-5e85-4af6-8505-bf88bba9b921/slide4.jpg)
Ex 4.1, 3 - Chapter 4 Class 12 Determinants
Last updated at August 17, 2026 by Teachoo
Finding determinant of a 2x2 matrix
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
Ex 4.1, 3 If A = [■8(1&2@4&2)] , then show that |2A| = 4|A| We need to prove |2A| = 4|A| Solving L.H.S |2A| First calculating 2 A 2A = 2 [■8(1&2@4&2)] = [■8( 2 × 1&2 × 2@ 2 × 4&2 × 2)] = [■8(𝟐&𝟒@𝟖&𝟒)] So, |𝟐𝑨| = |■8(2&4@8&4)| = 2(4) – 8 (4) = 8 – 32 = –24 Taking R.H.S 4|𝑨| As A = [■8(1&2@4&2)] So |𝑨| = |■8(1&2@4&2)| = 1 (2) – 4 (2) = 2 – 8 = –6 Hence, 4|𝑨| = 4 (−6) = −24 = L.H.S ∴ L.H.S = R.H.S Hence proved
