Last updated at May 29, 2018 by Teachoo

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Ex 3.2, 14 Show that (i) [■8(5&−1@6&7)] [■8(2&1@3&4)] ≠ [■8(2&1@3&4)] [■8(5&−1@6&7)] Taking L.H.S [■8(5&−1@6&7)]_(2×2) [■8(2&1@3&4)]_(2×2) = [■8(5×2+(−1)×3&5×1+(−1)×4@6×2+7×3&6×1+7×4)]_(2×2) = [■8(10−3&5−4@12+21&6+28)] = [■8(7&1@33&34)] Taking R.H.S [■8(2&1@3&4)]_(2×2) [■8(5&−1@6&7)]_(2×2) = [■8(2×5+1×6&2×(−1)+1×7@3×5+4×6&3×(−1)+4×7)]_(2×2) = [■8(10+6&−2+7@15+24&−3+28)] = [■8(16&5@39&25)] ≠ L.H.S Hence L.H.S. ≠ R.H.S. Hence proved. Ex 3.2, 14 Show that (ii) [■8(1&2&3@0&1&0@1&1&0)][■8(−1&1&0@0&−1&1@2&3&4)] ≠[■8(−1&1&0@0&−1&1@2&3&4)][■8(1&2&3@0&1&0@1&1&0)] Taking L.H.S. [■8(1&2&3@0&1&0@1&1&0)]_(3×3) [■8(−1&1&0@0&−1&1@2&3&4)]_(3×3) = [■8(1×(⤶7−1)+2×0+3×2&1×1+2×(−1)+3×3&1×0+2×1+3×4@0×(⤶7−1)+1×0+0×2&0×(⤶7−1)+1×0+0×2&0×0+1×1+0×4@1×(⤶7−1)+1×0+0×2&1×(⤶7−1)+1×0+0×2&1×0+1×1+0×4)]_(3×3) = [■8(−1+0+6&1−0+9&0+2+12@0+0+0&0−1+0&0+1+0@−1+0+0&1−1+0&0+1+0)] = [■8(5&8&14@0&−1&1@−1&0&1)] Taking R.H.S [■8(−1&1&0@0&−1&1@2&3&4)]_(3×3) [■8(1&2&3@0&1&0@1&1&0)]_(3×3) = [■8(1×1+1×0+0×1&−1×2+1×1+0×1&−1×3+1×0+0×0@0×1+(−1)×0+1×1&0×2+(⤶7−1)×1+1×1&0×3+(−1)×0+1×0@1×1+(3)×0+4×1&2×2+3×1+4×1&2×3+(3)×0+4×0)]_(3×3) = [■8(−1+0+0&−2+1+0&−3+0+0@0+0+1&0−1+1&0+0+0@2+0+4&4+3+4&6+0+0)] = [■8(−1&−1&−3@1&0&0@6&11&6)] ≠ L.H.S ∴ L.H.S. ≠ R.H.S Hence proved

Chapter 3 Class 12 Matrices

Concept wise

- Formation and order of matrix
- Types of matrices
- Equal matrices
- Addition/ subtraction of matrices
- Statement questions - Addition/Subtraction of matrices
- Multiplication of matrices
- Statement questions - Multiplication of matrices
- Solving Equation
- Finding unknown - Element
- Finding unknown - Matrice
- Transpose of a matrix
- Symmetric and skew symmetric matrices
- Proof using property of transpose
- Inverse of matrix using elementary transformation
- Proof using mathematical induction

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.