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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 16 If A = [■8(1&1&−1@2&0&3@3&−1&2)], B = [■8(1&3@0&2@−1&4)] and C = [■8(1&2&3&−4@2&0&−2&1)], find A(BC), (AB) C and show that (AB) C = A (BC) For A (BC) First Calculating BC BC = [■8(1&3@0&2@−1&4)]_(3 × 2) [■8(1&2&3&−4@2&0&−2&1)]_(2 × 4) = [■8(1(1)+3(2)&1(2)+3(0)&1(3)+3(−2)&1(−4)+3(1)@0(1)+2(2)&0(2)+2(0)&0(3)+2(−2)&0(−4)+2(1)@−1(1)+4(2)&−1(2)+4(0)&−1(3)+4(−2)&−1 (−4)+4(1))]_(𝟑 × 𝟒) = [■8(1+6&2+0&3−6&−4+3@0+4&0+0&0−4&0+2@−1+8&−2+0&−3−8&4+4)] = [■8(𝟕&𝟐&−𝟑&−𝟏@𝟒&𝟎&−𝟒&𝟐@𝟕&−𝟐&−𝟏𝟏&𝟖)] Now, Calculating A (BC) A (BC) = [■8(1&1&−1@2&0&3@3&−1&2)]_(3×3) [■8(7&2&−3&−1@4&0&−4&2@7&−2&−11&8)]_(3×4) = [■8(1(7)+1(4)+(−1)(7)&1(2)+1(0)+(−1)(−2)&1×(−3)+1×(−4)+(−1)×(−11)&1(−1)+1(2)+(−1)(8)@2(7)+0(4)+3(7)&2(2)+0(0)+3(−2)&2×(−3)+0×(−4)+3×(−11)&2(−1)+0(2)+3(8)@3(7)+(−1)(4)+2(7)&3(2)+(−1)(0)+2(−2)&3×(−3)+(−1)×(−4)+2×(−11)&3(−1)+(−1)(2)+2(8))]_(3×4) = [■8(7+4−7&2+0+2&−3−4+11&−1+2−8@14+0+21&4+0−6&−6+0−33&−2+0+24@21−4+14&6+0−4&−9+4−22&−3−2+16)] = [■8(𝟒&𝟒&𝟒&−𝟕@𝟑𝟓&−𝟐&−𝟑𝟗&𝟐𝟐@𝟑𝟏&𝟐&−𝟐𝟕&𝟏𝟏)] For (AB) C First calculating (AB) AB = [■8(1&1&−1@2&0&3@3&−1&2)]_(3 × 3) [■8(1&3@0&2@−1&4)]_(3 × 2) = [■8(1(1)+1(0)+(−1)(−1)&1(3)+1(2)+(−1)(4)@2(1)+0(0)+3(−1)&2(3)+0(2)+3(4)@3(1)+(−1)(0)+2(−1)&3(3)+(−1)(2)+2(4))]_(3 × 4) = [■8(1+0+1&3+2−4@2+0−3&6+0+12@3+0−2&9−2+8)] = [■8(𝟐&𝟏@−𝟏&𝟏𝟖@𝟏&𝟏𝟓)] Now, calculating (AB)C (AB)C = [■8(2&1@−1&18@1&15)]_(3×2) [■8(1&2&3&−4@2&0&−2&1)]_(2×4) = [■8(2(1)+1(2)&2(2)+1(0)&2(3)+1(−2)&2(−4)+1(1)@−1(1)+18(2)&−1(2)+18(0)&−1(3)+18(−2)&−1(−4)+18(1)@1(1)+15(2)&1(2)+15(0)&1(3)+15(−2)&1(−4)+15(1) )]_(3×4) = [■8(2+2&4+0&6−2&−8+1@−1+36&−2+0&−3−36&4+18@1+30&2+0&3−30&−4+15)] = [■8(𝟒&𝟒&𝟒&−𝟕@𝟑𝟓&−𝟐&−𝟑𝟗&𝟐𝟐@𝟑𝟏&𝟐&−𝟐𝟕&𝟏𝟏)] = A(BC) ∴ (AB) C = A (BC) Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.