Example 17 - Calculate AC, BC, (A + B)C. Also, verify - Examples - Examples

part 2 - Example 17 - Examples - Serial order wise - Chapter 3 Class 12 Matrices
part 3 - Example 17 - Examples - Serial order wise - Chapter 3 Class 12 Matrices
part 4 - Example 17 - Examples - Serial order wise - Chapter 3 Class 12 Matrices

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