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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo