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Example 14 - Chapter 4 Class 12 Determinants
Last updated at May 29, 2023 by Teachoo
Inverse of two matrices and verifying properties
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
Example 14 If A = 2 3 1 4 and B = 1 2 1 3 , then verify that (AB)-1 = B-1 A-1 Taking L .H.S (AB) 1 First calculating AB AB = 2 3 1 4 1 2 1 3 = 2 1 +3( 1) 2 2 +3(3) 1 1 +( 4)( 1) 1 2 + 4 3 = 2 3 4+9 1+4 2 12 = 1 5 5 14 Now, (AB)-1 = 1 |AB| adj (AB) exists if |AB| 0 |AB| = 1 5 5 14 = (-1)(-14) 5(5) = 14 25 = 11 Since |AB| 0 (AB)-1 exists AB = 1 5 5 14 adj (AB) = 1 5 5 14 = 14 5 5 1 Now, (AB) 1 = 1 |AB| adj (AB) Putting values = 1 11 14 5 5 1 = 1 11 14 5 5 1 Taking R.H.S B-1A-1 First Calculating B 1 B = 1 2 1 3 B = 1 |B| adj (B) exists if |B| 0 |B| = 1 2 1 3 = 3 2 = 1 Since |B| 0 , B-1 exist B = 1 2 1 3 adj (B) = 1 2 1 3 = 3 2 1 1 Now, (B) 1 = 1 |B| adj (B) Putting values = 1 1 3 2 1 1 = 3 2 1 1 Finding A-1 A-1 = 1 |A| adj (A) exists if |A| 0 |A| = 2 3 1 4 = 2 ( 4) 1( 3) = 8 3 = 11 Since |A| 0 , A 1 exists A = 2 3 1 4 adj (A) = 2 3 1 4 = 4 3 1 2 Now, A-1 = 1 |A| adj (A) = 1 11 4 3 1 2 = 1 11 4 3 1 2 Thus, B-1A-1 = 3 2 1 1 1 11 4 3 1 2 = 1 11 3 2 1 1 4 3 1 2 = 1 11 3 4 +2(1) 3 3 +2( 2) 1 4 +1(1) 1 3 +1( 2) = 1 11 12+2 9 4 4+1 3 2 = 1 11 14 5 5 1 = L.H.S L.H.S = R.H.S Hence proved
