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Ex 4.5, 9 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Find solution of equations- Equations given
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
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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 4.5, 9 Solve system of linear equations, using matrix method. 4x – 3y = 3 3x – 5y = 7 The system of equation is 4x – 3y = 3 3x – 5y = 7 Writing equation as AX = B [■8(4&−3@3&−5)] [■8(𝑥@𝑦)] = [■8(3@7)] Hence A = [■8(4&−3@3&−5)], X = [■8(𝑥@𝑦)] & B = [■8(3@7)] Calculating |A| |A| = |■8(4&−3@3&−5)| = 4(–5) – 3 (–3) = –20 + 9 = – 11 Since |A| ≠ 0 ∴ System of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) A = [■8(4&−3@3&−5)] adj A = [■8(4&−3@3&−5)] = [■8(−5&3@−3&4)] Now, A-1 = 1/(|A|) adj A A-1 = 1/(−11) [■8(−5&3@−3&4)] = 1/11 [■8(5&−3@3&−4)] Thus, X = A-1 B [■8(𝑥@𝑦)] = 1/11 [■8(5&−3@3&−4)] [■8(3@7)] [■8(𝑥@𝑦)] = 1/11 [■8(5(3)+(⤶7−3)7@3(3)+(−4)7)] = 1/11 [■8(15−21@9−28)] [■8(𝑥@𝑦)] = [((−6)/11)¦((−19)/11)] Hence, x = (−𝟔)/𝟏𝟏 & y = (−𝟏𝟗)/𝟏𝟏
