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Ex 4.4, 13 - Chapter 4 Class 12 Determinants
Last updated at May 29, 2023 by Teachoo
Finding inverse when Equation of matrice 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
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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 4.4, 13 If A = [■8(3&1@−1&2)] show that A2 – 5A + 7I = O. Hence find A–1. Calculating A2 A2 = A.A = [■8(3&1@−1&2)] [■8(3&1@−1&2)] =[■8(3(3)+1(−1)&3(1)+1(2)@−1(3)+2(−1)&−1(1)+2(2))] = [■8(9−1&3+2@−3−2&−1+4)] = [■8(8&5@−5&3)] Solving L.H.S A2 – 5A + 7I = [■8(8&5@−5&3)] – 5 [■8(3&1@−1&2)] + 7 [■8(1&[email protected]&1)] = [■8(8&5@−5&3)] – [■8(5(3)&5(1)@5(−1)&5(2))] + [■8(7(1)&7(0)@7(0)&7(1))] = [■8(8&5@−5&3)] – [■8(15&5@−5&10)] + [■8(7&[email protected]&7)] = [■8(8−15+7&5−5+0@−5−(−5)+0&3−10+7)] = [■8(8−15+7&5−5+0@−5+5+0&3−10+7)] = [■8(0&[email protected]&0)] = O Thus, A2 – 5A + 7I = O Hence proved Finding A–1 A2 – 5A + 7I = O Pre multiplying A-1 both sides A-1 (A2 – 5A + 7I ) = A-1 O A-1 . A2 – 5A-1A + 7A-1 = O A-1 AA – 5(A-1 A) + 7A-1 = O (A-1A)A – 5 (A-1 A) + 7 (A-1 I) = O IA – 5I + 7A-1 = O A – 5I + 7 A-1 = 0 7A-1 = 5I – A A-1 = 𝟏/𝟕 (5I – A) (AA-1 = I ) (IA = A) Putting values A-1 = 1/7 (5[■8(1&[email protected]&1)]−[■8(3&1@−1&2)]) = 1/7 ([■8(5&[email protected]&5)]−[■8(3&1@−1&2)]) = 1/7 ([■8(5−3&0−[email protected]−(−1)&5−2)]) = 1/7 [■8(2&−[email protected]&3)] Thus, A-1 = 𝟏/𝟕 [■8(𝟐&−𝟏@𝟏&𝟑)]
