Ex 4.5, 7 - 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, 7 Solve system of linear equations, using matrix method. 5x+ 2y = 4 7x + 3y = 5 The system of equations is 5x + 2y = 4 7x + 3y = 5 Writing equation as AX = B [■8(5&2@7&3)] [■8(𝑥@𝑦)] = [■8(4@5)] Hence A = [■8(5&2@7&3)], X = [■8(𝑥@𝑦)] & B = [■8(4@5)] Calculating |A| |A| = |■8(5&2@7&3)| = 5(3) – 7(2) = 15 – 14 = 1 ≠ 0 Since |A|≠ 0, System of equations is consistent & has a unique solution Now, AX = B Solution is X = A-1 B Calculating A−1 A-1 = 1/(|A|) adj (A) A = [■8(5&2@7&3)] adj A = [■8(5&2@7&3)] = [■8(3&−2@−7&5)] Now, A-1 = 1/(|A|) adj A Putting values = 1/1 [■8(3&−2@−7&5)] = [■8(3&−2@−7&5)] Now X = A-1 B [■8(𝑥@𝑦)] = [■8(3&−2@−7&5)] [■8(4@5)] [■8(𝑥@𝑦)] = [■8(3(4)+(⤶7−2)5@−7(4)+5(5))] [■8(𝑥@𝑦)] = [■8(2@−3)] Hence, x = 2 & y = –3
