Ex 3.2

Chapter 3 Class 12 Matrices
Serial order wise

### Transcript

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)] Solving 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)] Solving 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 Thus, 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)] Solving 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&0×1+1×(⤶7−1)+0×3&0×0+1×1+0×4@1×(⤶7−1)+1×0+0×2&1×1+1×(⤶7−1)+0×3&1×0+1×1+0×4)]_(3×3) = [■8(−1+0+6&1−2+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)] Solving 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@2×1+(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