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


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Example 17 If A = [■8(0&6&7@−6&0&8@7&−8&0)], B = [■8(0&1&1@1&0&2@1&2&0)] , C = [█(2@−2@3)] Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC Calculating AC AC = [■8(0&6&7@−6&0&8@7&8&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+6(−2)+7(3)@−6(2)+0(−2)+8(3)@7(2)+8(−2)+0(3))]_(3 × 1) = [█(0−12+21@−12+0+24@14+16+0)] = [█(𝟗@𝟏𝟐@𝟑𝟎)] Calculating BC BC = [■8(0&1&1@1&0&2@1&2&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+1(−2)+1(3)@1(2)+0(−2)+2(3)@1(2)+2(−2)+0(3))]_(3 × 1) = [█(0−2+3@2+0+6@2−4+0)] = [█(𝟏@𝟖@−𝟐)] Calculating AC + BC AC + BC =[█(9@12@30)] + [█(1@8@−2)] = [█(9+1@12+8@30−2)] = [█(10@20@28)] Calculating (A + B) C First calculating A + B A + B = [■8(0&6&7@−6&0&8@7&−8&0)] +[■8(0&1&1@1&0&2@1&2&0)] = [■8(0+0&6+1&7+1@−6+1&0+0&8+2@7+1&−8+2&0+0)] = [■8(𝟎&𝟕&𝟖@−𝟓&𝟎&𝟏𝟎@𝟖&−𝟔&𝟎)] Now, Calculating (A + B) C (A + B) C = [■8(0&7&8@−5&0&10@8&−6&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+7(−2)+8(3)@−5(2)+0(−2)+10(3)@8(2)+(−6)(−2)+0(3))]_(3 × 1) = [█(0−14+24@−10+0+30@16+12+0 )] = [█(𝟏𝟎@𝟐𝟎@𝟐𝟖)] = AC + BC ∴ L.H.S. = R.H.S. 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.