Ex 3.2, 14 - Chapter 3 Class 12 Matrices (Term 1)

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